Having difficulty with a proof - distance from a point to a line

**Kakariki** · June 6th 2010, 05:52 AM

Hey!

I am having trouble proving something, and I was hoping someone here could help me out!

Question
Prove that the distance from a point Q in space to a line through a point P with direction vector d is equal to $\frac{|PQ \times d|}{|d|}$ .

Solution
What I have learned about this topic so far is: to find the smallest distance from a point to a line you find the equation of a line that passes through the point and is perpendicular to the line.
I really have no idea where to start with this proof, any help is greatly appreciated!

**Plato** · June 6th 2010, 06:02 AM

To get the cross-product symbol:
[tex]PQ\times d[/tex] gives $PQ\times d$

**Kakariki** · June 6th 2010, 06:33 AM

Originally Posted by Plato

To get the cross-product symbol:
[tex]PQ\times d[/tex] gives $PQ\times d$

Thanks for the info! Any advice you can give with regards to the question I am having difficulty with?

**p0oint** · June 6th 2010, 06:34 AM

And does P lies on the line with direction d?

**vincisonfire** · June 6th 2010, 06:46 AM

Make a drawing, you will find that the distance is $\left| PQ sin(\theta) \right|$ .
Hence, play with $PQ\times d$ …

**Contributor** · June 6th 2010, 06:48 AM

It should be obvious if you notice that $|PQ \times d|$ is the area of a parallelogram and $|d|$ is the length of one of its sides. A plot will help to see the solution.

**Plato** · June 6th 2010, 08:15 AM

Originally Posted by Kakariki

Thanks for the info! Any advice you can give with regards to the question I am having difficulty with?

We know that $\sin (\phi ) = \frac{{\left\| {\overrightarrow {QP} \times d} \right\|}}{{\left\| {\overrightarrow {QP} } \right\|\left\| d \right\|}}$ where $\phi$ is the angle between $\overrightarrow {QP}$ and the line.
From that we see $D(P;\ell ) = \left\| {\overrightarrow {QP} } \right\|\sin (\phi ) = \frac{{\left\| {\overrightarrow {QP} \times d} \right\|}}{{\left\| d \right\|}}$

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