# Ask Uncle Colin: A Pair of Vectors

Dear Uncle Colin,

The question says "Under what circumstances does $|\mathbf{a} + \mathbf{b} | = | \mathbf{a} | + | \mathbf{b} |$ and I don’t have a clue. Can you help?

Sines? Cosines? Angle Leaves A Residue

Hi, SCALAR, and thanks for your message!

I have several ways to approach this, but I think the two best ones are algebraic and geometric.

## Algebra: square it

Since $| \mathbf{x} |$ means $\sqrt{ \mathbf{x}\cdot \mathbf{x} }$, squaring both sides seems like the way to go.

On the left:

$\left( |\mathbf{a} + \mathbf{b} |\right)^2 = \left(\mathbf{a} + \mathbf{b}\right)\cdot \left(\mathbf{a} + \mathbf{b} \right)$

$\dots = \mathbf{a}\cdot\mathbf{a} + 2\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot \mathbf{b}$.

On the right:

$\left(|\mathbf{a}| + |\mathbf{b}|\right)^2 = \mathbf{a}\cdot\mathbf{a} + \mathbf{b}\cdot\mathbf{b} + 2|\mathbf{a}||\mathbf{b}|$.

So, getting rid of the $\mathbf{a}\cdot\mathbf{a}$s and $\mathbf{b}\cdot\mathbf{b}$s, we have:

$\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|$

By the definition of the dot product, $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos(\theta)$, so $\cos(\theta) = 1$.

That means the angle between $\mathbf{a}$ and $\mathbf{b}$ is 0 - the two vectors are parallel.

## Geometry: think about it

Because of the way vectors are added together, $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{a} + \mathbf{b}$ can form the three sides of a triangle, and their magnitudes are the lengths of those sides.

If the lengths of the first two sides are the same as the last side, the ‘triangle’ is more of a straight line! Vectors $\mathbf{a}$ and $\mathbf{b}$ must be going in the same direction.

Hope that helps!

- Uncle Colin

* Edited 2020-02-12 to correct a variable name. Thanks, RM3!

## 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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##### 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.