Ask Uncle Colin: A STEP vectors problem

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 and Uncle Colin will do what he can.

Dear Uncle Colin,

I'm struggling with this STEP question. The first two parts are fine -- equality holds when there is some constant $k$ for which $a = kx$, $b = ky$ and $c=kz$, and part (i) follows directly from the original inequality.

Screen Shot 2016-07-20 at 22.26.03

I can get an answer to part (ii) -- $\colvecthree {p}{q}{r} = \colvecthree{24}{6}{1}$ -- but that's from luck, not from using the inequality. (I tried using $\colvecthree {1}{4}{9}$, but didn't get anywhere. How would you go about it?

-- Students Can't Always Look At Rationales

That would have been my first thought as well, SCALAR! However, there's a clue (under the devious STEP-setter definition of "something that's not really a clue unless you know what you're looking for") in that all of the elements on the left of the first equation are squares. That suggests, to a certain kind of mind, that $\colvecthree{p}{2q}{3r}$ might be the relevant one. This suggestion is strengthened by noticing that the second equation can be written as $\colvecthree {8}{4}{1} \cdot \colvecthree {p}{2q}{3r} = 243$.

Let's let $\bb a = \colvecthree {p}{2q}{3r}$ and $\bb b = \colvecthree {8}{4}{1}$, and look at the original inequality, which says that $\left( \bb a \cdot \bb b \right)^2 \le \left| \bb a \right|^2 \left| \bb b \right|^2$.

Here, the left hand side is $(243)^2$, which I'm not going to work out because a) I can see it's $3^{10}$ and that's all I care about and b) I'm going to divide it by stuff shortly.

The right hand side is $\left| \bb a \right|^2 \left| \bb b \right|^2$; however, the first factor is 729 (from what we're told) and the second is 81 (by working out the modulus of $\colvecthree {8}{4}{1}$).

That tells us that $3^{10} \le 3^6 \times 3^4$. Well, we sorta knew that already. What we didn't know was that the two were equal -- which means $\bb a$ is a multiple of $\bb b$, as established before.

The easiest way I can see to find out what multiple is to notice that we already know $\left| \bb{a} \right|^2 = 729$ and $\left| \bb{b} \right|^2 = 81$, so if $\bb{a} = k \bb{b}$, then $729 = k^2 \times 81$ and $k=\pm 3$. (It has to be positive, since $\bb a \cdot \bb b = k \bb b \cdot \bb b = k \left| \bb b \right|^2$ is.)

So, since we know $k=3$, we know that $\colvecthree {p}{2q}{3r} = \colvecthree {24}{12}{3}$, so $p=24$, $q=6$ and $r=1$. A quick check in the original equations shows that these do, indeed work.


-- Uncle Colin

* Post edited 2016-07-20 to put the question in! Thanks to Andy for pointing out that it was missing.


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.


2 comments on “Ask Uncle Colin: A STEP vectors problem

  • Andy

    Where’s the question Colin?

    • Colin

      Er… good question, well presented! (I’ve put it in now. Sorry for the confusion.)

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter