00:01
In this exercise, we want to find a plane that passes through the points.
00:04
P, 6, 2, 1, q, 5, negative 3, negative 1, and r, which is 7, 0.
00:17
However, we know we have the equation for a point and a normal vector, but we don't know how to do this with three points.
00:26
So, we will use these three points to find a normal vector, and then we can use any of these three as a point that the plane passes through with that normal vector to find the equation.
00:38
So to do that, we will use the cross product.
00:42
Let's first calculate these two vectors that we can use to find the cross product.
00:48
Let's go with rp, which is equal to 6 minus 7, negative 2 minus 0, and 1 minus 0, which gives us negative 1.
01:00
Negative 2 and 1 and let's also find the vector rq which is equal to 5 minus 6 excuse me minus 7 negative 3 minus 0 and negative 1 minus 0 which finally gives us negative 2 negative 3 negative 1 so these are two that lie on the plane that we're looking for.
01:30
So their cross -product is a vector that is perpendicular to that plane, or rather to those two vectors and therefore that plane as well...