Thread: Distance between a point and a line using vector geometry and projections

1. Distance between a point and a line using vector geometry and projections

I'm stuck on a question:

Find the point on the line

x = [1,2,3]+m[2,1,-2] (where x is a vector in R3 and m is a scalar)

closest to b=[3,2,1] (where b is a vector in R3).

Also find the distance from b to the line.

I'm guessing that I have to use the projection formula to start off with but I'm not sure which two vectors to use in the equation...

Thank you!

2. Re: Distance between a point and a line using vector geometry and projections

Originally Posted by JDoops
Find the point on the line
x = [1,2,3]+m[2,1,-2] (where x is a vector in R3 and m is a scalar)
closest to b=[3,2,1] (where b is a vector in R3).
Also find the distance from b to the line.
Suppose that $P$ not on the line $\ell: Q+tD$
The distance from $P$ to $\ell$ is $\dfrac{\|\overrightarrow {QP}\times D\|}{\| D\|}$

To find the point on $\ell$ closest to $P$, write the equation of the plane
$D\cdot\overrightarrow {QR}=0$ where $\vec{R}=<x,y,z>$.
Then find where the line intersects that plane.

3. Re: Distance between a point and a line using vector geometry and projections

Do you understand that the shortest distance for a point to a line is along the perpendicular to that line? The line x = [1,2,3]+m[2,1,-2] has direction vector [2, 1, -2] and the plane perpendicular to that, containing the point [3, 2, 1], is given by [2, 1, -2].[x- 3, y- 2, z- 1]= 2(x- 3)+ (y- 2)- 2(z- 1)= 0 or 2x+ y- 2z- 6= 0. The line [1, 2, 3]+ m[2, 1, -2] which is the same as x= 1+ 2m, y= 2+ m, z= 3- 2m, crosses that plane where 2(1+ 2m)+ (2+ m)- 2(3- 2,)- 6= 0. Solve that equation for m to find the point where the line crosses the plane, then find the distance between that point and [3, 2, 1].