<s>[INST] Finding a perpendicular vector This is given: $P = (-1,1,0),\; Q = (1,5,6)\; \text{and} \; R = (3,-1,4).$
My question is:
Find the values of $x$ (where x is a real number) for which $PR + xQR\;$ is perpendicular to $PR$.
So far, what I have done is used the dot product of two vectors and equated 
$\vec {u} \cdot \vec{v} = 0$ and my result was $x = -3$.

Am I missing any other values? Thank you.
 [/INST] You are correct:
$$PR = (4, -2, 4)\\
QR = (2, -6, -2)\\
PR + xQR = (4 + 2x, -2-6x, 4-2x)\\
PR\cdot (PR + xQR) = 4(4+2x) - 2(-2 -6x) + 4(4-2x) = \\=16+8x+4+12x+16-8x=36+12x$$
So the solution is $28+12x=0$ or $x=-\frac{36}{12} = -3$.
Alternatively, you are solving the equation $$PR\cdot(PR-x\cdot QR) = 0$$
You can rewrite that into $$PR\cdot PR = -x\cdot (PR\cdot QR)$$
Now, since $PR\cdot QR$ is a real number, you get
$$x=-\frac{PR\cdot PR}{PR\cdot QR}.$$
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