I know that projecting onto a vector means finding the component of another vector in the direction of that vector, so I can use the formula for the projection of $\mathbf{v}$ onto $\mathbf{u}$, which is $\frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}.$ To find the matrix $\mathbf{P}$ that represents this projection for any $\mathbf{v}$, I can use the standard basis vectors $\mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and find the projection of each of them onto $\mathbf{u}.$ Then I can write $\mathbf{P}$ as a matrix whose columns are those projections. So, to find the projection of $\mathbf{i}$ onto $\mathbf{u}$, I use the formula and get $\frac{\mathbf{u} \cdot \mathbf{i}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}.$ To simplify this, I need to compute the dot products. $\mathbf{u} \cdot \mathbf{i} = 4 \cdot 1 + (-7) \cdot 0 = 4.$ $\mathbf{u} \cdot \mathbf{u} = 4 \cdot 4 + (-7) \cdot (-7) = 65.$ So, the projection of $\mathbf{i}$ onto $\mathbf{u}$ is $\frac{4}{65} \mathbf{u}.$ Similarly, to find the projection of $\mathbf{j}$ onto $\mathbf{u}$, I use the formula and get $\frac{\mathbf{u} \cdot \mathbf{j}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}.$ $\mathbf{u} \cdot \mathbf{j} = 4 \cdot 0 + (-7) \cdot 1 = -7.$ So, the projection of $\mathbf{j}$ onto $\mathbf{u}$ is $\frac{-7}{65} \mathbf{u}.$ Now, I can write $\mathbf{P}$ as a matrix whose first column is $\frac{4}{65} \mathbf{u}$ and whose second column is $\frac{-7}{65} \mathbf{u}.$ This means that $\mathbf{P} = \frac{1}{65} \begin{pmatrix} 4 \cdot 4 & 4 \cdot (-7) \\ 4 \cdot (-7) & (-7) \cdot (-7) \end{pmatrix}.$ I can simplify this by multiplying out the numbers and getting $\mathbf{P} = \frac{1}{65} \begin{pmatrix} 16 & -28 \\ -28 & 49 \end{pmatrix}.$ Now, to find the determinant of $\mathbf{P}$, I use the formula for the determinant of a 2x2 matrix, which is $\det \mathbf{P} = \mathbf{P}_{11} \mathbf{P}_{22} - \mathbf{P}_{12} \mathbf{P}_{21}.$ Plugging in the values, I get $\det \mathbf{P} = \frac{1}{65} \cdot \frac{1}{65} \cdot (16 \cdot 49 - (-28) \cdot (-28)).$ I can simplify this by multiplying out the numbers and getting $\det \mathbf{P} = \frac{1}{4225} \cdot (784 - 784).$ This means that $\det \mathbf{P} = 0.$ 
Let $\mathbf{P}$ be the matrix for projecting onto the vector $\begin{pmatrix} 4 \\ -7 \end{pmatrix}.$  Find $\det \mathbf{P}.$