Dear Uncle Colin,

I’m told that $v + 3w$ is perpendicular to $7v - 5w$ and that $v-4w$ is perpendicular to $7v - 2w$, and I have to find the angle between $v$ and $w$. I know it’s something to do with the dot product, but I’ve not been able to make it shake out. Can you help?

- Student Can’t Arrange Linear Algebra Right

Hi, SCALAR, and thanks for your message!

I think you’re right, the dot product is your friend here. Let’s write down what we know:

  • $(v + 3w) \cdot (7v - 5w) = 0$, or $7 v\cdot v + 16 v \cdot w - 15 w \cdot w = 0$, because the dot product commutes;
  • $(v-4w)\cdot (7v - 2w) = 0$, or $7v \cdot v - 30 v \cdot w + 8 w \cdot w = 0$.

Now, the definition of the dot product is that $a \cdot b = | a | | b | \cos(\theta)$, where $\theta$ is the angle between $a$ and $b$.

We can write that as $\cos(\theta) = \frac{a \cdot b}{\sqrt{(a \cdot a)(b \cdot b)}}$.

OK: I’m fed up of writing dot products, so I’m going to let $V = v \cdot v$, $W = w \cdot w$ and $X = v \cdot w$.

We have (after a little rearrangement):

  • $16X = 15W - 7V$
  • $30X = 8W + 7V$

And we’re looking for $\frac{X}{\sqrt{VW}}$, which I’ll call $C$.

A rabbit from the hat

It’s not really a rabbit. I’m going to divide both equations by $\sqrt{VW}$ and get:

  • $16C = 15 \sqrt{\frac{W}{V}} - 7\sqrt{\frac{V}{W}}$
  • $30C = 8 \sqrt{\frac{W}{V}} + 7\sqrt{\frac{V}{W}}$.

A bit ugly, and I’m conscious I’m introducing a lot of letters, and there’s another on its way: let $Y = \sqrt{\frac{W}{V}}$:

  • $16C = 15Y - \frac{7}{Y}$
  • $30C = 8Y + \frac{7}{Y}$

If nothing else, we’ve at least got it down to two variables! In this case, I think it’s easiest to eliminate $C$ and substitute back:

  • $240C = 225Y - \frac{105}{Y}$
  • $240C = 64Y + \frac{56}{Y}$

So $0 = 161Y - \frac{161}{Y}$

So, after all of that, $Y = \pm 1$. In fact, it’s a square root, so it has to be $1$.

That gives us $16C = 8$, so $C = \frac{1}{2}$ and the (acute) angle between the vectors is $\frac{\pi}{3}$ radians.

Given how nice the numbers are, I can’t help but wonder if there’s a more straightforward, less alphabetti spaghetti, route to the answer.

I hope that helps!

- Uncle Colin