So my question has two parts:
a) Let L be a line given by y=2x, find the projection of $\vec{x}$=$\begin{bmatrix}5\\3\end{bmatrix}$ onto the line L.
So, for this one:
proj$_L$($\vec{x}$) = $\frac{\vec{x}\bullet \vec{y}}{\vec{y}\bullet \vec{y}}$$\times \vec{y}$ = $\frac{(\begin{bmatrix}5\\3\end{bmatrix} \bullet \begin{bmatrix}2\\1\end{bmatrix}}{(\begin{bmatrix}2\\1\end{bmatrix} \bullet \begin{bmatrix}2\\1\end{bmatrix}} ) \times \begin{bmatrix}2\\1\end{bmatrix}$ = $\frac{13}{5} \times \begin{bmatrix}2\\1\end{bmatrix}$ = \begin{bmatrix}5.2\\2.6\end{bmatrix}
b) using the above, find the sitance between L and the terminal point of x.
Here is where I am stuck… my instinct is to just do:
$\begin{bmatrix}5\\3\end{bmatrix} – \begin{bmatrix}5.2\\2.6\end{bmatrix}$ = $\begin{bmatrix}-.2\\.4\end{bmatrix}$
but I'm sure this is incorrect… how would I solve this?
Best Answer
Yes, yes, almost done. You need the length of this distance vector, use Pythagorean theorem.
One moment, your line is $y=2x$, then it rather contains $\pmatrix{1\\2}$ than $\pmatrix{2\\1}$ (and its normalvector is $\pmatrix{2\\-1}$)..