00:01
Okay, let's find the intersection of the planes.
00:02
Let's call z is equal to t, okay? then it becomes x minus 3y is equal to 19 plus 3t.
00:13
3x minus y is equal to 13 plus 3t.
00:20
And from here, solve the systems of equations, right? multiply this by negative 3.
00:26
And you will get x is equal to 20 over 8 plus 6 over 8t.
00:36
Or simplify this to 5 over 2 plus 3 over 4t.
00:43
And then put that back and find y as negative 11 over 2 minus 3 over 4t, okay? so the line we have is this.
00:56
The parametric equation for the line is 5 over 2 plus 3 over 4t as x.
01:06
Negative 11 over 2 minus 3 over 4t, that's the y.
01:11
And z is this.
01:13
So the point here, you know the line contains a point and a vector, right? this is the point and the vector of this line is 3 over 4, negative 3 over 4 and 1.
01:32
Now, we also have this point given on top, right? so what we're going to do is we're going to write a line between these two points and use that line and then this vector to find the normal of the plane.
01:47
So i'm going to call this p1, p2...