00:01
For this problem, we are asked to determine the angle to the nearest degree between each pair of vectors, where for part a, we have the vector a equals 1 minus root 2, root 2 negative 1, and b equals 1 -1, where that means that we have that b is a vector just in the x, y, plane, but we can add a 0 to make sure that everything essentially works out nicely here.
00:22
So we know that theta is going to be the inverse cosine of a .b over the magnitude of a, times the magnitude of b.
00:33
So first we'll find a .b.
00:35
So that will be 1 minus root 2 times 1, so just 1 minus root 2, plus root 2, so plus root 2.
00:46
So that's just going to equal 1.
00:48
Then for the magnitude of a, we'll have the square root of 1 minus root 2 squared, which would be 1 plus 2 root 2, or actually that should be 1 minus 2 root 2.
01:02
One minus two root two plus root two squared, so that's going to be plus two.
01:10
Then plus root two squared, so that's going to be plus two.
01:13
Then we'll have plus one.
01:15
So let's see here...