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

  1. #1
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    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!
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  2. #2
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    Re: Distance between a point and a line using vector geometry and projections

    Quote Originally Posted by JDoops View Post
    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.
    Thanks from JDoops
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  3. #3
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    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].
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