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Algebra

Spatial Vectors 2

Author: oLahav

Vectors… in… Space….!

Welcome to a mini-math series about spatial vectors. In part 1, we've introduced vectors and put them in space. Now in part 2 we will see what we can do with them.

How cool does that sound?

Seriously now, it's vector time

We know vectors have 2 important features- size and direction. But how can we figure this out using our coordinate vector system?

It's actually pretty easy. In the 2-D case, say our vector is (3,4), like so:

The Pythagorean Theorem tells us the length of our vector is 5. So in general, for a vector (a,b), $s=\sqrt{a ^2+b ^2}$ . The direction, looking at the positive x axis as 0 degrees, would just be 53 degrees. And in general, $\theta=\tan ^{ - 1} \frac{b}{a}$ .

In 3-D, it's hard to think about direction, so we won't. But you can still figure out the size of your vector (a,b,c) in a similar manner: $s=\sqrt{a ^2+b ^2+c ^2}$ . Cool.

You'll quickly note that our base vectors, which represent our axes- in 3D (1,0,0) for x, (0,1,0) for y and (0,0,1) for z- have size 1. All vectors of size 1 represent unit vectors, and as they're all unique they each represent a certain direction in space.

The nice thing is, there are 2 elementary operations that work on vectors. The first is addition. From physics vectors you'll know addition looks like this:

However, using coordinate vectors, it's much easier. It's pretty clear from the diagram that if our vectors A is (40,0) and B is (0,30), A+B = (40,30).

In general, for v_1=(a_1,b_1) , and v_2=(a_2,b_2) , $v_{1+2}=(a_1+a_2, b_1+b_2)$ . In 3D the same principle applies. so I'm going to save myself from bothering with writing the long equation again.

Anyway, the other basic operation is scalar multiplication (AKA resizing). Look at this:

We can change the size of a vector by multiplying it with a scalar number. In general, in 3D, s * V = (s * a, s * b, s * c) . This also works if s is negative, but in that case our vector reverses direction:

However, note that you shouldn't multiply a vector by a 0. This would result in the zero-vector (0,0,0) which isn't very useful for anything.

Now comes an important part- if, in 2D, 2 vectors are not scalar multiples of each other, they can be used with addition and scalar multiplication to create any other vector on that space. This is simply because:

If v_1=(x_1,x_2), v_2=(y_1,y_2) , then for any V=(a,b) we can find some r,s such that a=rx_1+sy_1 and b=rx_2+sy_2 - 2 linear equations in 2 unknowns (s and r, we know all other values) can be easily solved, unless the two original vectors are scalar multiples of each other.

Now, in that case, the 2 vectors are called linearly independent and they form a basis for the 2D space. Of course, we have an analogous concept for this in 3D, but it involves planes- if 3 vectors are not all on the same plane in 3D, they're linearly independent and can be used as a basis to add up to any vector in that 3D space.

And that's a great!

I think this is enough for simple vector operations.

In part 3, we'll learn all about vector products and their applications. How fun would that be?