00:01
So in this question, they say, i want to find a parametric equation for the line of intersection of the planes 3x minus y plus 3z equals negative 1 and negative 4x plus 2y minus 4z equals 2.
00:16
So the first thing we have to do is come up with a point that lies on both these planes and hence on the line of intersection.
00:25
To do that, let's suppose for the sake of argument that x is equal to 0 to begin.
00:32
If x is equal to zero, what i get is a system of equations, negative y plus 3z equals negative 1 and 2y minus 4z equals 2.
00:48
Now, what i'm going to do is multiply this top equation by 2 so that i have negative 2 plus 6z equals negative 2.
01:01
And if i then add those equations together, what do we get? well, we get the 2y is canceling out.
01:10
I'm getting positive 2z equal 0, so that i have z equals 0.
01:19
Now, if z is equal to 0, i can plug back in to one of my equations.
01:25
For example, 2y minus 4z equals 2, getting 2y equals 2 and y equals 1.
01:34
So that i have a point, namely the point 010, which satisfies the equations of both of these planes.
01:45
Now, i need to get a direction vector for my line...