00:01
In the question we have to find the equation of the plane and it is given that one direction vector is p1 p2.
00:15
So p1 p2 is equal to p2 -p1 which is equal to 3 -1, 2 -2, 1 -3 and which is equal to 2 ,0 ,2.
00:34
Since the desired plane is perpendicular to the given plane, the normal n1 of the given plane is also a directional vector of the desired plane.
00:44
So n1 is equal to 4, -1 ,2 and the normal vector n of the desired plane is orthogonal to both directional ratios.
00:57
So we will take the cross product.
00:59
So n will be equal to p1 p2 multiplied by n1 which will be equal to 2 ,0, -2 multiplied by 4, -1 ,2 which will be equal to minus 2, minus 12, minus 2.
01:22
Any non -zero multiple of n is also a normal vector to the desired plane.
01:27
So divide by minus 2.
01:30
So n will be equal to 1 ,6 ,1.
01:37
Now with a point in the plane and the normal vector to the plane, we can write an equation for the plane...