00:01
This question wants us to do some vector arithmetic.
00:04
It gives us two vectors.
00:06
We have u, which is equal to 2a, a, and v, which is equal to negative a, negative 2a.
00:17
This question wants us to find the magnitude of 2u over the magnitude of v, minus 3v over the magnitude of u.
00:32
So there's a lot of steps to this problem.
00:33
Let's dive right in.
00:35
I'm going to start by finding the magnitude of both of these.
00:38
So the magnitude of u is of course going to be the square root of 2a squared plus a squared.
00:49
What that's going to be is the squared of 4a squared plus a squared or 5a squared.
00:57
So that'll leave us with a root 5.
01:01
So that's the magnitude of u.
01:04
Let's try the magnitude of v, and we'll see something pretty similar.
01:09
Again, we'll do negative a squared plus negative 2a squared.
01:15
That's going to turn into the square root of a squared plus 4a squared, and that's exactly the same thing.
01:22
So we'll be a squared of 5a squared, which becomes a root 5.
01:26
So both magnitudes are the same.
01:30
Now what i want to do is find this chunk here.
01:34
So it's going to be 2 u over the magnitude of v.
01:40
Let's plug in what we have.
01:42
We know that that's going to be equal to.
01:45
2 times u, which we know is 2a, a, over the magnitude of v, which we know is a root 5.
01:55
So let's simplify this as much as we can.
01:57
I'm going to distribute the 2 in, and actually we are dividing here by a, so i'm going to take out the a's from the inside.
02:08
What i mean is i'm going to distribute the two, factor out the a's.
02:11
It'll be a times 4 -2 over a -r -5, and all that it is, it becomes 1 over root 5 for 2, because those a -s are going to cancel out.
02:29
So now let's say we've got, let's do the other section...