(a) Find the orthogonal projection of $(-1, 0, 8)$ onto the normal vector to the plane $x-2y+z=0$.
Is this question saying to find the orthogonal projection in other words? The way the question is phrased "onto the normal vector to the plane" is confusing me..
(b) Find the distance from the point $(-1, 0, 8)$ to the plane $x-2y+z=0$. Answer: The distance from the point $(-1, 0, 8)$ to the plane is the length of the projection onto the normal direction of any vector $v$ which connects a point on the plane to the point $(-1, 0 , 8)$. Since the plane contains the origin, we can choose $v = (-1, 0, 8)$ and compute the projection.
I don't get what this is saying.. is the question asking us to find the $u$ if $v=w+u$? Also, how come we can choose $(-1, 0, 8)$ just because the plane contains the origin?