00:01
In this problem, we want to find the equation of a plane that contains the point 2, minus 2, minus 1, and the line of intersection of the plane 2x plus y minus 3z equal to minus 23, and x minus y plus 2z equal to 4.
00:18
So this problem is easy to solve if we draw a diagram.
00:22
So we have two planes that intersect at some line.
00:33
So this is the line of intersection right here.
00:35
And the plane equation that we want to find needs to contain this line.
00:44
It also needs to contain some arbitrary point, well not arbitrary, some given point p.
00:54
So our plane looks something like this.
01:03
It'll contain the line of intersection of the two planes that are given to us in the problem statement, and it'll contain the point p.
01:16
So planes are described by their normal vector, such that a plane equation is given by ax plus by plus cz equal to some constant d, where the coefficients a, b, and c correspond to a vector that is normal to our plane.
01:41
So the goal is to try to find this normal vector to our plane.
01:46
A good choice would be to find some point that is in our two planes of intersect.
01:55
So let's say we find a point here that is on our line of intersection.
01:58
We can find a vector u1 connecting this point p to our given point that is in our plane.
02:06
And if we can find a directional vector of our line equation, then the cross product of these two vectors will give us the normal vector we want that'll describe our plane.
02:20
So now the goal is to find vector u and v here, where u will be from a point that is on our line to a point to the plane, and v will be the vector, the directional vector of our line of intersection.
02:43
So to find the directional vector of the line of intersection, we need to note that this will correspond to the cross product of the two normal vectors of our two given planes.
02:55
So we have n1 for our first plane equal to 2, 1, minus 3.
03:05
We have n2 equal to 1, minus 1, 2.
03:12
So the cross product of these two vectors will give us the directional vector of the line of intersection.
03:20
So let's calculate this cross product.
03:22
We want to calculate the determinant of vector ijk in the first row, vector n1 in the second row, and vector n2 in our last row, 1, minus 1.
03:37
And calculating this cross product by separating into an addition of three smaller determinants, we obtain the vector 1, 7.
03:50
So this is our directional vector of our line of intersection.
03:53
So now we just found v...