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

1. ## 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 $\displaystyle \frac{|PQ \times d|}{|d|}$.

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!

2. To get the cross-product symbol:
$$PQ\times d$$ gives $\displaystyle PQ\times d$

3. Originally Posted by Plato
To get the cross-product symbol:
$$PQ\times d$$ gives $\displaystyle PQ\times d$
Thanks for the info! Any advice you can give with regards to the question I am having difficulty with?

4. And does P lies on the line with direction d?

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

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

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