OK, so you’ve got to grips with the SUVAT equations, you’re on top of resolving forces, you understand that $F=ma$ and you have M1 under control… only for them to start throwing $\bi$s and $\bj$s around. Who ordered those?

Maybe you have a vague recollection of vectors from GCSE - although it’s quite possible you learnt the two tricks you were expected to do at that level and promptly forgot about them; in fact, your GCSE knowledge probably won’t help all that much (except to help you notice that vectors are what the bold letters represent.)

### There’s good news

A good deal of what you need to do for M1 vector questions simply involves treating $\bi$ and $\bj$ as constants - you can get a long way without worrying about “this is east and this is north”. You can - almost - use the SUVAT equations and $\bb{F} = m\bb{a}$ as they stand ((I’ve bolded the $F$ and the $a$. Those are vectors, because forces and acceleration have direction, while mass doesn’t.)) For example, if you know the mass of a particle is 4kg and its acceleration is $-3\bi + 7\bj$, you can simply multiply the two together to get the force: $\bb{F} = m\bb{a} = 4(-3\bi + 7\bj) = -12\bi + 28\bj$. Easy as.

Similarly, you can use the SUVAT equations (apart from one) just the same way: if you know a particle’s acceleration is $\bb{a} = 2\bi - 3\bj$, has initial speed $\bb{u} = -7\bi + 24\bj$ and travels for 2 seconds, its displacement is simply $\bb{s} = \bb{u}t + \frac{1}{2} \bb{a}t^2$ ((All of the SUVAT variables are vectors apart from time.)).

That gives $\bb{s} = (-7\bi + 24\bj)\times 2 + \frac{1}{2} \left(2\bi -3\bj\right)\times 2^2 = (-14\bi + 48\bj) + (4\bi - 6\bj) = -10\bi + 42\bj$ (by grouping the $\bi$s together and the $\bj$s together.)

Sometimes you’ll have variables in there too - often a $t$ - but you work with them just like normal variables.

### What’s the SUVAT equation you can’t use?

Good question: it’s the one where you try to square vectors: $v^2 = u^2 + 2as$. There is a vector version, but you need to know something about scalar products first - the dot product you come across in C4. Because there are two ways to multiply vectors, it doesn’t make sense to square a vector, but you can find the scalar product of a vector with itself. The last SUVAT equation becomes:

$\bb{v} \cdot \bb{v} = \bb{u} \cdot \bb{u} + 2 \bb{a} \cdot \bb{s}$

To find a dot product, you simply multiply the $\bi$ terms together and write them down; multiply the $\bj$ terms together and write them down; then add the two up ((In C4, there may also be a $\bk$ term - that doesn’t change anything much.)). That means something like $(3\bi + 4\bj)\cdot(5\bi - 12\bj) = 15 - 48 = -33$.

You don’t need to know this for M1, but since you asked, I thought it’d be rude not to answer. How about something more useful?

### What about speed, distance and angles?

Now, that’s more like it - and we can’t put off working out what $\bi$ and $\bj$ mean any longer. They’re orthonormal vectors, which means they’re at right angles to each other, and are both one unit long. Right angles are Very Useful, because they help us build triangles.

There are two (sensible) ways to express any two-dimensional vector: as “across and up”, like you did at GCSE; or as “magnitude and direction”, which needs a bit of trigonometry. Luckily, it’s GCSE trigonometry.

The magnitude of any vector (‘how long it is’) is easy to work out using Pythagoras’s Theorem: you simply square the $\bi$ terms, square the $\bj$ terms, add them up and square root them ((Eagle-eyed readers will spot that the magnitude is $\sqrt{\bb{x} \cdot \bb{x}}$.)) The magnitude of $8\bi -15\bj$ is $\sqrt{8^2 + (-15)^2 } = \sqrt{64 + 225} = \sqrt{289} = 17$.

To find the direction of a vector, you need to draw a triangle. Treat the $\bi$ term as ‘how far to the right’ and then the $\bj$ term as ‘how far up’ - obviously, negative numbers mean left or down. Draw a line from where you started to where you finished, and figure out the angle using your $\tan^{-1}$ button. You may need to convert this into a bearing; that may involve adding or taking away your angle from a reference angle (north is 000º or 360º, east is 090º, south is 180º and west is 270º).

### And the other way?

If you’re given a speed and an angle, for instance, you can turn this into a velocity by drawing a triangle, too. (The same trick works for distance and angle into a displacement vector, of from any magnitude and direction into a vector).

Draw a line of the given length in the appropriate direction - no need to measure, it’s a sketch. That’s the hypotenuse of your triangle. You now need to draw a horizontal line from one end of the vector and a vertical line from the other, so that they meet. Bingo! A right angle.

If you know the direction of the line, you can work out the angles in the triangle; once you have that, it’s simply SOHCAHTOA or (if you prefer) sine rule to get the remaining sides. The horizontal one is the $\bi$ term and the vertical one is the $\bj$ term.

For instance, if you have a speed of 5 m/s at a bearing of 053º, you might draw a triangle with its right angle in the bottom right. The angle at the bottom left would be 37º and the angle at the top right would be 53º. The horizontal side is $5 \cos(37^º) \simeq 4$ and the vertical side is $5 \sin(37^º) \simeq 3$, so the vector is $4\bi + 3\bj$.

I think I’ve covered the basics of M1 vectors in this post - but let me know if there’s anything I’ve missed!

* Edited 2016-06-03 to correct a direction. Thanks to Pravin Dodia for catching it!

* Edited 2017-04-17 to get my $\bi$s and $\bj$s the right way way around. Thanks to Javeria for pointing it out!