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Thread: Distance between point and line (vectors)

  1. #1
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    Distance between point and line (vectors)

    Find the distance from the point A(7,6,10) to the line passing through B(3,7,0) and C(4,9,-2).

    I know that the vector is \overrightarrow{BC} = \left \langle 1,2,-2 \right \rangle and that the equation of the line is thus \mathbf{x} = (3,7,0) + t(1,2,-2) t\epsilon \mathbb{R} but I dont understand how you can find the distance between A and this line?
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    What you can try is to find the vector direction of AB. Then, find the angle CBA by using the dot product of the vectors.

    Have a sketch, BC is a line, AB is another line. Draw the angle CBA. Drop a perpendicular line from A to line CB.

    Now, the perpendicular line, AB and BM, where M is the point where the perpendicular meets BC is a triangle.

    Have now a right angled triangle, an angle and you can find the length AB. Use

    sin(M\hat{B}A) = \frac{|AM|}{|AB|}
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    Quote Originally Posted by SyNtHeSiS View Post
    Find the distance from the point A(7,6,10) to the line passing through B(3,7,0) and C(4,9,-2).
    Given a line \ell :P + tD and that R\notin \ell then the distance from R to \ell is d(R;\ell ) = \left\| {\overrightarrow {PR}  - \frac{{\overrightarrow {PR}  \cdot D}}{{D \cdot D}}D} \right\| = \dfrac{{\left\| {\overrightarrow {PR}  \times D} \right\|}}{{\left\| D \right\|}}.
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