Ask Uncle Colin: Perpendicular vectors

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

I'm struggling a bit with my C4 vectors. Most of it is fine, except when I have to find a point $P$ on a given line such that $\vec{AP}$ is perpendicular to the line, for some known $A$. How do I figure that out?

— Any Vector Insight Lavishly Appreciated

Hi, AVILA, and thanks for your message!

There are a couple of approaches I like for this, both of which involve the dot product in slightly different ways.

The perpendicular way

This is the way I was taught. Suppose you've got the equation of a line as $\mathbf{r} = \colvecthree{x_0}{y_0}{z_0} + \lambda \colvecthree{\Delta_x}{\Delta_y}{\Delta_z}$, and a known point $A$ at $\colvecthree{X}{Y}{Z}$. For any point $P$ on the line, $\vec{AP} = \colvecthree{X-x_0}{Y-y_0}{Z-z_0} – \lambda_P \colvecthree{\Delta_x}{\Delta_y}{\Delta_z}$, for some value of $\lambda_P$.

For that to be perpendicular to the line, the scalar (dot) product between $\vec{AP}$ and the direction vector of the line, $\colvecthree{\Delta_x}{\Delta_y}{\Delta_z}$, must be zero. So, all you need to do is solve: $\left(\colvecthree{X-x_0}{Y-y_0}{Z-z_0} – \lambda_P \colvecthree{\Delta_x}{\Delta_y}{\Delta_z}\right)\cdot \colvecthree{\Delta_x}{\Delta_y}{\Delta_z} = 0$ for $\lambda_P$, the only thing you don't know.

You can try to be clever about rearranging and simplifying, but I wouldn't bother: stick in the numbers and get an answer; then put $\lambda_P$ back into the equation of the line to find point $P$.

The trigonometry way

A way I like slightly better, as I can see the geometry behind it, is to make a right-angled triangle between a known point $Q$ on the line1, known point $A$ and point $P$. It helps if you normalise the direction vector of the line: divide it by its modulus and call the result $\mathbf{\hat v}$.

You know the length of the hypotenuse: it's simply $|\vec{QA}|$.

You can easily figure out the cosine of the angle $A\hat QP$: call the direction vector of the line $\mathbf{v}$ and it's $\cos(\theta) = \frac{\vec{QA} \cdot \mathbf{\hat v}}{|\vec{QA}|}$ .

Then the distance $|\vec{QP}|$ is then $|\vec{QA}| \cos(\theta)$, using standard right-angled trigonometry.

This is a place where simplification helps: if you notice the $|\vec{QA}|$ on the bottom of the $\cos(\theta)$ fraction, you can see that the distance you want is $d = \vec{QA} \cdot \mathbf{\hat {v}}$.

Now you just need to go this far along the line from $Q$ – which means you need to add on $d$ lots of $\mathbf {\hat v}$.

Your final answer for point $P$ is $\mathbf{q} + \left(\vec{QA} \cdot \mathbf{\hat{v}}\right) \mathbf{\hat{v}}$, where $\mathbf{q}$ is the vector $\vec{OQ}$.

Hope that helps!

— Uncle Colin

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. if you're not given one, you can find one by sticking in any value you like for $\lambda$; my favourite is 0. []

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