1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ You are very correct. Just proceed as you've written. $\endgroup$ Feb 6 '18 at 5:57
  • $\begingroup$ I am lost in getting the vector that I am supposed to subtract the projection from... $\endgroup$
    – x1124970
    Feb 6 '18 at 6:03
1
$\begingroup$

Perpendicular
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: $$|l| = |b|\cdot \cos(\alpha) = |b|\frac{\vec{b}\cdot \vec{n}}{|b|\cdot |n|} = \frac{\vec{b}\cdot \vec{n}}{|n|} = 4,\quad\text{ and }\quad \frac{1}{|n|}\vec{n} = \frac{1}{3}\cdot\begin{pmatrix}{2 \\ -2\\ 1}\end{pmatrix}$$ $$\vec{l} = |l|\cdot \left(\frac{1}{|n|}\cdot\vec{n}\right) = 4\cdot\frac{1}{3}\cdot \begin{pmatrix}{2 \\ -2\\ 1}\end{pmatrix} = \begin{pmatrix}{\frac{8}{3} \\ -\frac{8}{3}\\ \frac{4}{3}}\end{pmatrix}$$ $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$enter image description here

Parallel
And the vector $\vec{t}$ ist parallel to the plane $E$. This is also the projection of $\vec{b}$ on $E$:

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$enter image description here $$\vec{t} = \vec{b}-\vec{l} = \begin{pmatrix}{\quad 3 - \frac{8}{3}\\ -1 + \frac{8}{3}\\\quad 4 - \frac{4}{3}}\end{pmatrix} = \begin{pmatrix}{\frac13\\\frac53\\\frac83 }\end{pmatrix} $$

$\endgroup$
2
  • $\begingroup$ shouldn't the equation for the parallel vector be t = b-l? $\endgroup$
    – x1124970
    Feb 6 '18 at 22:50
  • $\begingroup$ @x1124970 Yes, you are right, I have updated the answer. Thanks $\endgroup$ Feb 7 '18 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.