Question

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

Answer 1

Explanation:

Here,

( 1 2) , ( 3 -8)

As a scalar ,

1 + 3 =0

2 + ( -8) =0

or, 4 -6 = 0

Answer 2

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

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

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

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

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

$\mathrm{Simplify}$

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