00:01
So in this first question, we're considering the points below.
00:04
P is the point 202, q is negative 214, and r is 426.
00:10
In part a, we want to find a non -zero vector that is orthogonal to the plane through the points pq and r.
00:19
So we're going to first write the vectors pq and pr and then take their cross product.
00:25
So pq is the vector that goes down four units in the x direction.
00:32
Up 1 unit in the y direction, and up 2 units in the z direction.
00:39
While the vector pr goes up 2 in the x direction, up 2 in the y direction, up 4 in the z direction.
00:50
And so now we take this cross product.
00:55
So we're going to have ijk across the top row, negative 412 across the middle row, and 2 to 4 across the bottom row, giving me i times 4.
01:10
4 minus 4.
01:13
And then it's minus j times negative 16 minus 4.
01:21
And then it's plus k times negative 8 minus 2.
01:28
And so that's giving us the vector.
01:31
Excuse me there.
01:33
Sorry about that weird jump.
01:35
That's giving me the vector.
01:37
0 in the first component, negative 20, but you negate that to get positive 20 in the second component, and negative 10 in the third component.
01:52
So 0, 20, negative 10.
01:57
If you then want the area of triangle pqr, you take half of the magnitude of pq cross pr.
02:09
So this is one half times the square root of 0, plus 400 plus 100, that's one half square root of 500.
02:22
This does simplify a little bit because square of 500 is 10 radical 5, so we're getting 5 radical 5.
02:34
So the area of that triangle would be 5 radical 5...