00:01
When we take the cross product of vectors a and b, we will get a vector, i'll call it c, that is perpendicular to both a and b.
00:14
Okay, so we'll take the cross product of a and b, and the vector that results will then take the magnitude of.
00:24
So let's find the cross product.
00:26
I'll use the determinant method.
00:28
In the first row we have i hat, j hat, k hat.
00:33
In the second row, we have the components of a, which are 3, 9, and negative 3.
00:39
In the third row, we have the components of b, which are negative 5, positive 5, and 6.
00:46
Now to take the determinant, we multiply 9 times 6, which is 54, and we subtract negative 3 times 5, which is negative 15.
01:01
This is the component.
01:03
Then we subtract 3 times 6, which is 18, minus negative 3 times negative 5, which is positive 15.
01:16
This is the j component.
01:20
And finally, we have plus 3 times 5, which is 15, minus 9 times negative 5, which is negative 45...