# Projection of the vector $\vec{b}=\left(\begin{smallmatrix}3\\-1\\4\end{smallmatrix}\right)$ in some given direction

Find the projection of the vector $\vec{b}=\begin{pmatrix} 3 \\ -1\\ 4 \end{pmatrix}$ in the direction

$(a)$ perpendicular to the plane: $$2x-2y+z=5.$$ $(b)$ parallel to the plane: $$2x-2y+z=5.$$

For the part $(a)$ projecting $\vec{b}=\begin{pmatrix}3 \\ -1\\ 4\end{pmatrix}$ to $\vec{a}=\begin{pmatrix}2 \\ -2\\ 1\end{pmatrix}$ will give me the perpendicular projection right? For part $(b)$ I know I have to subtract what I found in a from a vector to find the projection of the vector parallel to the plane, but I'm kinda lost.

• You are very correct. Just proceed as you've written. Feb 6 '18 at 5:57
• I am lost in getting the vector that I am supposed to subtract the projection from... Feb 6 '18 at 6:03

The plane $E = 2x-2y+z=5$ is defined by the normal vector $\vec{n}=\begin{pmatrix}{2 \\ -2\\ 1}\end{pmatrix}$ and the projection of $\vec{b}$ on $\vec{n}$, we define it as $\vec{l}$, is obtained by $\,|l|\cdot \left(\frac{1}{|n|}\vec{n}\right)\,$ (See unit vector)
It follows: $$|l| = |b|\cdot \cos(\alpha) = |b|\frac{\vec{b}\cdot \vec{n}}{|b|\cdot |n|} = \frac{\vec{b}\cdot \vec{n}}{|n|} = 4,\quad\text{ and }\quad \frac{1}{|n|}\vec{n} = \frac{1}{3}\cdot\begin{pmatrix}{2 \\ -2\\ 1}\end{pmatrix}$$ $$\vec{l} = |l|\cdot \left(\frac{1}{|n|}\cdot\vec{n}\right) = 4\cdot\frac{1}{3}\cdot \begin{pmatrix}{2 \\ -2\\ 1}\end{pmatrix} = \begin{pmatrix}{\frac{8}{3} \\ -\frac{8}{3}\\ \frac{4}{3}}\end{pmatrix}$$ $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
And the vector $\vec{t}$ ist parallel to the plane $E$. This is also the projection of $\vec{b}$ on $E$:
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$ $$\vec{t} = \vec{b}-\vec{l} = \begin{pmatrix}{\quad 3 - \frac{8}{3}\\ -1 + \frac{8}{3}\\\quad 4 - \frac{4}{3}}\end{pmatrix} = \begin{pmatrix}{\frac13\\\frac53\\\frac83 }\end{pmatrix}$$