Why is the residual vector y minus the projection perpendicular to the line spanned by v?

First, find the residual by subtracting the projection $(8,4)$ from $y=(7,6)$, resulting in $(-1,2)$. Then, compute the dot product of the projection $(8,4)$ and the residual $(-1,2)$.

Conditions: $y = (7,6)$; Projection is $(8,4)$; Residual is $(-1,2)$

First, calculate the dot products $y \cdot u = 20$ and $u \cdot u = 5$. The scalar coefficient is $20/5 = 4$.

Conditions: $y = (7,6)$; $u = (2,1)$

The orthogonal projection of $y$ onto the line spanned by a nonzero vector $v$ is given by the formula $\frac{y \cdot v}{v \cdot v} v$. This formula scales the vector $v$ by the ratio of the dot product of $y$ and $v$ to the dot product of $v$ with itself.

Conditions: The vector $v$ must be nonzero.; The projection is onto the line spanned by $v$.

Calculate the projection $p$ of $y$ onto a line spanned by $v$. Calculate the residual $r = y - p$.

Conditions: $v$ is a nonzero vector defining the line.; $y$ is the vector to be decomposed.