1.4E: Exercises for The Cross Product
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For exercises 1-4, the vectors \(\vecs{u}\) and \(\vecs{v}\) are given.
a. Find the cross product \(\vecs{u}\times\vecs{v}\) of the vectors \(\vecs{u}\) and \(\vecs{v}\) . Express the answer in component form.
b. Sketch the vectors \(\vecs{u}, \, \vecs{v}\) , and \(\vecs{u}\times\vecs{v}\) .
1) \(\quad \vecs{u}=⟨2,0,0⟩, \quad \vecs{v}=⟨2,2,0⟩\)
- Answer
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\(a. \vecs{u}\times\vecs{v}=⟨0,0,4⟩;\)
\(b.\)
2) \(\quad \vecs{u}=⟨3,2,−1⟩, \quad \vecs{v}=⟨1,1,0⟩\)
3) \(\quad \vecs{u}=2\mathbf{\hat i}+3\mathbf{\hat j}, \quad \vecs{v}=\mathbf{\hat j}+2\mathbf{\hat k}\)
- Answer
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\( a. \vecs{u}\times\vecs{v}=⟨6,−4,2⟩;\)
\(b.\)
4) \(\quad \vecs{u}=2\mathbf{\hat j}+3\mathbf{\hat k}, \quad \vecs{v}=3\mathbf{\hat i}+\mathbf{\hat k}\)
5) Simplify \((\mathbf{\hat i}×\mathbf{\hat i}−2\mathbf{\hat i}×\mathbf{\hat j}−4\mathbf{\hat i}×\mathbf{\hat k}+3\mathbf{\hat j}×\mathbf{\hat k})×\mathbf{\hat i}.\)
- Answer
- \(−2\mathbf{\hat j}−4\mathbf{\hat k}\)
6) Simplify \(\mathbf{\hat j}×(\mathbf{\hat k}×\mathbf{\hat j}+2\mathbf{\hat j}×\mathbf{\hat i}−3\mathbf{\hat j}×\mathbf{\hat j}+5\mathbf{\hat i}×\mathbf{\hat k}).\)
In exercises 7-10, vectors \(\vecs{u}\) and \(\vecs{v}\) are given. Find unit vector \(\vecs{w}\) in the direction of the cross product vector \(\vecs{u}×\vecs{v}\) . Express your answer using standard unit vectors.
7) \(\quad \vecs{u}=⟨3,−1,2⟩, \quad \vecs{v}=⟨−2,0,1⟩\)
- Answer
- \(\vecs{w}=−\frac{\sqrt{6}}{18}\mathbf{\hat i}−\frac{7\sqrt{6}}{18}\mathbf{\hat j}−\frac{\sqrt{6}}{9}\mathbf{\hat k}\)
8) \(\quad \vecs{u}=⟨2,6,1⟩, \quad \vecs{v}=⟨3,0,1⟩\)
9) \(\quad \vecs{u}=\vecd{AB}, \quad \vecs{v}=\vecd{AC},\) where \(A(1,0,1),\, B(1,−1,3)\), and \(C(0,0,5)\)
- Answer
- \(\vecs{w}=−\frac{4\sqrt{21}}{21}\mathbf{\hat i}−\frac{2\sqrt{21}}{21}\mathbf{\hat j}−\frac{\sqrt{21}}{21}\mathbf{\hat k}\)
10) \(\quad \vecs{u}=\vecd{OP}, \quad \vecs{v}=\vecd{PQ},\) where \(P(−1,1,0)\) and \(Q(0,2,1)\)
11) Determine the real number \(α\) such that \(\vecs{u}\times\vecs{v}\) and \(\mathbf{\hat i}\) are orthogonal, where \(\vecs{u}=3\mathbf{\hat i}+\mathbf{\hat j}−5\mathbf{\hat k}\) and \(\vecs{v}=4\mathbf{\hat i}−2\mathbf{\hat j}+α\mathbf{\hat k}.\)
- Answer
- \(α=10\)
12) Show that \(\vecs{u}\times\vecs{v}\) and \( 2\mathbf{\hat i}−14\mathbf{\hat j}+2\mathbf{\hat k}\) cannot be orthogonal for any α real number, where \(\vecs{u}=\mathbf{\hat i}+7\mathbf{\hat j}−\mathbf{\hat k}\) and \(\vecs{v}=α\mathbf{\hat i}+5\mathbf{\hat j}+\mathbf{\hat k}\).
13) Show that \(\vecs{u}\times\vecs{v}\) is orthogonal to \(\vecs{u}+\vecs{v}\) and \(\vecs{u}−\vecs{v}\), where \(\vecs{u}\) and \(\vecs{v}\) are nonzero vectors.
14) Show that \(\vecs{v}\times\vecs{u}\) is orthogonal to \( (\vecs{u}⋅\vecs{v})(\vecs{u}+\vecs{v})+\vecs{u}\), where \(\vecs{u}\) and \(\vecs{v}\) are nonzero vectors.
15) Calculate the determinant \( \begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\1&−1&7\\2&0&3\end{vmatrix}\).
- Answer
- \( −3\mathbf{\hat i}+11\mathbf{\hat j}+2\mathbf{\hat k}\)
16) Calculate the determinant \( \begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\0&3&−4\\1&6&−1\end{vmatrix}\).
For exercises 17-18, the vectors \(\vecs{u}\) and \(\vecs{v}\) are given. Use determinant notation to find vector \(\vecs{w}\) orthogonal to vectors \(\vecs{u}\) and \(\vecs{v}\) .
17) \(\quad \vecs{u}=⟨−1, 0, e^t⟩, \quad \vecs{v}=⟨1, e^{−t}, 0⟩,\) where \(t\) is a real number
- Answer
- \(\vecs{w}=⟨−1, e^t, −e^{−t}⟩\)
18) \(\quad \vecs{u}=⟨1, 0, x⟩, \quad \vecs{v}=⟨\frac{2}{x},1, 0⟩,\) where \(x\) is a nonzero real number
19) Find vector \( (\vecs{a}−2\vecs{b})×\vecs{c},\) where \( \vecs{a}=\begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\2&−1&5\\0&1&8\end{vmatrix}, \vecs{b}=\begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\0&1&1\\2&−1&−2\end{vmatrix},\) and \(\vecs{c}=\mathbf{\hat i}+\mathbf{\hat j}+\mathbf{\hat k}.\)
- Answer
- \( −26\mathbf{\hat i}+17\mathbf{\hat j}+9\mathbf{\hat k}\)
20) Find vector \( \vecs{c}×(\vecs{a}+3\vecs{b}),\) where \( \vecs{a}=\begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\5&0&9\\0&1&0\end{vmatrix}, \vecs{b}=\begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\0&−1&1\\7&1&−1\end{vmatrix},\) and \(\vecs{c}=\mathbf{\hat i}−\mathbf{\hat k}.\)
21) [T] Use the cross product \(\vecs{u}\times\vecs{v}\) to find the acute angle between vectors \(\vecs{u}\) and \(\vecs{v}\), where \(\vecs{u}=\mathbf{\hat i}+2\mathbf{\hat j}\) and \(\vecs{v}=\mathbf{\hat i}+\mathbf{\hat k}.\) Express the answer in degrees rounded to the nearest integer.
- Answer
- \( 72°\)
22) [T] Use the cross product \(\vecs{u}\times\vecs{v}\) to find the obtuse angle between vectors \(\vecs{u}\) and \(\vecs{v}\), where \(\vecs{u}=−\mathbf{\hat i}+3\mathbf{\hat j}+\mathbf{\hat k}\) and \(\vecs{v}=\mathbf{\hat i}−2\mathbf{\hat j}.\) Express the answer in degrees rounded to the nearest integer.
23) Use the sine and cosine of the angle between two nonzero vectors \(\vecs u\) and \(\vecs v\) to prove Lagrange’s identity: \(\|\vecs{u}\times\vecs{v}\|^2=\|\vecs{u}\|^2\|\vecs{v}\|^2−(\vecs{u}⋅\vecs{v})^2\).
24) Verify Lagrange’s identity \(\|\vecs{u}\times\vecs{v}\|^2=\|\vecs{u}\|^2\|\vecs{v}\|^2−(\vecs{u}⋅\vecs{v})^2\) for vectors \(\vecs{u}=−\mathbf{\hat i}+\mathbf{\hat j}−2\mathbf{\hat k}\) and \(\vecs{v}=2\mathbf{\hat i}−\mathbf{\hat j}.\)
25) Nonzero vectors \(\vecs{u}\) and \(\vecs{v}\) are called collinear if there exists a nonzero scalar \(α\) such that \( \vecs{v}=α\vecs{u}\). Show that \(\vecs{u}\) and \(\vecs{v}\) are collinear if and only if \( \vecs{u}\times\vecs{v}=0.\)
26) Nonzero vectors \(\vecs{u}\) and \(\vecs{v}\) are called collinear if there exists a nonzero scalar \(α\) such that \( \vecs{v}=α\vecs{u}\). Show that vectors \( \vecd{AB}\) and \(\vecd{AC}\) are collinear, where \(A(4,1,0), \, B(6,5,−2),\) and \(C(5,3,−1).\)
27) Find the area of the parallelogram with adjacent sides \(\vecs{u}=⟨3,2,0⟩\) and \(\vecs{v}=⟨0,2,1⟩\).
- Answer
- \(7\)
28) Find the area of the parallelogram with adjacent sides \(\vecs{u}=\mathbf{\hat i}+\mathbf{\hat j}\) and \(\vecs{v}=\mathbf{\hat i}+\mathbf{\hat k}.\)
29) Consider points \(A(3,−1,2),\, B(2,1,5),\) and \(C(1,−2,−2).\)
a. Find the area of parallelogram \(ABCD\) with adjacent sides \(\vecd{AB}\) and \( \vecd{AC}\).
b. Find the area of triangle \(ABC\).
c. Find the distance from point \(A\) to line \(BC\).
- Answer
- a. \(5\sqrt{6};\) b. \(\frac{5\sqrt{6}}{2};\) c. \(\frac{5\sqrt{6}}{\sqrt{59}} =\frac{5\sqrt{354}}{59} \)
30) Consider points \(A(2,−3,4),\, B(0,1,2),\) and \(C(−1,2,0).\)
a. Find the area of parallelogram \(ABCD\) with adjacent sides \( \vecd{AB}\) and \( \vecd{AC}\).
b. Find the area of triangle \(ABC\).
c. Find the distance from point \(B\) to line \(AC.\)
In exercises 31-32, vectors \(\vecs{u}\) , \(\vecs{v}\) , and \(\vecs{w}\) are given.
a. Find the triple scalar product \(\vecs{u}⋅(\vecs{v}×\vecs{w})\) .
b. Find the volume of the parallelepiped with the adjacent edges \(\vecs{u}\) , \(\vecs{v}\) , and \(\vecs{w}\) .
31) \(\quad \vecs{u}=\mathbf{\hat i}+\mathbf{\hat j}, \quad \vecs{v}=\mathbf{\hat j}+\mathbf{\hat k},\) and \(\quad \vecs{w}=\mathbf{\hat i}+\mathbf{\hat k}\)
- Answer
- \( a. 2; \quad b. 2\) units 3
32) \(\quad \vecs{u}=⟨−3,5,−1⟩, \quad \vecs{v}=⟨0,2,−2⟩,\) and \(\quad \vecs{w}=⟨3,1,1⟩\)
33) Calculate the triple scalar products \(\vecs{v}⋅(\vecs{u}×\vecs{w})\) and \(\vecs{w}⋅(\vecs{u}×\vecs{v}),\) where \(\vecs{u}=⟨1,1,1⟩, \vecs{v}=⟨7,6,9⟩,\) and \(\vecs{w}=⟨4,2,7⟩.\)
- Answer
- \(\vecs{v}⋅(\vecs{u}×\vecs{w})=−1, \quad \vecs{w}⋅(\vecs{u}×\vecs{v})=1\)
34) Calculate the triple scalar products \(\vecs{w}⋅(\vecs{v}×\vecs{u})\) and \(\vecs{u}⋅(\vecs{w}×\vecs{v}),\) where \(\vecs{u}=⟨4,2,−1⟩, \vecs{v}=⟨2,5,−3⟩,\) and \(\vecs{w}=⟨9,5,−10⟩.\)
35) Find vectors \(\vecs{a},\, \vecs{b}\), and \(\vecs{c}\) with a triple scalar product given by the determinant \( \begin{vmatrix}1&2&3\\0&2&5\\8&9&2\end{vmatrix}\). Determine their triple scalar product.
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- \(\vecs{a}=⟨1,2,3⟩, \quad \vecs{b}=⟨0,2,5⟩, \quad \vecs{c}=⟨8,9,2⟩; \quad \vecs{a}⋅(\vecs{b}×\vecs{c})=−9\)
36) The triple scalar product of vectors \(\vecs{a},\,\vecs{b}\), and \(\vecs{c}\) is given by the determinant \( \begin{vmatrix}0&−2&1\\0&1&4\\1&−3&7\end{vmatrix}\). Find vector \(\vecs{a}−\vecs{b}+\vecs{c}.\)
37) Consider the parallelepiped with edges \( OA,OB,\) and \( OC\), where \( A(2,1,0),B(1,2,0),\) and \( C(0,1,α).\)
a. Find the real number \( α>0\) such that the volume of the parallelepiped is \( 3\) units 3 .
b. For \( α=1,\) find the height \(h\) from vertex \(C\) of the parallelepiped. Sketch the parallelepiped.
- Answer
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\( a. \, α=1; \quad b. \, h=1\) unit,
38) Consider points \( A(α,0,0),B(0,β,0),\) and \( C(0,0,γ)\), with \( α, β\), and \( γ\) positive real numbers.
a. Determine the volume of the parallelepiped with adjacent sides \( \vecd{OA}, \vecd{OB},\) and \( \vecd{OC}\).
b. Find the volume of the tetrahedron with vertices \( O,A,B,\) and \( C\). (Hint: The volume of the tetrahedron is \( 1/6\) of the volume of the parallelepiped.)
c. Find the distance from the origin to the plane determined by \( A,B,\) and \( C\). Sketch the parallelepiped and tetrahedron.
39) Let \( u,v,\) and \( w\) be three-dimensional vectors and \(c\) be a real number. Prove the following properties of the cross product.
a. \(\vecs u×\vecs u=\vecs 0\)
b. \(\vecs u×(\vecs v+\vecs w)=(\vecs u×\vecs v)+(\vecs u×\vecs w)\)
c. \( c(\vecs u×\vecs v)=(c\vecs u)×\vecs v=\vecs u×(c\vecs v)\)
d. \( \vecs u⋅(\vecs u×\vecs v)=\vecs 0\)
40) Show that vectors \(\vecs u=⟨1,0,−8⟩,\,\vecs v=⟨0,1,6⟩\), and \(\vecs w=⟨−1,9,3⟩\) satisfy the following properties of the cross product.
a. \(\vecs u×\vecs u=\vecs 0\)
b. \(\vecs u×(\vecs v+\vecs w)=(\vecs u×\vecs v)+(\vecs u×\vecs w)\)
c. \( c(\vecs u×\vecs v)=(c\vecs u)×\vecs v=\vecs u×(c\vecs v)\)
d. \(\vecs u⋅(\vecs u×\vecs v)=\vecs 0\)
41) Nonzero vectors \(\vecs u,\,\vecs v\), and \(\vecs w\) are said to be linearly dependent if one of the vectors is a linear combination of the other two. For instance, there exist two nonzero real numbers \( α\) and \( β\) such that \(\vecs w=α\vecs u+β\vecs v\). Otherwise, the vectors are called linearly independent. Show that \(\vecs u,\vecs v\), and \(\vecs w\) could be placed on the same plane if and only if they are linear dependent.
42) Consider vectors \(\vecs u=⟨1,4,−7⟩,\,\vecs v=⟨2,−1,4⟩,\,\vecs w=⟨0,−9,18⟩\), and \(\vecs p=⟨0,−9,17⟩.\)
a. Show that \(\vecs u,\,\vecs v\), and \(\vecs w\) can be placed on the same plane by using their triple scalar product
b. Show that \(\vecs u,\,\vecs v\), and \(\vecs w\) can be placed on the same plane by using the definition that there exist two nonzero real numbers \( α\) and \( β\) such that \( w=αu+βv.\)
c. Show that \(\vecs u,\,\vecs v\), and \(\vecs p\) are linearly independent—that is, none of the vectors is a linear combination of the other two.
43) Consider points \( A(0,0,2), B(1,0,2), C(1,1,2),\) and \( D(0,1,2).\) Are vectors \( \vecd{AB}, \vecd{AC},\) and \( \vecd{AD}\) linearly dependent (that is, one of the vectors is a linear combination of the other two)?
- Answer
- Yes, \( \vecd{AD}=α\vecd{AB}+β\vecd{AC},\) where \( α=−1\) and \( β=1.\)
44) Show that vectors \( \mathbf{\hat i}+\mathbf{\hat j}, \mathbf{\hat i}−\mathbf{\hat j},\) and \( \mathbf{\hat i}+\mathbf{\hat j}+\mathbf{\hat k}\) are linearly independent—that is, there exist two nonzero real numbers \(α\) and \(β\) such that \(\mathbf{\hat i}+\mathbf{\hat j}+\mathbf{\hat k}=α(\mathbf{\hat i}+\mathbf{\hat j})+β(\mathbf{\hat i}−\mathbf{\hat j}).\)
45) Let \(\vecs u=⟨u_1,u_2⟩\) and \(\vecs v=⟨v_1,v_2⟩\) be two-dimensional vectors. The cross product of vectors \(\vecs u\) and \(\vecs v\) is not defined. However, if the vectors are regarded as the three-dimensional vectors \( \tilde{\vecs u}=⟨u_1,u_2,0⟩\) and \( \tilde{\vecs v}=⟨v_1,v_2,0⟩\), respectively, then, in this case, we can define the cross product of \( \tilde{\vecs u}\) and \( \tilde{\vecs v}\). In particular, in determinant notation, the cross product of \( \tilde{\vecs u}\) and \( \tilde{\vecs v}\) is given by
\( \tilde{\vecs u}×\tilde{\vecs v}=\begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\u_1&u_2&0\\v_1&v_2&0\end{vmatrix}\).
Use this result to compute \( (\cos θ\,\mathbf{\hat i}+\sin θ\,\mathbf{\hat j})×(\sin θ\,\mathbf{\hat i}−\cos θ\,\mathbf{\hat j}),\) where \( θ\) is a real number.
- Answer
- \( −\mathbf{\hat k}\)
46) Consider points \( P(2,1), Q(4,2),\) and \( R(1,2).\)
a. Find the area of triangle \( PQR\).
b. Determine the distance from point \( R\) to the line passing through \( P\) and \( Q\).
47) Determine a vector of magnitude \( 10\) perpendicular to the plane passing through the x-axis and point \( P(1,2,4).\)
- Answer
- \( ⟨0,±4\sqrt{5},2\sqrt{5}⟩\)
48) Determine a unit vector perpendicular to the plane passing through the z -axis and point \( A(3,1,−2).\)
49) Consider \(\vecs u\) and \(\vecs v\) two three-dimensional vectors. If the magnitude of the cross product vector \(\vecs u×\vecs v\) is \( k\) times larger than the magnitude of vector \(\vecs u\), show that the magnitude of \(\vecs v\) is greater than or equal to \( k\), where \( k\) is a natural number.
50) [T] Assume that the magnitudes of two nonzero vectors \(\vecs u\) and \(\vecs v\) are known. The function \( f(θ)=‖\vecs u‖‖\vecs v‖\sin θ\) defines the magnitude of the cross product vector \(\vecs u×\vecs v,\) where \( θ∈[0,π]\) is the angle between \(\vecs u\) and \(\vecs v\).
a. Graph the function \( f\).
b. Find the absolute minimum and maximum of function \( f\). Interpret the results.
c. If \( ‖\vecs u‖=5\) and \( ‖\vecs v‖=2\), find the angle between \(\vecs u\) and \(\vecs v\) if the magnitude of their cross product vector is equal to \( 9\).
51) Find all vectors \(\vecs w=⟨w_1,w_2,w_3⟩\) that satisfy the equation \( ⟨1,1,1⟩×\vecs w=⟨−1,−1,2⟩.\) Hint: You should be able to write all components of \(\vecs w\) in terms of one of the constants \(w_1,w_2,\) or \(w_3\).
- Answer
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Writing all components in terms of the constant \(w_3\), one way to represent these vectors is: \(\vecs w=⟨w_3−1,w_3+1,w_3⟩,\) where \( w_3\) is any real number.
Note that we could use any parameter we wish here. We could set \(w_3 = a\). Then \(\vecs w=⟨a−1,a+1,a⟩\) would also represent these vectors.
52) Solve the equation \(\vecs w×⟨1,0,−1⟩=⟨3,0,3⟩,\) where \(\vecs w=⟨w_1,w_2,w_3⟩\) is a nonzero vector with a magnitude of \( 3\).
53) [T] A mechanic uses a 12-in. wrench to turn a bolt. The wrench makes a \( 30°\) angle with the horizontal. If the mechanic applies a vertical force of \( 10\) lb on the wrench handle, what is the magnitude of the torque at point \( P\) (see the following figure)? Express the answer in foot-pounds rounded to two decimal places.
- Answer
- 8.66 ft-lb
54) [T] A boy applies the brakes on a bicycle by applying a downward force of 20 lb on the pedal when the 6-in. crank makes a \( 40°\) angle with the horizontal (see the following figure). Find the torque at point \( P\). Express your answer in foot-pounds rounded to two decimal places.
55) [T] Find the magnitude of the force that needs to be applied to the end of a 20-cm wrench located on the positive direction of the \(y\)-axis if the force is applied in the direction \( ⟨0,1,−2⟩\) and it produces a \( 100\) N·m torque to the bolt located at the origin.
- Answer
- \(250\sqrt{5}\) N \(\approx 559\) N
56) [T] What is the magnitude of the force required to be applied to the end of a 1-ft wrench at an angle of \( 35°\) to produce a torque of \( 20\) N·m?
57) [T] The force vector \(\vecs F\) acting on a proton with an electric charge of \( 1.6×10^{−19}\,C\) (in coulombs) moving in a magnetic field \(\vecs B\) where the velocity vector \(\vecs v\) is given by \(\vecs F=1.6×10^{−19}(\vecs v×\vecs B)\) (here, \(\vecs v\) is expressed in meters per second, \(\vecs B\) is in tesla [T], and \(\vecs F\) is in newtons [N]). Find the force that acts on a proton that moves in the \(xy\)-plane at velocity \(\vecs v=10^5\mathbf{\hat i}+10^5\mathbf{\hat j}\) (in meters per second) in a magnetic field given by \(\vecs B=0.3\mathbf{\hat j}\).
- Answer
- \(\vecs F=4.8×10^{−15}\,kN\)
58) [T] The force vector \(\vecs F\) acting on a proton with an electric charge of \( 1.6×10^{−19}\,C\) moving in a magnetic field \(\vecs B\) where the velocity vector \(\vecs v\) is given by \(\vecs F=1.6×10^{−19}(\vecs v×\vecs B)\) (here, \(\vecs v\) is expressed in meters per second, \(\vecs B\) in \( T\), and \(\vecs F\) in \( N\)). If the magnitude of force \(\vecs F\) acting on a proton is \( 5.9×10^{−17}\,N\) and the proton is moving at the speed of 300 m/sec in magnetic field \(\vecs B\) of magnitude 2.4 T, find the angle between velocity vector \(\vecs v\) of the proton and magnetic field \(\vecs B\). Express the answer in degrees rounded to the nearest integer.
59) [T] Consider \(\vecs r(t)=⟨\cos t,\,\sin t,\,2t⟩\) the position vector of a particle at time \( t∈[0,30]\), where the components of \(\vecs r\) are expressed in centimeters and time in seconds. Let \( \vecd{OP}\) be the position vector of the particle after \( 1\) sec.
a. Determine unit vector \(\vecs B(t)\) (called the binormal unit vector) that has the direction of cross product vector \(\vecs v(t)×\vecs a(t),\) where \(\vecs v(t)\) and \(\vecs a(t)\) are the instantaneous velocity vector and, respectively, the acceleration vector of the particle after \( t\) seconds.
b. Use a CAS to visualize vectors \(\vecs v(1),\,\vecs a(1)\), and \(\vecs B(1)\) as vectors starting at point \( P\) along with the path of the particle.
- Answer
-
a. \(\vecs B(t)=⟨\frac{2\sqrt{5}\sin t}{5},−\frac{2\sqrt{5}\cos t}{5},\frac{\sqrt{5}}{5}⟩;\)
b.
60) A solar panel is mounted on the roof of a house. The panel may be regarded as positioned at the points of coordinates (in meters) \( A(8,0,0), B(8,18,0), C(0,18,8),\) and \( D(0,0,8)\) (see the following figure).
a. Find vector \(\vecs n=\vecd{AB}×\vecd{AD}\) perpendicular to the surface of the solar panels. Express the answer using standard unit vectors. Note that the magnitude of this vector should give us the area of rectangle \(ABCD\).
b. Assume unit vector \(\vecs s=\frac{1}{\sqrt{3}}\mathbf{\hat i}+\frac{1}{\sqrt{3}}\mathbf{\hat j}+\frac{1}{\sqrt{3}}\mathbf{\hat k}\) points toward the Sun at a particular time of the day and the flow of solar energy is \(\vecs F=900\vecs s\) (in watts per square meter [\( W/m^2\)]). Find the predicted amount of electrical power the panel can produce, which is given by the dot product of vectors \(\vecs F\) and \(\vecs n\) (expressed in watts).
c. Determine the angle of elevation of the Sun above the solar panel. Express the answer in degrees rounded to the nearest whole number. (Hint: The angle between vectors \(\vecs n\) and \(\vecs s\) and the angle of elevation are complementary.)
Contributors
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .


10.5: Miller v. Alabama 567 US ___ (2012)
NOTE: Where it is feasible, a syllabus (headnote) will be released, as is being done in connection with this case, at the time the opinion is issued. The syllabus constitutes no part of the opinion of the Court but has been prepared by the Reporter of Decisions for the convenience of the reader. See United States v. Detroit Timber & Lumber Co., 200 U. S. 321 .
SUPREME COURT OF THE UNITED STATES
MILLER v. ALABAMA
certiorari to the court of criminal appeals of alabama
No. 10–9646. Argued March 20, 2012—Decided June 25, 2012 [ 1 ]
In each of these cases, a 14-year-old was convicted of murder and sentenced to a mandatory term of life imprisonment without the possibility of parole. In No. 10−9647, petitioner Jackson accompanied two other boys to a video store to commit a robbery; on the way to the store, he learned that one of the boys was carrying a shotgun. Jackson stayed outside the store for most of the robbery, but after he entered, one of his co-conspirators shot and killed the store clerk. Arkansas charged Jackson as an adult with capital felony murder and aggravated robbery, and a jury convicted him of both crimes. The trial court imposed a statutorily mandated sentence of life imprisonment without the possibility of parole. Jackson filed a state habeas petition, arguing that a mandatory life-without-parole term for a 14-year-old violates the Eighth Amendment. Disagreeing, the court granted the State’s motion to dismiss. The Arkansas Supreme Court affirmed.
In No. 10−9646, petitioner Miller, along with a friend, beat Miller’s neighbor and set fire to his trailer after an evening of drinking and drug use. The neighbor died. Miller was initially charged as a juvenile, but his case was removed to adult court, where he was charged with murder in the course of arson. A jury found Miller guilty, and the trial court imposed a statutorily mandated punishment of life without parole. The Alabama Court of Criminal Appeals affirmed, holding that Miller’s sentence was not overly harsh when compared to his crime, and that its mandatory nature was permissible under the Eighth Amendment.
Held: The Eighth Amendment forbids a sentencing scheme that mandates life in prison without possibility of parole for juvenile homicide offenders. Pp. 6−27.
(a) The Eighth Amendment’s prohibition of cruel and unusual punishment “guarantees individuals the right not to be subjected to excessive sanctions.” Roper v. Simmons, 543 U. S. 551 . That right “flows from the basic ‘precept of justice that punishment for crime should be graduated and proportioned’ ” to both the offender and the offense. Ibid.
Two strands of precedent reflecting the concern with proportionate punishment come together here. The first has adopted categorical bans on sentencing practices based on mismatches between the culpability of a class of offenders and the severity of a penalty. See, e.g., Kennedy v. Louisiana, 554 U. S. 407 . Several cases in this group have specially focused on juvenile offenders, because of their lesser culpability. Thus, Roper v. Simmons held that the Eighth Amendment bars capital punishment for children, and Graham v. Florida, 560 U. S. ___, concluded that the Amendment prohibits a sentence of life without the possibility of parole for a juvenile convicted of a nonhomicide offense. Graham further likened life without parole for juveniles to the death penalty, thereby evoking a second line of cases. In those decisions, this Court has required sentencing authorities to consider the characteristics of a defendant and the details of his offense before sentencing him to death. See, e.g., Woodson v. North Carolina, 428 U. S. 280 (plurality opinion). Here, the confluence of these two lines of precedent leads to the conclusion that mandatory life without parole for juveniles violates the Eighth Amendment.
As to the first set of cases: Roper and Graham establish that children are constitutionally different from adults for sentencing purposes. Their “ ‘lack of maturity’ ” and “ ‘underdeveloped sense of responsibility’ ” lead to recklessness, impulsivity, and heedless risk-taking. Roper, 543 U. S., at 569. They “are more vulnerable . . . to negative influences and outside pressures,” including from their family and peers; they have limited “contro[l] over their own environment” and lack the ability to extricate themselves from horrific, crime-producing settings. Ibid. And because a child’s character is not as “well formed” as an adult’s, his traits are “less fixed” and his actions are less likely to be “evidence of irretrievabl[e] deprav[ity].” Id., at 570. Roper and Graham emphasized that the distinctive attributes of youth diminish the penological justifications for imposing the harshest sentences on juvenile offenders, even when they commit terrible crimes.
While Graham’s flat ban on life without parole was for nonhomicide crimes, nothing that Graham said about children is crime-specific. Thus, its reasoning implicates any life-without-parole sentence for a juvenile, even as its categorical bar relates only to nonhomicide offenses. Most fundamentally, Graham insists that youth matters in determining the appropriateness of a lifetime of incarceration without the possibility of parole. The mandatory penalty schemes at issue here, however, prevent the sentencer from considering youth and from assessing whether the law’s harshest term of imprisonment proportionately punishes a juvenile offender. This contravenes Graham’s (and also Roper’s) foundational principle: that imposition of a State’s most severe penalties on juvenile offenders cannot proceed as though they were not children.
Graham also likened life-without-parole sentences for juveniles to the death penalty. That decision recognized that life-without-parole sentences “share some characteristics with death sentences that are shared by no other sentences.” 560 U. S., at ___. And it treated life without parole for juveniles like this Court’s cases treat the death penalty, imposing a categorical bar on its imposition for nonhomicide offenses. By likening life-without-parole sentences for juveniles to the death penalty, Graham makes relevant this Court’s cases demanding individualized sentencing in capital cases. In particular, those cases have emphasized that sentencers must be able to consider the mitigating qualities of youth. In light of Graham’s reasoning, these decisions also show the flaws of imposing mandatory life-without-parole sentences on juvenile homicide offenders. Pp. 6−17.
(b) The counterarguments of Alabama and Arkansas are unpersuasive. Pp. 18–27.
(1) The States first contend that Harmelin v. Michigan, 501 U. S. 957 , forecloses a holding that mandatory life-without-parole sentences for juveniles violate the Eighth Amendment. Harmelin declined to extend the individualized sentencing requirement to noncapital cases “because of the qualitative difference between death and all other penalties.” Id., at 1006 (Kennedy, J., concurring in part and concurring in judgment). But Harmelin had nothing to do with children, and did not purport to apply to juvenile offenders. Indeed, since Harmelin, this Court has held on multiple occasions that sentencing practices that are permissible for adults may not be so for children. See Roper, 543 U. S. 551 ; Graham, 560 U. S ___.
The States next contend that mandatory life-without-parole terms for juveniles cannot be unconstitutional because 29 jurisdictions impose them on at least some children convicted of murder. In considering categorical bars to the death penalty and life without parole, this Court asks as part of the analysis whether legislative enactments and actual sentencing practices show a national consensus against a sentence for a particular class of offenders. But where, as here, this Court does not categorically bar a penalty, but instead requires only that a sentencer follow a certain process, this Court has not scrutinized or relied on legislative enactments in the same way. See, e.g., Sumner v. Schuman, 483 U. S. 66 .
In any event, the “objective indicia of society’s standards,” Graham, 560 U. S., at ___, that the States offer do not distinguish these cases from others holding that a sentencing practice violates the Eighth Amendment. Fewer States impose mandatory life-without-parole sentences on juvenile homicide offenders than authorized the penalty (life-without-parole for nonhomicide offenders) that this Court invalidated in Graham. And as Graham and Thompson v. Oklahoma, 487 U. S. 815 , explain, simply counting legislative enactments can present a distorted view. In those cases, as here, the relevant penalty applied to juveniles based on two separate provisions: One allowed the transfer of certain juvenile offenders to adult court, while another set out penalties for any and all individuals tried there. In those circumstances, this Court reasoned, it was impossible to say whether a legislature had endorsed a given penalty for children (or would do so if presented with the choice). The same is true here. Pp. 18–25.
(2) The States next argue that courts and prosecutors sufficiently consider a juvenile defendant’s age, as well as his background and the circumstances of his crime, when deciding whether to try him as an adult. But this argument ignores that many States use mandatory transfer systems. In addition, some lodge the decision in the hands of the prosecutors, rather than courts. And even where judges have transfer-stage discretion, it has limited utility, because the decisionmaker typically will have only partial information about the child or the circumstances of his offense. Finally, because of the limited sentencing options in some juvenile courts, the transfer decision may present a choice between a light sentence as a juvenile and standard sentencing as an adult. It cannot substitute for discretion at post-trial sentencing. Pp. 25−27.
No. 10−9646, 63 So. 3d 676, and No. 10−9647, 2011 Ark. 49, ___ S. W. 3d ___, reversed and remanded.
Kagan, J., delivered the opinion of the Court, in which Kennedy, Ginsburg, Breyer, and Sotomayor, JJ., joined. Breyer, J., filed a concurring opinion, in which Sotomayor, J., joined. Roberts, C. J., filed a dissenting opinion, in which Scalia, Thomas, and Alito, JJ., joined. Thomas, J., filed a dissenting opinion, in which Scalia, J., joined. Alito, J., filed a dissenting opinion, in which Scalia, J., joined.
Notes
Contributors and Attributions
- Located at : https://supreme.justia.com/cases/federal/us/567/10-9646/ . License : Public Domain: No Known Copyright


40.2: Acne Drugs
By the end of this section, you should be able to:
- 40.2.1 Identify the characteristics of drugs used to treat acne.
- 40.2.2 Explain the indications, actions, adverse reactions, and interactions of drugs used to treat acne.
- 40.2.3 Describe nursing implications of drugs used to treat acne.
- 40.2.4 Explain the client education related to drugs used to treat acne.
Acne vulgaris (acne) is a skin condition in which sebum —dead skin cells—and bacteria accumulate in a skin pore. The pore becomes clogged and allows bacterial growth, which causes the surrounding skin to erupt in skin lesions or vesicles. Although acne is typically considered a condition of adolescents and young adults, occasionally it may continue into adulthood, especially in female clients. Oily skin, hormonal changes, family history, and some medications may cause or exacerbate acne. Acne presents as multiple pustules, papules, or nodules on the skin, particularly on the face but also on the back, upper arms, and trunk (Sutaria et al., 2023). Figure 40.3 shows the development of acne within a pore.
Adverse Effects and Contraindications
Common adverse effects associated with these treatments include hypersensitivity reactions, yeast infections, and gastrointestinal intolerance. Tetracyclines are also associated with skin rashes, tooth damage and discoloration, and altered bone growth and development. Tetracycline hydrochloride has more specific precautions and contraindications (see Table 40.2) (DailyMed, Tetracycline Hydrochloride , 2023).
Table 40.2 is a drug prototype table for systemic acne drugs featuring tetracycline hydrochloride. It lists drug class, mechanism of action, adult and pediatric dosage , indications, therapeutic effects, drug and food interactions, adverse effects, and contraindications.
|
Drug Class
Antibiotic (gram-negative and gram-positive) Mechanism of Action Bacteriostatic action to reduce bacterial colonization Inhibits protein synthesis |
Drug Dosage
Oral Adults: 1000 mg daily in divided doses for acute acne; maintenance for acne prevention: 125–500 mg daily. Children ≥ 8 years : 25–50 mg/kg daily, divided into 4 equal doses. Children <8 years: Do not administer. Topical 3% ointment: Apply twice daily in a thin layer over the entire affected area. |
|
Indications
Acne and other skin infections Sexually transmitted infections (syphilis, chlamydia, gonorrhea) Therapeutic Effects Reduces pustules that form in acne by eliminating the bacteria that cause lesions |
Drug Interactions
Antacids Iron Sodium bicarbonate Zinc penicillin Anticoagulants Food Interactions Dairy products |
|
Adverse Reactions
Gastrointestinal upset Esophageal irritation Enterocolitis Black hairy tongue Yeast infections Tooth discoloration and dental caries Skin rashes that may progress to exfoliative dermatitis Renal and liver toxicities Anemia Thrombocytopenia Neutropenia Severe photosensitivity Anaphylaxis |
Contraindications
Hypersensitivity Children younger than 8 years Pregnancy and lactation Caution: Decreases effectiveness of hormonal contraceptives—barrier contraceptives may be needed |
Topical Acne Drugs
Topical acne medications are often used in combination with systemic acne medications to provide better management of acne. Topical therapy is applied directly to the skin that is affected by the presence of acne lesions. Azelaic acid 15% and benzoyl peroxide 2.5% are used in both acne and rosacea to eliminate bacteria and decrease inflammation as well as to reduce oil on the skin. Salicylic acid, another topical treatment, assists in acne treatment by exfoliating dead skin cells to reduce pore blockages, reducing inflammation, and decreasing sebum secretion. Adapalene 1% topical ointment, when applied to the areas affected by acne, helps to differentiate skin cells and reduces the accumulation of cells in skin pores. Tazarotene, another topical treatment for acne, is thought to assist with reducing fine lines and discolorations in the skin. It is thought that this medication may also play a role in the reduction of scarring from acne lesions. Tretinoin 0.025% ointment, a retinoid, benefits clients undergoing treatment for severe acne. The actions of topical retinoids are not fully understood but are thought to increase cell mitosis and decrease cell cohesion to prevent clogging of skin pores (DailyMed, Tretinoin, 2019).
Link to Learning
The Psychosocial Impact of Skin Disorders
Although the physical effects of acne can be devastating, the psychosocial impacts—anxiety/depression, self-esteem, stigma, body image, social support, social interaction, sexual life, social acceptance and optimism, and objective factors—also must be considered. This NursingOpen article discusses the psychologic impact of acne and other skin disorders (Zhang et al., 2021).
Table 40.3 lists common topical acne drugs and typical routes and dosing for adult and pediatric clients .
| Drug | Routes and Dosage Ranges |
|---|---|
|
Azelaic acid
( Azelex , Finacea , Finevin ) |
Adults:
Apply a thin layer topically twice daily.
Children: Safety and effectiveness have not been established in pediatric clients. |
|
Benzoyl peroxide
( Benzac ) |
Adults:
Apply a thin layer to the affected area 1–3 times daily.
Children: Safety and effectiveness have not been established in pediatric clients. |
|
Adapalene
( Differin ) |
Adults and children ≥12 years:
Apply a thin layer topically once daily, in the evening.
Children <12 years: Safety and effectiveness have not been established. |
|
Tazarotene
( Tazorac ) |
Adults:
Apply a thin layer topically once daily, in the evening.
Children: Safety and efficacy have not been established in clients under the age of 12. |
|
Salicylic acid
( Salex , Virasal ) |
Adults:
Apply a thin layer topically 1–3 times daily.
Children >2 years: Follow health care provider recommendations. Not recommended for use in children under 2 years. |
|
Tretinoin
( Retin-A , Atralin ) |
Adults and children ≥12 years: Apply a thin layer topically once daily, at night. |
Adverse Effects and Contraindications
Adverse effects for topical acne medications most often involve localized skin redness, scaling, itching, and burning. These may worsen when adapalene is used concomitantly with other substances that cause irritation or dry skin. For this reason, contraindications include the use of other non-approved substances, soaps, and cleansers that cause skin irritation. Other contraindications are sun or other ultraviolent light and the use of occlusive dressings—those that do not allow air or moisture to get in or out.
Table 40.4 is a drug prototype table for topical acne drugs featuring adapalene. It lists drug class, mechanism of action, adult and pediatric dosage , indications, therapeutic effects, drug and food interactions, adverse effects, and contraindications.
|
Drug Class
Retinoid, anticomedogenic Mechanism of Action Decreases inflammation and clogging of pores in acne |
Drug Dosage
Adults and children ≥12 years: Apply a thin layer topically once daily, in the evening. Children <12 years: Safety and effectiveness have not been established. |
|
Indications
Acne and other skin infections Therapeutic Effects Decreases formation of acne lesions |
Drug Interactions
Soaps, creams, and cosmetics that have a drying effect or contain alcohol or astringents Food Interactions No significant interactions |
|
Adverse Reactions
Skin irritation Redness Itching Flakiness |
Contraindications
Hypersensitivity Dry, flaky, or open skin |
Special Considerations
Topical Acne Drugs
Adolescents using topical acne medications are at high risk of discontinuing treatment prematurely because of improved symptoms, side effects, and the need to use multiple products or treatments. Research suggests that monotherapy, treatment with oral medications, and/or anticipatory guidance on side effects can increase an adolescent client’s adherence to the prescribed therapy.
(Source: Habeshian & Cohan, 2020)
Nursing Implications
The nurse should do the following for clients who are taking acne drugs:
- Assess and thoroughly document skin condition baseline.
- Assess and monitor the client for adverse effects, drug and food interactions, and contraindications.
- Evaluate and document the client’s response to current treatment and overall improvement of skin condition.
- Assess the psychosocial impact of acne on individual clients.
- Provide client teaching regarding the drug and when to call the health care provider. See below for client teaching guidelines.
Client Teaching Guidelines
The client taking a systemic acne medication should:
- Take prescribed medications exactly as directed.
- Report side effects such as skin redness, excessive dryness, or peeling to their health care provider.
- Use birth control methods as instructed and avoid pregnancy.
- Refrain from breastfeeding.
- Follow up with appointments as instructed.
The client taking a topical acne medication should:
- Take prescribed medications exactly as directed.
- Report side effects such as skin redness, excessive dryness, or peeling to their health care provider.
- Cleanse skin thoroughly before applying medication.
- Apply only a thin layer of medications to the affected areas only.
- Avoid contact with eyes and mucous membranes.
- Wash hands well before and after application.
The client taking a topical acne medication should not:
- Cover the medication with occlusive dressings.
FDA Black Box Warning
Systemic Acne Medications
Embryo-fetal loss and malformations may occur when systemic acne drugs are administered to a pregnant client.


What you’ll learn to do: use psychodynamic theories (like those from Freud and Erikson) to explain development
Many people sometimes feel intimidated by theory; even the phrase, “Now we are going to look at some theories…” may elicit some blank stares or yawns. But, don’t tune out quite yet! Theories are valuable tools for understanding human behavior; they are proposed explanations for the “how” and “whys” of development. In this first section, we’ll examine some of the most persistent theories, developed by Sigmund Freud over a hundred years ago. While some of Freud’s ideas have since been debunked, others have lasted and continue to shape the way we think about development.
Contributors and Attributions
CC licensed content, Original
-
Introduction to Psychodynamic Theories.
Authored by
: Sonja Ann Miller for Lumen Learning.
Provided by
: Lumen Learning.
License
:
CC BY: Attribution
CC licensed content, Shared previously


20.3D: Web, Network, and Ring of Life Models
- Describe the web, network, and ring of life models of phylogenetic trees
The recognition of the importance of Horizontal gene transfer (HGT), especially in the evolution of prokaryotes, has caused some to propose abandoning the classic “tree of life” model. In 1999, a phylogenetic model that resembles a web or a network more than a tree was proposed. The hypothesis is that eukaryotes evolved not from a single prokaryotic ancestor, but from a pool of many species that were sharing genes by HGT mechanisms. Some individual prokaryotes were responsible for transferring the bacteria that caused mitochondrial development in the new eukaryotes, whereas other species transferred the bacteria that gave rise to chloroplasts. This model is often called the “web of life.” In an effort to save the tree analogy, some have proposed using the Ficus tree with its multiple trunks as a phylogenetic tree to represent the evolutionary role for HGT.
Others have proposed abandoning any tree-like model of phylogeny in favor of a ring structure. The ” ring of life ” is a phylogenetic model where all three domains of life evolved from a pool of primitive prokaryotes. Using the conditioned reconstruction algorithm, it proposes a ring-like model in which species of all three domains (Archaea, Bacteria, and Eukarya) evolved from a single pool of gene-swapping prokaryotes. This structure is proposed as the best fit for data from extensive DNA analyses; the ring model is the only one that adequately takes HGT and genomic fusion into account. However, phylogeneticists remain highly skeptical of this model.
In summary, the “tree of life” model proposed by Darwin must be modified to include HGT. This does not mean a tree, web, or a ring will correlate completely to an accurate description of phylogenetic relationships of life. A consequence of the new thinking about phylogenetic models is the idea that Darwin’s original conception of the phylogenetic tree is too simple, but made sense based on what was known at that time.
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- clonal. Provided by : Wiktionary. Located at : http://en.wiktionary.org/wiki/clonal . License : CC BY-SA: Attribution-ShareAlike
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- OpenStax College, Biology. October 23, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44593/latest...ol11448/latest . License : CC BY: Attribution
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Key Points
- A phylogenetic model that resembles a web or a network was proposed since eukaryotes evolved not from a single prokaryotic ancestor, but from a pool of many species that were sharing genes by HGT mechanisms.
- A phylogenetic model that resembles a ring was proposed in which species of all three domains, Archaea, Bacteria, and Eukarya, evolved from a single pool of gene-swapping prokaryotes.
- Phylogenetic models will continue to evolve as phylogeneticists remain highly skeptical of the current tree, web, and ring models.
Key Terms
- web of life : a phylogenetic model that resembles a web or a network more than a tree
- ring of life : a phylogenetic model where all three domains of life (Archaea, Bacteria, and Eukarya) evolved from a pool of primitive prokaryotes


46.3B: The Water (Hydrologic) Cycle
- Explain the path of the hydrologic cycle and its importance
Water is the basis of all living processes.  Thus, most land animals need a supply of fresh water to survive. However, when examining the stores of water on earth, 97.5 percent of it is non-potable salt water. Of the remaining water, 99 percent is locked underground as water or as ice. Thus, less than 1 percent of fresh water is easily accessible from lakes and rivers. Many living things, such as plants, animals, and fungi, are dependent on the small amount of fresh surface water supply, a lack of which can have massive effects on ecosystem dynamics. Humans, of course, have developed technologies to increase water availability, such as digging wells to harvest groundwater, storing rainwater, and using desalination to obtain drinkable water from the ocean. Although this pursuit of drinkable water has been ongoing throughout human history, the supply of fresh water is still a major issue in modern times.
Water cycling is extremely important to ecosystem dynamics as it has a major influence on climate and, thus, on the environments of ecosystems. For example, when water evaporates, it takes up energy from its surroundings, cooling the environment. When it condenses, it releases energy, warming the environment. The evaporation phase of the cycle purifies water, which then replenishes the land with fresh water. The flow of liquid water and ice transports minerals across the globe. It is also involved in reshaping the geological features of the earth through processes including erosion and sedimentation. The water cycle is also essential for the maintenance of most life and ecosystems on the planet. Most of the water on earth is stored for long periods in the oceans, underground, and as ice. Residence time is a measure of the average time an individual water molecule stays in a particular reservoir. A large amount of the earth’s water is locked in place in these reservoirs as ice, beneath the ground, and in the ocean, and, thus, is unavailable for short-term cycling (only surface water can evaporate).
There are various processes that occur during the cycling of water, which include the following:
- evaporation/ sublimation
- condensation/precipitation
- subsurface water flow
- surface runoff/snowmelt
- streamflow
The water cycle is driven by the sun’s energy as it warms the oceans and other surface waters. This leads to the evaporation (water to water vapor) of liquid surface water and the sublimation (ice to water vapor) of frozen water, which deposits large amounts of water vapor into the atmosphere. Over time, this water vapor condenses into clouds as liquid or frozen droplets, which is eventually followed by precipitation (rain or snow), returning water to the earth’s surface. Rain eventually percolates into the ground, where it may evaporate again (if it is near the surface), flow beneath the surface, or be stored for long periods. More easily observed is surface runoff: the flow of fresh water either from rain or melting ice. Runoff can then make its way through streams and lakes to the oceans or flow directly to the oceans themselves. Rain and surface runoff are major ways in which minerals, including carbon, nitrogen, phosphorus, and sulfur, are cycled from land to water.
Key Points
- Water cycling affects the climate, transports minerals, purifies water, and replenishes the land with fresh water.
- Water with a longer residence time, such as water in oceans and glaciers, is not available for short-term cycling, which occurs via evaporation.
- Surface water evaporates (water to water vapor) or sublimates (ice to water vapor), which deposits large amounts of water vapor into the atmosphere.
- Water vapor in the atmosphere condenses into clouds and is eventually followed by precipitation, which returns water to the earth’s surface.
- Rain percolates into the ground, where it may evaporate or enter bodies of water.
- Surface runoff enters oceans directly or via streams and lakes.
Key Terms
- residence time : the average time a particular molecule of water will remain in a body of water
- condensation : the conversion of a gas to a liquid; the condensate so formed
- surface runoff : overland flow of excess water (with or without accumulated contaminants) that cannot be absorbed by the ground as infiltration
- Evaporation : the process of a liquid converting to the gaseous state
- sublimation : the transition of a substance from the solid phase directly to the vapor state such that it does not pass through the intermediate liquid phase


Educators need to develop innovative models of teaching and delivery methods tailored to emerging technologies. Future learning systems should contain intelligent agents to duplicate one-to-one tutoring. Multiple intelligent agents could also monitor learners’ progress, and cater to individual needs and styles. Intelligent learning systems will allow learners to be more active and will place more responsibility on them in the learning process. Research is needed on how to empower learners to learn on their own and how to activate learners’ metacognitive skills.
Content will be designed as small chunks in the form of information and learning objects. This will allow intelligent agents to prescribe the most appropriate materials based on learner’s learning style, progress, and needs. The intelligent agents will assemble these chunks into a larger instructional sequence so that learners can achieve the learning outcomes of the lesson. More work is needed on how to develop learning objects and how to tag them for easy retrieval by intelligent agents.
Future technologies will use intelligent agents to assemble courses and modules of instruction immediately by accessing learning objects from repositories. Because of the changing nature of content, models are needed to develop learning materials in as short time as possible using techniques similar to rapid application development (Lohr et al., 2003). Smart learning systems in emerging technologies will be able to assemble unique courses for each learner, based on the learner’s prior knowledge, learning preferences, and needs.
Pervasive computing is making it possible for computing power to be included everywhere, thanks to tiny microprocessors and wireless access. As a result, educators must design for pervasive computing where learners will access learning materials using everyday objects and environments. For example, learners might be able to access course materials using kitchen appliances, or their clothing.
The trend in hardware development is towards virtual devices, such as the virtual keyboard and virtual screen. With these devices, learners are able to turn on the device, use it, and then turn it off. For example, for input into a computer, a learner can press a button to turn on a virtual keyboard on a temporary surface, use it, then turn it off. When developing learning materials for emerging technologies, educators must design for delivery on these virtual devices.


2.19: Introduction to Bach and Handel
This section contains materials on the two composers considered by many to be the greatest composers of the Baroque and among the greatest composers of all time. Both these composers are known for applying their considerable genius to existing genres rather than establishing a genre as Monteverdi and Corelli did. In some cases, such as with Bach’s church cantatas, their compositions elevated the genre to a level of technical and artistic mastery that has been rarely duplicated since. Our readings will focus on the biographies of these two composers and their contributions to particular musical genres.
This section includes the following pages:
- Slideshow: The Late Baroque
-
Johann Sebastian Bach
- Chorale
- Bach’s Cantatas
- Cantata 140, Wachet auf, ruft uns die Stimme
- Cello Suites
- Fugue
- Fugue in G minor, “Little” BWV 578
- Brandenburg Concertos
-
George Frideric Handel
- Messiah
- Water Music
Contributors and Attributions
- Authored by : Elliott Jones. Provided by : Santa Ana College. Located at : http://www.sac.edu . License : CC BY: Attribution
- Provided by : Lumen Learning. License : CC BY: Attribution


9.20: Gross Margin Percentage
- Analyze the gross margin percentage of sample retailers
Gross margin = net sales – cost of goods sold. However, to express this as a percentage, gross margin is divided by net sales (gross margin dollars / net sales). This allows the retailer to assess what portion of sales were spent on COGS to generate the sales and what portion remains to cover operating expenses. That is:
- 1 – gross margin percent = the portion of sales spent as COGS
- 1 – (COGS / net sales) = gross margin percent
Contributors and Attributions
CC licensed content, Original
- Gross Margin Percentage. Authored by : Patrick Williams. Provided by : Lumen Learning. License : CC BY: Attribution


1.6: The Equivalence Principle (Part 2)
Gravitational Red-shifts
Earlier, we saw experimental evidence that the rate of flow of time changes with height in a gravitational field. We can now see that this is required by the equivalence principle.
By the equivalence principle, there is no way to tell the difference between experimental results obtained in an accelerating laboratory and those found in a laboratory immersed in a gravitational field. 30 In a laboratory accelerating upward, a photon emitted from the floor and would be Doppler-shifted toward lower frequencies when observed at the ceiling, because of the change in the receiver’s velocity during the photon’s time of flight. The effect is given by
\[\dfrac{\Delta E}{E} = \dfrac{\Delta f}{f} = \dfrac{ay}{c^{2}}\]
where \(a\) is the lab’s acceleration, \(y\) is the height from floor to ceiling, and \(c\) is the speed of light.
Note
Problem 4 verifies, in one specific example, that this way of stating the equivalence principle is implied by the one stated earlier.
Exercise \(\PageIndex{1}\)
Self-check: Verify this statement.
By the equivalence principle, we find that when such an experiment is done in a gravitational field g, there should be a gravitational effect on the energy of a photon equal to \(\frac{\Delta E}{E} = \frac{gy}{c^{2}}\). Since the quantity gy is the gravitational potential (gravitational energy per unit mass), the photon’s fractional loss of energy is the same as the (Newtonian) loss of energy experienced by a material object of mass m and initial kinetic energy mc 2 .
The interpretation is as follows. Classical electromagnetism requires that electromagnetic waves have inertia. For example, if a plane wave strikes an ohmic surface, as in Figure 1.5.14, the wave’s electric field excites oscillating currents in the surface. These currents then experience a magnetic force from the wave’s magnetic field, and application of the right-hand rule shows that the resulting force is in the direction of propagation of the wave. Thus the light wave acts as if it has momentum. The equivalence principle says that whatever has inertia must also participate in gravitational interactions. Therefore light waves must have weight, and must lose energy when they rise through a gravitational field.
Exercise \(\PageIndex{2}\)
Self-check: Verify the application of the right-hand rule described above.
Further interpretation:
- The quantity \(mc^2\) is famous, even among people who don’t know what \(m\) and \(c\) stand for. This is the first hint of where it comes from. The full story is given in section 4.2.
- The relation p = \(\frac{E}{c}\) between the energy and momentum of a light wave follows directly from Maxwell’s equations, by the argument above; however, we will see in section 4.2 that according to relativity this relation must hold for any massless particle
- What we have found agrees with Niels Bohr’s correspondence principle, which states that when a new physical theory, such as relativity, replaces an older one, such as Newtonian physics, the new theory must agree with the old one under the experimental conditions in which the old theory had been verified by experiments. The gravitational mass of a beam of light with energy \(E\) is \(\frac{E}{c^{2}}\), and since c is a big number, it is not surprising that the weight of light rays had never been detected before Einstein trying to detect it.
- This book describes one particular theory of gravity, Einstein’s theory of general relativity. There are other theories of gravity, and some of these, such as the Brans-Dicke theory , do just as well as general relativity in agreeing with the presently available experimental data. Our prediction of gravitational Doppler shifts of light only depended on the equivalence principle, which is one ingredient of general relativity. Experimental tests of this prediction only test the equivalence principle; they do not allow us to distinguish between one theory of gravity and another if both theories incorporate the equivalence principle.
- If an object such as a radio transmitter or an atom in an excited state emits an electromagnetic wave with a frequency f, then the object can be considered to be a type of clock. We can therefore interpret the gravitational red-shift as a gravitational time dilation: a difference in the rate at which time itself flows, depending on the gravitational potential. This is consistent with the empirical results presented in Section 1.2.
Example 3: Chiao's paradox revisited
The equivalence principle says that electromagnetic waves have gravitational mass as well as inertial mass, so it seems clear that the same must hold for static fields. In Chiao’s paradox, the orbiting charged particle has an electric field that extends out to infinity. When we measure the mass of a charged particle such as an electron, there is no way to separate the mass of this field from a more localized contribution. The electric field “falls” through the gravitational field, and the equivalence principle, which is local, cannot guarantee that all parts of the field rotate uniformly about the earth, even in distant parts of the universe. The electric field pattern becomes distorted, and this distortion causes a radiation reaction which back-reacts on the particle, causing its orbit to decay.
The Pound-Rebka Experiment
The 1959 Pound-Rebka experiment at Harvard 31 was one of the first high-precision, relativistic tests of the equivalence principle to be carried out under controlled conditions, and in this section we will discuss it in detail.
When y is on the order of magnitude of the height of a building, the value of \(\frac{\Delta E}{E} = \frac{gy}{c^{2}}\) is ∼ 10 −14 , so an extremely high-precision experiment is necessary in order to detect a gravitational red-shift. A number of other effects are big enough to obscure it entirely, and must somehow be eliminated or compensated for. These are listed below, along with their orders of magnitude in the experimental design finally settled on by Pound and Rebka.
| (1) Classical Doppler broadening due to temperature . Thermal motion causes Doppler shifts of emitted photons, corresponding to the random component of the emitting atom’s velocity vector along the direction of emission. | ∼ 10 −6 |
| (2) The recoil Doppler shift. When an atom emits a photon with energy E and momentum \(p = \frac{E}{c}\), conservation of momentum requires that the atom recoil with momentum \(p = − \frac{E}{c}\) and energy \(\frac{p^{2}}{2m}\). This causes a downward Doppler shift of the energy of the emitted photon. A similar effect occurs on absorption, doubling the problem. | ∼ 10 −12 |
| (3) Natural line width . The Heisenberg uncertainty principle says that a state with a half-life \(\tau\) must have an uncertainty in its energy of at least ∼ \(\frac{h}{\tau}\), where h is Planck’s constant. | ∼ 10 −12 |
| (4) Special-relativistic Doppler shift due to temperature . Section 1.2 presented experimental evidence that time flows at a different rate depending on the motion of the observer. Therefore the thermal motion of an atom emitting a photon has an effect on the frequency of the photon, even if the atom’s motion is not along the line of emission. The equations needed in order to calculate this effect will not be derived until section 2.2; a quantitative estimate is given in example 13. For now, we only need to know that this leads to a temperature-dependence in the average frequency of emission, in addition to the broadening of the bell curve described by effect (1) above. | ∼ 10 −14 per degree C |
The most straightforward way to mitigate effect (1) is to use photons emitted from a solid. At first glance this would seem like a bad idea, since electrons in a solid emit a continuous spectrum of light, not a discrete spectrum like the ones emitted by gases; this is because we have N electrons, where N is on the order of Avogadro’s number, all interacting strongly with one another, so by the correspondence principle the discrete quantum-mechanical behavior must be averaged out. But the protons and neutrons within one nucleus do not interact much at all with those in other nuclei, so the photons emitted by a nucleus do have a discrete spectrum. The energy scale of nuclear excitations is in the keV or MeV range, so these photons are x-rays or gamma-rays. Furthermore, the timescale of the random vibrations of a nucleus in a solid are extremely short. For a velocity on the order of 100 m/s, and vibrations with an amplitude of ∼ 10 −10 m, the time is about 10 −12 s. In many cases, this is much shorter than the half-life of the excited nuclear state emitting the gamma-ray, and therefore the Doppler shift averages out to nearly zero.
Effect (2) is still much bigger than the 10 −14 size of the effect to be measured. It can be avoided by exploiting the Mössbauer effect, in which a nucleus in a solid substance at low temperature emits or absorbs a gamma-ray photon, but with significant probability the recoil is taken up not by the individual nucleus but by a vibration of the atomic lattice as a whole. Since the recoil energy varies as \(\frac{p^{2}}{2m}\), the large mass of the lattice leads to a very small dissipation of energy into the recoiling lattice. Thus if a photon is emitted and absorbed by identical nuclei in a solid, and for both emission and absorption the recoil momentum is taken up by the lattice as a whole, then there is a negligible energy shift. One must pick an isotope that emits photons with energies of about 10-100 keV. X-rays with energies lower than about 10 keV tend to be absorbed strongly by matter and are difficult to detect, whereas for gamma-ray energies \(\gtrsim\) 100 keV the M¨ossbauer effect is not sufficient to eliminate the recoil effect completely enough.
If the Mössbauer effect is carried out in a horizontal plane, resonant absorption occurs. When the source and absorber are aligned vertically, p, gravitational frequency shifts should cause a mismatch, destroying the resonance. One can move the source at a small velocity (typically a few mm/s) in order to add a Doppler shift onto the frequency; by determining the velocity that compensates for the gravitational effect, one can determine how big the gravitational effect is.
The typical half-life for deexcitation of a nucleus by emission of a gamma-ray with energy E is in the nanosecond range. To measure an gravitational effect at the 10 −14 level, one would like to have a natural line width, (3), with \(\frac{\Delta E}{E}\) \(\lesssim\) 10 −14 , which would require a half-life of \(\gtrsim\) 10 \(\mu\)s. In practice, Pound and Rebka found that other effects, such as (4) and electron-nucleus interactions that depended on the preparation of the sample, tended to put nuclei in one sample “out of tune” with those in another sample at the 10 −13 -10 −12 level, so that resonance could not be achieved unless the natural line width gave \(\frac{\Delta E}{E} \gtrsim\) 10 −12 . As a result, they settled on an experiment in which 14 keV gammas were emitted by 57 Fe nuclei (Figure 1.5.17) at the top of a 22-meter tower, and absorbed by 57 Fe nuclei at the bottom. The 100-ns half-life of the excited state leads to \(\frac{\Delta E}{E}\) ∼ 10 −12 . This is 500 times greater than the gravitational effect to be measured, so, as described in more detail below, the experiment depended on high-precision measurements of small up-and-down shifts of the bell-shaped resonance curve.
The absorbers were seven iron films isotopically enhanced in 57 Fe, applied directly to the faces of seven sodium-iodide scintillation detectors (bottom of Figure 1.5.16). When a gamma-ray impinges on the absorbers, a number of different things can happen, of which we can get away with considering only the following: (a) the gamma-ray is resonantly absorbed in one of the 57 Fe absorbers, after which the excited nucleus decays by re-emission of another such photon (or a conversion electron), in a random direction; (b) the gamma-ray passes through the absorber and then produces ionization directly in the sodium iodide crystal. In case b, the gamma-ray is detected. In case a, there is a 50% probability that the re-emitted photon will come out in the upward direction, so that it cannot be detected. Thus when the conditions are right for resonance, a reduction in count rate is expected. The M¨ossbauer effect never occurs with 100% probability; in this experiment, about a third of the gammas incident on the absorbers were resonantly absorbed.
The choice of y = 22 m was dictated mainly by systematic errors. The experiment was limited by the strength of the gamma-ray source. For a source of a fixed strength, the count rate in the detector at a distance y would be proportional to y −2 , leading to statistical errors proportional to \(1/ \sqrt{count\; rate} \propto y\). Since the effect to be measured is also proportional to y, the signal-to-noise ratio was independent of y. However, systematic effects such as (4) were easier to monitor and account for when y was fairly large. A lab building at Harvard happened to have a 22-meter tower, which was used for the experiment. To reduce the absorption of the gammas in the 22 meters of air, a long, cylindrical mylar bag full of helium gas was placed in the shaft, p.
The resonance was a bell-shaped curve with a minimum at the natural frequency of emission. Since the curve was at a minimum, where its derivative was zero, the sensitivity of the count rate to the gravitational shift would have been nearly zero if the source had been stationary. Therefore it was necessary to vibrate the source up and down, so that the emitted photons would be Doppler shifted onto the shoulders of the resonance curve, where the slope of the curve was large. The resulting asymmetry in count rates is shown in Figure 1.5.18. A further effort to cancel out possible systematic effects was made by frequently swapping the source and absorber between the top and bottom of the tower.
For y = 22.6 m, the equivalence principle predicts a fractional frequency shift due to gravity of 2.46 × 10 −15 . Pound and Rebka measured the shift to be (2.56 ± 0.25) × 10 −15 . The results were in statistical agreement with theory, and verified the predicted size of the effect to a precision of 10%.
Figure \(\PageIndex{19}\) - Pound and Rebka at the top and bottom of the tower.
References
31 Phys. Rev. Lett. 4 (1960) 337


4.1: Prelude to Further Applications of Trigonometry
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The world’s largest tree by volume, named General Sherman, stands \(274.9\) feet tall and resides in Northern California. Just how do scientists know its true height? A common way to measure the height involves determining the angle of elevation, which is formed by the tree and the ground at a point some distance away from the base of the tree. This method is much more practical than climbing the tree and dropping a very long tape measure.
Figure \(\PageIndex{1}\): General Sherman, the world’s largest living tree. (credit: Mike Baird, Flickr)
In this chapter, we will explore applications of trigonometry that will enable us to solve many different kinds of problems, including finding the height of a tree. We extend topics we introduced in Trigonometric Functions and investigate applications more deeply and meaningfully.


2.4: The Dot Product of Two Vectors, the Length of a Vector, and the Angle Between Two Vectors
The Dot Product of Two Vectors
The length of a vector or the angle between two vectors \(\vec{u}\boldsymbol{=}\left\langle u_x,\left.u_y\right\rangle \right.\) and \(\vec{v}\boldsymbol{=}\left\langle v_x,\left.v_y\right\rangle \right.\) can be found using the dot product.
The dot product of vectors \(\vec{u}=\left\langle u_x, u_y\right\rangle\) and \(\vec{v}=\left\langle v_x, v_y\right\rangle\) is a scalar (real number) and is defined to be \[\vec{u} \cdot \vec{v}=u_x v_x+u_y v_y \nonumber \]
Since \(u_x,\ u_y,\ v_x\) and \(v_y\) are real numbers, you can see that the dot product is itself a real number and not a vector.
To compute the dot product of the vectors \(\vec{u}\boldsymbol{=}\left\langle 5,\left.2\right\rangle \right.\) and \(\vec{v}\boldsymbol{=}\left\langle 3,\left.4\right\rangle \right.\)
Solution
We compute: \[\overrightarrow{u}\bullet \overrightarrow{v}\boldsymbol{=}5\bullet 3\mathrm{+\ }2\bullet 4=15+8=23 \nonumber \]
Since the dot product is a scalar, it follows the properties of real numbers.
- \(\vec{u} \cdot \vec{v}=\vec{v} \cdot \vec{u}\)the dot product is commutative
- \(\vec{u} \cdot(\vec{v}+\vec{w})=\vec{u} \cdot \vec{v}+\vec{u} \cdot \vec{w}\) the dot product distributes over vector addition
- \(\vec{u} \cdot \overrightarrow{0}=0\),the dot product with the zero vector \(\overrightarrow{0}\), is the scalar 0.
- \(\vec{u} \cdot \vec{u}=\|\vec{u}\|^2\)
Compute the dot product \(\overrightarrow{u}\bullet \left(\overrightarrow{v}+\overrightarrow{w}\right)=\overrightarrow{u}\bullet \overrightarrow{v}+\overrightarrow{u}\bullet \overrightarrow{w}\) , where \(\overrightarrow{u}=\left\langle 5,\left.-2\right\rangle \right.\) , \(\overrightarrow{v}=\left\langle 6,\left.4\right\rangle \right.\) , and \(\overrightarrow{w}=\left\langle -3,\left.7\right\rangle \right.\) .
Solution
\[\begin{align} \vec{u} \cdot(\vec{v}+\vec{w})&=\vec{u} \cdot \vec{v}+\vec{u} \cdot \vec{w} \nonumber\\
\vec{u} \cdot(\vec{v}+\vec{w})&=\langle 5,-2\rangle \cdot\langle 6,4\rangle+\langle 5,-2\rangle \cdot\langle-3,7\rangle \nonumber \\
&=(5 \cdot 6+(-2) \cdot 4)+(5 \cdot(-3)+(-2) \cdot 7) \nonumber \\
&=30-8-15-14 \nonumber \\
&=-7 \nonumber
\end{align} \nonumber \]
The Length of a Vector
The length (magnitude) of a vector you know is given by \(\left\|\overrightarrow{v}\right\|=\sqrt{{v_x}^2+{v_y}^2}\) . The length can also be found using the dot product. If we dot a vector \(\overrightarrow{v}=\left\langle v_x,\left.v_y\right\rangle \right.\) with itself, we get
\[\overrightarrow{v}\bullet \overrightarrow{v}=\left\langle v_x,\left.v_y\right\rangle \right.\bullet \left\langle v_x,\left.v_y\right\rangle \right. \nonumber \] \[\overrightarrow{v}\bullet \overrightarrow{v}=v_x\bullet v_x+v_y\bullet v_y \nonumber \] \[\overrightarrow{v}\bullet \overrightarrow{v}= {v_x}^2+\ {v_y}^2 \nonumber \]
By Vector Property 4, \(\overrightarrow{v}\bullet \overrightarrow{v}=\ {\left\|\overrightarrow{v}\right\|}^2\) . This gives \({\left\|\overrightarrow{v}\right\|}^2={v_x}^2+{v_y}^2\) .
Taking the square root of each side produces
\[\sqrt{{\|\overrightarrow{v}\|}^2}=\sqrt{{v_x}^2+{v_y}^2} \nonumber \]
\[\|\overrightarrow{v}\|=\sqrt{{v_x}^2+{v_y}^2} \nonumber \]
Which is the length of the vector \(\overrightarrow{v}\) .
The dot product of a vector \(\vec{v}=\left\langle v_x, v_y\right\rangle\) with itself gives the length of the vector.
\[\begin{equation}
\|\vec{v}\|=\sqrt{v_x^2+v_y^2}
\end{equation} \nonumber \]
Use the dot product to find the length of the vector \(\overrightarrow{v}=\left[ \begin{array}{c} 2 \\ 6 \end{array} \right]\) .
Solution
In this case, \(v_x=2\) and \(v_y=6.\)
Using \(\|\overrightarrow{v}\|=\sqrt{{v_x}^2+{v_y}^2}\) , we get \[\|\overrightarrow{v}\|=\sqrt{2^2+6^2} \nonumber \] \[\|\overrightarrow{v}\|=\sqrt{40} \nonumber \] \[\|\overrightarrow{v}\|=\sqrt{4\bullet 10} \nonumber \] \[\|\overrightarrow{v}\|=\sqrt{4}\bullet \sqrt{10} \nonumber \] \[\|\overrightarrow{v}\|=2\sqrt{10} \nonumber \]
The length of the vector \(\overrightarrow{v}=\left[ \begin{array}{c} 2 \\ 6 \end{array} \right]\) is \(2\sqrt{10}\) units.
The Angle between two Vectors
The dot product and elementary trigonometry can be used to find the angle \(\theta\) between two vectors.
If \(\theta\) is the smallest nonnegative angle between two non-zero vectors \(\vec{u}\) and \(\vec{v}\) then
\[\begin{equation}
\cos \theta=\frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|} \; \text { or } \; \theta=\cos ^{-1} \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|}
\end{equation} \nonumber \]
\[\begin{equation}
\text { where } \; 0 \leq \theta \leq 2 \pi \; \text { and } \; \|\vec{u}\|=\sqrt{u_x^2+u_y{ }^2} \; \text { and } \; \|\vec{v}\|=\sqrt{v_x^2+v_y{ }^2}
\end{equation} \nonumber \]
Find the angle between the vectors \(\overrightarrow{u}=\left\langle 5,\left.-3\right\rangle \right.\) and \(\overrightarrow{v}=\left\langle 2,\left.4\right\rangle \right.\) .
Solution
Using \(\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{\overrightarrow{u}\bullet \overrightarrow{v}}{\|\overrightarrow{u}\|\bullet \|\overrightarrow{v}\|}\) , we get
\[\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{\left\langle 5,\left.-3\right\rangle \right.\bullet \left\langle 2,\left.4\right\rangle \right.}{\sqrt{5^2+{\left(-3\right)}^2}\bullet \sqrt{2^2+4^2}} \nonumber \]
\[\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{5\bullet 2+\left(-3\right)\bullet 4}{\sqrt{25+9}\bullet \sqrt{4+16}} \nonumber \]
\[\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{-2}{\sqrt{34}\bullet \sqrt{20}} \nonumber \]
\[\theta =94.4 \nonumber \]
We conclude that the angle between these two vectors is close to 94.4 \(\mathrm{{}^\circ}\) .
Using Technology
We can use technology to find the angle \(\theta\) between two vectors.
Go to www.wolframalpha.com.
To find the angle between the vectors \(\overrightarrow{u}=\left\langle 5,\left.-3\right\rangle \right.\) and \(\overrightarrow{v}=\left\langle 2,\left.4\right\rangle \right.\) , enter angle between the vectors \(\mathrm{<}\) 5, -3 \(\mathrm{>}\) and \(\mathrm{<}\) 2, 4 \(\mathrm{>}\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(\theta =94.4\) , rounded to one decimal place.
Try These
Find the dot product of the vectors \(\vec{u}=\langle-2,3\rangle \) and \(\vec{v}=\langle 5,-1\rangle \).
- Answer
-
\(\overrightarrow{u}\bullet \overrightarrow{v}=-13\)
Find the dot product of the vectors \(\vec{u}=\langle-4,6\rangle \) and \(\vec{v}=\langle 3,2\rangle \).
- Answer
-
\(\overrightarrow{u}\bullet \overrightarrow{v}=0\)
Find the length of the vector \(\vec{u}=\langle 4,-7\rangle \).
- Answer
-
\(\sqrt{65}\)
Find the length of the vectors \(\vec{v}=\langle 0,5\rangle \).
- Answer
-
\(5\)
Find the angle between the vectors \(\vec{u}=\langle-2,3\rangle \) and \(\vec{v}=\langle 5,-1\rangle \).
- Answer
-
\(135{}^\circ\)
Find the angle between the vectors \(\vec{u}=\langle -4,6\rangle \) and \(\vec{v}=\langle 3,2\rangle \).
- Answer
-
\(90{}^\circ\)


Main Concepts
Identify the main concepts in your research question by selecting nouns important to the meaning of your question. Leave out words that don’t help the search, such as adjectives, adverbs, prepositions and, usually, verbs. Nouns that you would use to tag your research question so you could find it later are likely to be its main concepts.
Finding the main concepts in a research question is a lot like finding the main idea in an essay or story. Often the main idea is in the first paragraph, but not always. Sometimes it’s in a later paragraph or even in the conclusion. The same is true with research questions—the main concepts can be at the beginning, middle, or end. Stick to the nouns and only what’s necessary, not already implied. Don’t read in concepts that are not really there. Be alert to words that may have connotations other than the concept you are interested in. For instance, if you identify depression as a main idea, beware that the search engine won’t automatically know whether you mean depression as a psychological state or as a condition of the economy or as a weather characteristic.
Example: How are birds affected by wind turbines?
The main concepts are birds and wind turbines. Avoid terms like affect (except the noun) and effect as search terms, even when you’re looking for studies that report effects or effectiveness.
Example: What lesson plans are available for teaching fractions?
The main concepts are lesson plans and fractions. Stick to what’s necessary. For instance, don’t include: children—nothing in the research question suggests the lesson plans are for children; teaching—teaching isn’t necessary because lesson plans imply teaching; available—available is not necessary.
Sometimes your research question itself can seem complicated. Make sure you’ve stated the question as precisely as possible (as you learned in Research Questions). Then apply our advice for identifying main concepts as usual.
EXAMPLE: Does the use of mobile technologies by teachers and students in the classroom distract or enhance the educational experience?
Acceptable main concepts are teaching methods and mobile technology. Another possibility is mobile technologies and education.
Watch out for overly broad terms. For example, don’t include:
-
Educational experience (it misses mobile technology).
-
Classroom distractions (too broad because there are distractions that have nothing to do with technology).
-
Technology (too broad because the question is focused on mobile technology).


11: Soil Systems
By the end of the chapter you should be able to:
- Create and label a diagram of horizon development process.
- Explain how soil color relates to soil properties.
- Define soil texture, structure and bulk density, and explain their importance to soil development.
- Use the soil texture triangle to determine soil texture class.
- Explain basic soil chemistry and relate it to soil fertility.
- Construct, label, and describe the horizons in a typical soil profile.
- Describe each of the soil development factors and how they influence properties of soils.
- Describe the soil forming (pedogenic) processes and their effects on soil properties.
- Describe the major soil orders and explain their general distribution.
Soil is crucial to the functioning of the Earth system. Soil is the foundation upon which vegetation grows and home to a diverse community of organisms. The global mosaic of world soils is a reflection of the geographic variation of climate, vegetation, and geological processes. Soil is a resource easily degraded or destroyed by misuse. Once damaged, it take hundreds of years to regenerate and become productive again.


28.9B: Chromosomal Structural Rearrangements
Structural rearrangements of chromosomes include both inversions and translocations, which may have detrimental effects on an organism.
- Describe the various types of structural rearrangements of chromosomes and how they can impact an organism
Key Points
- A chromosome inversion is the detachment, 180° rotation, and reinsertion of part of a chromosome; this may have no effect on the organism, but if the inversion occurs within a gene or moves a gene away from its regulatory elements it can have an adverse effect.
- Pericentric inversions include the centromere, while paracentric inversions occur outside of the centromere; a pericentric inversion can change the length of the chromosome arms above and below the centromere.
- A pericentric inversion on chromsome 18 appears to have been involved in the evolution of humans.
- A translocation occurs when a segment of a chromosome dissociates and reattaches to a different, nonhomologous chromosome and can be benign or detrimental; in reciprocal translocations, there is no gain or loss of genetic information, so these are usually benign.
Key Terms
- inversion : a segment of DNA in the context of a chromosome that is reversed in orientation relative to a reference karyotype or genome
- translocation : a transfer of a chromosomal segment to a new position, especially on a nonhomologous chromosome
Chromosomal Structural Rearrangements
Cytologists have characterized numerous structural rearrangements in chromosomes, but chromosome inversions and translocations are the most common. Both are identified during meiosis by the adaptive pairing of rearranged chromosomes with their former homologs to maintain appropriate gene alignment. If the genes carried on two homologs are not oriented correctly, a recombination event could result in the loss of genes from one chromosome and the gain of genes on the other. This would produce aneuploid gametes.
Chromosome Inversions
A chromosome inversion is the detachment, 180° rotation, and reinsertion of part of a chromosome. Inversions may occur in nature as a result of mechanical shear, or from the action of transposable elements (special DNA sequences capable of facilitating the rearrangement of chromosome segments with the help of enzymes that cut and paste DNA sequences). Unless they disrupt a gene sequence, inversions only change the orientation of genes and are likely to have milder effects than aneuploid errors. However, altered gene orientation can result in functional changes because regulators of gene expression could be moved out of position with respect to their targets, causing aberrant levels of gene products.
An inversion can be pericentric and include the centromere, or paracentric and occur outside of the centromere. A pericentric inversion that is asymmetric about the centromere can change the relative lengths of the chromosome arms, making these inversions easily identifiable.
Inversions can be pericentric or paracentric : Pericentric inversions include the centromere, and paracentric inversions do not. A pericentric inversion can change the relative lengths of the chromosome arms; a paracentric inversion cannot.
When one homologous chromosome undergoes an inversion, but the other does not, the individual is described as an inversion heterozygote. To maintain point-for-point synapsis during meiosis, one homolog must form a loop, and the other homolog must mold around it. Although this topology can ensure that the genes are correctly aligned, it also forces the homologs to stretch and can be associated with regions of imprecise synapsis.
Inversion heterozygotes : When one chromosome undergoes an inversion, but the other does not, one chromosome must form an inverted loop to retain point-for-point interaction during synapsis. This inversion pairing is essential to maintaining gene alignment during meiosis and to allow for recombination.
Not all structural rearrangements of chromosomes produce nonviable, impaired, or infertile individuals. In rare instances, such a change can result in the evolution of a new species. In fact, a pericentric inversion in chromosome 18 appears to have contributed to the evolution of humans. This inversion is not present in our closest genetic relatives, the chimpanzees. Humans and chimpanzees differ cytogenetically by pericentric inversions on several chromosomes and by the fusion of two separate chromosomes in chimpanzees that correspond to chromosome two in humans.
The pericentric chromosome 18 inversion is believed to have occurred in early humans following their divergence from a common ancestor with chimpanzees approximately five million years ago. Researchers characterizing this inversion have suggested that approximately 19,000 nucleotide bases were duplicated on 18p, and the duplicated region inverted and reinserted on chromosome 18 of an ancestral human.
A comparison of human and chimpanzee genes in the region of this inversion indicates that two genes—ROCK1 and USP14—that are adjacent on chimpanzee chromosome 17 (which corresponds to human chromosome 18) are more distantly positioned on human chromosome 18. This suggests that one of the inversion breakpoints occurred between these two genes. Interestingly, humans and chimpanzees express USP14 at distinct levels in specific cell types, including cortical cells and fibroblasts. Perhaps the chromosome 18 inversion in an ancestral human repositioned specific genes and reset their expression levels in a useful way. Because both ROCK1 and USP14 encode cellular enzymes, a change in their expression could alter cellular function. It is not known how this inversion contributed to hominid evolution, but it appears to be a significant factor in the divergence of humans from other primates.
Translocations
A translocation occurs when a segment of a chromosome dissociates and reattaches to a different, nonhomologous chromosome. Translocations can be benign or have devastating effects depending on how the positions of genes are altered with respect to regulatory sequences. Notably, specific translocations have been associated with several cancers and with schizophrenia. Reciprocal translocations result from the exchange of chromosome segments between two nonhomologous chromosomes such that there is no gain or loss of genetic information.
Reciprocal translocations do not involve loss of genetic information : A reciprocal translocation occurs when a segment of DNA is transferred from one chromosome to another, nonhomologous chromosome.
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2.3: Integers and Rationals
All natural elements of a field \(F,\) their additive inverses, and 0 are called the integral elements of \(F,\) briefly integers .
An element \(x \in F\) is said to be rational iff \(x=\frac{p}{q}\) for some integers \(p\) and \(q\) \((q \neq 0) ; x\) is irrational iff it is not rational.
We denote by \(J\) the set of all integers, and by \(R\) the set of all rationals, in \(F .\)
Every integer
\(p\) is also a
rational
since \(p\) can be written as \(p / q\) with \(q=1\)
Thus
\[R \supseteq J \supset N\]
In an ordered field,
\[N=\{x \in J | x>0\} .(\mathrm{Why} ?)\]
If \(a\) and \(b\) are integers ( or rationals ) in \(F,\) so are \(a+b\) and \(ab\).
- Proof
-
For integers, this follows from Examples (a) and (d) in Section 2; one only has to distinguish three cases:
(i) \(a, b \in N\);
(ii) \(-a \in N, b \in N\);
(iii) \(a \in N,-b \in N\).
The details are left to the reader (see Basic Concepts of Mathematics, Chapter \(2, § 7,\) Theorem 1\()\).
Now let \(a\) and \(b\) be rationals, say,
\[a=\frac{p}{q} \text{ and } b=\frac{r}{s}\]
where \(q s \neq 0 ;\) and \(q s\) and \(p r\) are \(i n t e g e r s\) by the first part of the proof (since \(p, q, r, s \in J ) .\)
\[a \pm b=\frac{p s \pm q r}{q s} \text{ and } a b=\frac{p r}{q s}\]
where \(q s \neq 0 ;\) and \(q s\) and \(p r\) are integers by the first part of the proof (since \(p, q, r, s \in J )\).
Thus \(a \pm b\) and \(a b\) are fractions with integral numerators and denominators. Hence, by definition, \(a \pm b \in R\) and \(a b \in R .\) \(\square\)
In any field \(F,\) the set \(R\) of all rationals is a field itself, under the operations defined in \(F,\) with the same neutral elements 0 and 1. Moreover, \(R\) is an ordered field if \(F\) is. ( We call \(R\) the rational subfield of \(F.)\)
- Proof
-
We have to check that \(R\) satisfies the field axioms.
The closure law 1 follows from Theorem 1.
Axioms 2, 3, and 6 hold for rationals because they hold for all elements of \(F ;\) similarly for Axioms 7 to 9 if \(F\) is ordered.
Axiom 4 holds in \(R\) because the neutral elements 0 and 1 belong to \(R ;\) indeed, they are integers, hence certainly rationals.
To verify Axiom 5, we must show that \(-x\) and \(x^{-1}\) belong to \(R\) if \(x\) does. If, however,
\[x=\frac{p}{q} \quad(p, q \in J, q \neq 0)\]
then
\[-x=\frac{-p}{q}\]
where again \(-p \in J\) by the definition of \(J ;\) thus \(-x \in R\).
If, in addition, \(x \neq 0,\) then \(p \neq 0,\) and
\[x=\frac{p}{q} \text{ implies } x^{-1}=\frac{q}{p} .(\mathrm{Why} ?)\]
Thus \(x^{-1} \in R\). \(\square\)
Note. The representation
\[x=\frac{p}{q} \quad(p, q \in J)\]
is not unique in general; in an ordered field, however, we can always choose \(q>0,\) i.e., \(q \in N(\) take \(p \leq 0\) if \(x \leq 0)\).
Among all such \(q\) there is a least one by Theorem 2 of \(\$ 85-6 .\) If \(x=p / q\), with this minimal \(q \in N,\) we say that the rational \(x\) is given in lowest terms.