Distance from Point on Plane to Origin I have a practice question that is asking:
Find the point on the plane 2x - 3y + z = 3 closest to the origin. What is the distance from
that point to the origin?
Here is my work so far (if you can check to see if it is incorrect, please let me know):
d = (2,-3,1)
Vector equation of line: 
x = p + td
x = (0,0,0) + t(2,-3,1)
Parametric form:
x=2t
y=-3t
z=t
Equation:
2x-3y+z=3
2(2t) - 3(-3t) + t = 3
4t + 9t + t = 3
14t = 3
t = 3/14
Therefore:x = 2(3/14), y = -3(3/14), z = 3/14
x = 6/14, y = -9/14, z = 3/14
Closest point to origin: P(3/7, -9/14, 3/14)
But now I have to find the distance from P(3/7, -9/14, 3/14) to the origin. Can someone help me with that? And also can you please check to see what I did to find the point on the plane closest to the origin is correct ? Would greatly appreciate the help. It is a practice question for an upcoming exam so I want to get this correct.
Thanks!
 A: The distance between $P = (\frac{3}{7}, \frac{-9}{14}, \frac{3}{14})$ and the origin is given simply as:
$$\begin{align}\text{dist} &= \sqrt{\left(\frac{3}{7}\right)^2 + \left(\frac{-9}{14}\right)^2 + \left(\frac{3}{14}\right)^2}\\
&= \frac{3}{\sqrt{14}}\end{align}$$
However, it is probably worth noting that one can calculate this distance without having to find the point on the plane nearest to the origin. This saves quite a lot of time in the exams, and can be used as a quick and convenient check against your answer.
Recall that the scalar product definition of  a plane says that the dot product of any vector on the plane with the normal of the plane gives a constant. In your case,
$$\vec{r}\cdot(2, -3, 1) = 3$$
But the dot product of any vector on a plane with the unit normal of the plane gives the length of projection of the vector along the normal, which is the shortest distance from the plane to the origin. We exploit this as follows:
$$\begin{align}\text{dist} &= \vec{r}\cdot\frac{(2, -3, 1)}{\sqrt{2^2 + (-3)^2 + 1^2}} = \frac{3}{\sqrt{14}}\\
&= \frac{3}{\sqrt{2^2 + (-3)^2 + 1^2}} \\
&= \frac{3}{\sqrt{14}}\end{align}$$
It is possible to extend this into finding the shortest between two arbitrary parallel planes in general.
A: That is a completely legitimate method of computing the point on a plane nearest to the origin.  Now that you have your point $P$, use the distance formula to find the distance from $P$ to the origin $(0,0,0)$.  Recall, the distance from $P = (x_1,y_1,z_1)$ to $Q = (x_2,y_2,z_2)$ is given by the formula:
$$
\text{dist}(P,Q) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}.
$$
Compare this to Pythagorean's theorem which describes the distance formula in 2-dimensions.
A: Find the perpendicular line to your line through the origin. Now figure out where it hits your line.  Take the distance from that point to the origin. Done.
