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

• June 6th 2010, 05:52 AM
Kakariki
Having difficulty with a proof - distance from a point to a line
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!
• June 6th 2010, 06:02 AM
Plato
To get the cross-product symbol:
$$PQ\times d$$ gives $PQ\times d$
• June 6th 2010, 06:33 AM
Kakariki
Quote:

Originally Posted by Plato
To get the cross-product symbol:
$$PQ\times d$$ gives $PQ\times d$

Thanks for the info! Any advice you can give with regards to the question I am having difficulty with?
• June 6th 2010, 06:34 AM
p0oint
And does P lies on the line with direction d?
• June 6th 2010, 06:46 AM
vincisonfire
Make a drawing, you will find that the distance is $\left| PQ sin(\theta) \right|$.
Hence, play with $PQ\times d$
• June 6th 2010, 06:48 AM
Contributor
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.
• June 6th 2010, 08:15 AM
Plato
Quote:

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 $D(P;\ell ) = \left\| {\overrightarrow {QP} } \right\|\sin (\phi ) = \frac{{\left\| {\overrightarrow {QP} \times d} \right\|}}{{\left\| d \right\|}}$