# Thread: Distance from a point to a line.

1. ## Distance from a point to a line.

How can I find

Distance from a point P=(-3,4,2) to a line Lx,y,z)=(3,-2,-1)+(1,-2,2)t

2. If $Q+tD$ is a line the the distance from the point $P$ to the line is $\frac{{\left| {\overrightarrow {QP} \times D} \right|}}{{\left\| D \right\|}}$.

3. ## How can I find q?

My answer was 3 square root 8.

Is this right?

4. I get $\sqrt{65}$.

5. Here's how to do it without that formula. Any plane perpendicular to the line (x, y, z)= )=(3,-2,-1)+(1,-2,2)t is for the form x- 2y+ 2z= C for some constant C. (-3, 4, 2) will be on that plane if -3- 2(4)+ 2(2)= -7= C.

That is, (-3, 4, 2) lies on the plane x- 2y+ 2z= -7 which is perpendicular to the given line. Now a point is on both the line and the plane (i.e. the line passes through the plane) when (3+ t)- 2(-2-t)+ 2(-1+ 2t)= -7. That gives 5+ 9t= -7, 9t= -12, t= -4/3. That tells us, then, that the line passes through the plane at $\left(-3-\frac{4}{3}, -2+ \frac{8}{3},-1-\frac{8}{3}\right)= \left(\frac{5}{3}, \frac{2}{3}, -\frac{11}{3}\right)$. The line from (-3, 4, 2) to that point is perpendicular to the line and so is the shortest distance to the line. And, of course, the distance form (-3, 4, 2) to $\left(\frac{5}{3}, \frac{2}{3}, -\frac{11}{3}\right)$ is $\sqrt{65}$ as Plato says.