Linear algebra: orthogonal projection?

Question

(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?

Answer 1

In the first part, they want you to first find the normal vector to the plane provided. Let this vector be $N$, and now find the orthogonal projection of $(-1,0,8)$ on $N$.

For the second part they want you to find the distance from a point to a plane. The distance from a point to a plane can be found by taking any vector $v$ from the plane to the point, and then projecting this vector $v$ onto a vector which is normal to the plane. Since the origin is in the plane $x-2y+z=0$, you can consider $v$ as the vector from the origin to the point. If the plane did not pass through the origin, you would have had to choose a different point on the plane first.

Hint: In the first part, you found the orthogonal projection of $(-1,0,8)$ onto a normal vector to the plane, so you can save yourself some work in the second part.

Answer 2

(a) "Onto the normal vector" means you need to find the projection NOT on the plane, but to it's orthogonal complement (which is spanned by $(1,-2,1)$).
(b) The distance is exactly the length of the projection (try this in the plane with a line and a point).

Answer 3

The plane $ax + by + cz = d$ has normal vector $n = ai + bj + ck$. A line perpendicular to a plane has the plane's normal as it direction vector. If your point is $(x_0, y_0, z_0)$, the line has parametric representation $x = x_0 + at$, $y = y_0 + bt$, $z = z_0 + ct$.

Once you do this, it's easy to find the intersection point and everything you seek.