# Distance between point and line (vectors)

• Sep 12th 2010, 07:07 AM
SyNtHeSiS
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?
• Sep 12th 2010, 07:14 AM
Unknown008
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

$sin(M\hat{B}A) = \frac{|AM|}{|AB|}$
• Sep 12th 2010, 08:49 AM
Plato
Quote:

Originally Posted by SyNtHeSiS
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 $d(R;\ell ) = \left\| {\overrightarrow {PR} - \frac{{\overrightarrow {PR} \cdot D}}{{D \cdot D}}D} \right\| = \dfrac{{\left\| {\overrightarrow {PR} \times D} \right\|}}{{\left\| D \right\|}}$.