Question

$scalar\:projection\:\begin{pmatrix}1&2\end{pmatrix},\:\begin{pmatrix}3&-8\end{pmatrix}$

Answer 1

$\mathrm{Computing\:the\:scalar\:projection\:of\:}\begin{pmatrix}3&-8\end{pmatrix}\mathrm{\:onto\:}\begin{pmatrix}1&2\end{pmatrix}:\quad -\frac{13\sqrt{5}}{5}$

$\mathrm{Finding\:the\:scalar\:projection\:of\:}\vec{b\:}\mathrm{\:onto\:}\vec{a\:}:\quad \:comp_{\vec{a\:}}\left(\vec{b\:}\right)=\frac{\vec{a\:}\cdot \vec{b\:}}{\left|\vec{a\:}\right|}$

$\vec{a\:}=\begin{pmatrix}1&2\end{pmatrix},\:\quad \vec{b\:}=\begin{pmatrix}3&-8\end{pmatrix}$

$\vec{a\:}\cdot \vec{b\:}:\quad \begin{pmatrix}1&2\end{pmatrix}\begin{pmatrix}3&-8\end{pmatrix}=-13$

$\left|\vec{a\:}\right|:\quad \left|\begin{pmatrix}1&2\end{pmatrix}\right|=\sqrt{5}$

$\frac{\vec{a\:}\cdot \vec{b\:}}{\left|\vec{a\:}\right|}=\frac{-13}{\sqrt{5}}$

$\mathrm{Simplify}$

$=-\frac{13\sqrt{5}}{5}$

Answer 2

projection of a on b is a vector a1 which is either null or parallel to b. More exactly: * a1 = 0 if θ = 90°, * a1 and b have the same direction if 0 ≤ θ < 90 degrees, * a1 and b have opposite directions if 90 degrees < θ ≤ 180 degrees.

scalar projection (1 2) (3 -8) = $-\frac{13\sqrt{5}}{5}$