INSTRUCTOR: Find an equation of the plane containing the lines L1 and L2, given by L1 equals x equals negative y equals z, and L2 is defined as x minus 3 over 2 equals y equals z minus 2. We know that the equation of a plane is given by a parentheses x minus x0 close parentheses plus b, multiply the quantity y minus y0, plus c times the quantity z minus z0 equals 0 if we have a normal vector that is given by a, b, c and a point P0 that is x0, y0, z0.

If we have those pieces of information, then we can write the equation of that plane. Two things that we need. Two things that we need. We need a normal vector and a point. And it needs to be a point on the plane given by these two lines. So really any point on either one of those lines will be in the setup.

So first thing I want to do is actually write-- or the second thing, I guess, is write each of these in their parametric forms. The first line is x equals t, y equals negative t, and z equals t. Pretty straightforward to write in that way.

And let's see. X for the second line, L2. X equals 2t plus 3, so 3 plus 2t, y equals t, and z equals 2 plus t. This will help me see a little more quickly what points are on this plane. And since either line can be used, I'd rather use the first one.

And if t is equal to 0 because parametric equations are defined that way, ranging from 0 to 1, that is, typically, if t is equal to 0, then I have the 0.000. So P0 is going to be the 0.000. At least that's what I'm going to use. You could use any point that's on this.

Now, in order to find the normal vector that we need here, normal vector a, b, c, what we can do is we can take the two direction vectors. So we've got the direction vector of the first line is 1 comma negative 1 comma 1, and the second line's direction vector, which will be 2, 1, 1-- remember, those are the slopes of the parametric equations. That's why I like using the parametric equations.

So if I take those two direction vectors and I find the cross product of those two, the cross product of two direction vectors is always normal. It's going to be orthogonal to both of those lines, which is kind of what we want. So let's go ahead and set up the determinant form here.

So I've got the determinant of i, j, k with my second row being 1, negative 1, 1, my third row being 2, 1, 1. That'll be equal to the determinant of my first minor matrix. That will be negative 1, 1, 1, 1 i minus the determinant of eliminating the j component, 1, 1, 2, 1 j, plus the determinant of eliminating the k component, 1, 1-- or sorry. 1, negative 1, 2, 1, determinant of that, k.

Now, that should come out to be negative 2i, once you find those individual determinants, negative 2i plus j plus 3k. So our normal vector, just going to go ahead and call it n, is the vector negative 2 comma 1 comma 3. Putting that together with what we know about the equation of a plane, given a point in a normal vector, we can write this as negative 2x plus y plus 3z equals 0 because our point was the origin, 0, 0, 0.