Ask Uncle Colin: A Plane-teaser

Dear Uncle Colin,

In a recent contest, I was told that $a$, $b$ and $c$ were real numbers such that $a-7b+8c=4$ and $8a + 4b -c = 7$. I had to find $a^2 – b^2 + c^2$ and couldn’t see a way in. Can you?

– Puzzle Lacks Answer – Nothing Evident!

Hello, PLANE, and thanks for your message!

There are at least a couple of ways to tackle this; I think one is much more reliable than the other.

Algebraic manipulation

Noticing there’s a bit of symmetry in the $a$ and $c$ terms, you might rewrite the equations as $a + 8c = 4 + 7b$ and $8a – c = 7 – 4b$. That’s already looking quite neat.

Square both equations to get $a^2 + 16ac + 64c^2 = 16 + 28b + 49b^2$ and $64a^2 – 16ac + c^2 = 49 – 28b + 16b^2$.

Adding these together will get rid of the $ac$ and $b$ terms, which gives $65a^2 + 65c^2 = 65 + 65b^2$, which means $a^2 – b^2 + c^2 = 1$.

That’s a nice use of the fact that 65 can be written as the sum of two squares in two different ways, incidentally – but it does rely on spotting a neat pattern.

Geometry

My approach would be more geometrical: the two equations define a pair of planes; these intersect in a line, and we can do maths with that.

The planes have normal vectors of $\colvecthree{1}{-7}{8}$ and $\colvecthree{8}{4}{-1}$, so their intersection has a direction parallel to their cross-product, which turns out to be $\colvecthree{-25}{65}{60}$. I prefer to factor out 5 to get $\colvecthree{-5}{13}{12}$, personally – it’s a direction vector and the factor is irrelevant.

Now we to find a point on both planes. It may be possible to find a nice one, but I would pick the more brutal method of “set one of the variables to 0 and see what comes out.” If $c=0$, then $a-7b=4$ and $8a+4b=7$.

If $a=4+7b$, then $8(4+7b)+4b=7$, or $60b=-25$ and $b=-\frac{5}{12}$. That makes $a=\frac{13}{12}$, so a point on the line is $\colvecthree{\frac{13}{12}}{-\frac{5}{12}}{0}$.

Now we can write the equation of the line explicitly: $\colvecthree{a}{b}{c} = \colvecthree{\frac{13}{12}}{-\frac{5}{12}}{0} + \lambda \colvecthree{-5}{13}{12}$.

Almost there:

$a^2 = \br{\frac{13}{12} – 5\lambda}^2 = \frac{169}{144} – \frac{130}{12}\lambda + 25\lambda^2$.
$b^2 = \br{\frac{-5}{12} + 13\lambda}^2 = \frac{25}{144}-\frac{130}{12}\lambda + 169\lambda^2$
$c^2 = 144\lambda^2$.

So, $a^2 – b^2 + c^2 = \frac{1}{144}\br{169-25+0} + \frac{1}{12}\br{-130 + 130} + \lambda^2 \br{25-169+144}$

The first term is 1, the second and third are both zero, so $a^2 – b^2 + c^2 = 1$.

In terms of elegance, I like the first way; in terms of reliability, I like the second.

Hope at least one of them 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.

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