 # Lety=\begin{bmatrix}3\\1\end{bmatrix}andu=\begin{bmatrix}8\\ York 2021-11-05 Answered

Let
$\left[\begin{array}{c}3\\ 1\end{array}\right]$
and $\left[\begin{array}{c}8\\ 6\end{array}\right]$
Compute the distance from y to the line through u and the origin.

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it hesgidiauE

Formula for orthogonal projection:
$\stackrel{^}{y}=\frac{y\cdot u}{u\cdot u}u$
Formula for distance:
$||y-\stackrel{^}{y}||$
$y\cdot u=\left(3,1\right)\cdot \left(8,6\right)=24+6=30$
$u\cdot u=\left(8,6\right)\cdot \left(8,6\right)=64+36=100$
$\stackrel{^}{y}=\frac{30}{100}u=\frac{3}{6}\left[\begin{array}{c}8\\ 6\end{array}\right]=\left[\begin{array}{c}24/10\\ 18/10\end{array}\right]$
$y-\stackrel{^}{y}=\left[\begin{array}{c}3\\ 1\end{array}\right]-\left[\begin{array}{c}24/10\\ 18/10\end{array}\right]=\left[\begin{array}{c}3/5\\ -4/5\end{array}\right]$
$||y-\stackrel{^}{y}||-\sqrt{\frac{9}{25}+\frac{16}{25}}=\sqrt{1}=1$