Write $\mathbf u =\begin{bmatrix}9\\-1\\2\end{bmatrix}$ as a sum of two vectors where one is orthogonal to the plane $x+2y-2z=0$ and the other is parallel to the same plane.
I've attempted to use the fact that the normal vector to the plane, $ \begin{bmatrix}1\\2\\-2\end{bmatrix}$, can be seen from the given plane equation above. We should be able to write $\mathbf u$ as
$$\mathbf u =\begin{bmatrix}9\\-1\\2\end{bmatrix} = s\begin{bmatrix}1\\2\\-2\end{bmatrix}-t\begin{bmatrix}a\\b\\c\end{bmatrix}.$$
How do I find $a,b,c$ in the vector that is parallel to the plane?
Best Answer
Vector projection is your best friend. By projecting $\vec{u}$ onto $\vec{n}:=\langle1,2,-2\rangle$, you capture ALL of the movement of $\vec{u}$ that is in the direction of $\vec{n}$; therefore, what is left over should be orthogonal.
Vector projection is most easily expressed when the target vector is a unit vector; so, let's compute $$ \hat{n}:=\frac{\vec{n}}{\|\vec{n}\|}=\begin{bmatrix}\frac{1}{3}\\\frac{2}{3}\\-\frac{2}{3}\end{bmatrix}. $$
Then we can compute $$ \text{proj}_{\vec{n}}(\vec{u})=\text{proj}_{\hat{n}}(\vec{u}):=(\vec{u}\cdot\hat{n})\,\hat{n}=1\cdot\begin{bmatrix}\frac{1}{3}\\\frac{2}{3}\\-\frac{2}{3}\end{bmatrix}=\begin{bmatrix}\frac{1}{3}\\\frac{2}{3}\\-\frac{2}{3}\end{bmatrix}. $$ Then, consider what happens if we 'remove' this component from $\vec{u}$: $$ \vec{u}-\text{proj}_{\hat{n}}(\vec{u})=\begin{bmatrix}\frac{26}{3}\\-\frac{5}{3}\\\frac{8}{3}\end{bmatrix}. $$ You can verify that this is orthogonal to $\vec{n}$ (or $\hat{n}$) via the dot product.