INSTRUCTOR: Find the area of the parallelogram PQRS with vertices P is 1, 1, 0, Q is 7, 1, 0, R is 9, 4, 2, and S is 3, 4, 2. First, we need to write some vectors so that we can actually do some work with this. So notice that PQRS, that's the order of the vertices.

So ultimately, this parallelogram is defined by two vectors, PQ and PS, or SP, if you'd rather. So if we can write vectors PQ and PS, we can then find the area of the parallelogram. All right. Well, let's see. To find PQ, we're going to take the terminal minus the initial points here.

And looking at P and Q as 1, 1, 0 and 7, 1, 0, we'll subtract those components. So 7 minus 1 is 6, 1 minus 1 is 0, and 0 minus 0 is 0. So PQ is the vector 6 comma 0 comma 0.

And then PS, PS we'll take those components. So S is 3 comma 4 comma 2, and P is 1 comma 1 comma 0. Subtracting those, we get the vector 2, 3, 2. 2 comma 3 comma 2.

Now, the definition of the area of a parallelogram with vectors is it's the magnitude-- it is the magnitude of, in our case, PQ cross PS. So if we find the cross product of these two and then find the magnitude of that vector, that will tell us the area of this parallelogram. So PQ cross PS.

Going to write this with determinate notation. I'll have a first row of i, j, k, second row of 6, 0, 0, and a third as 2, 3, 2. I find our minor matrices. So if we eliminate the i call, our minor matrix is 0, 0, 3, 2.

That tells the i component of our cross product minus-- if we eliminate the j column, we are left with the minor matrix of 6, 0, 2, 2 times j plus the minor matrix of for the k component, eliminate the k column. Going to be 6, 0, 2, 3.

Now, we'll find the determinant of each of those submatrices. That also tells us our i, j, and k components respectively. Now, let's see. The first one, the i component, is 0. So this will be 0i.

Second, we'll have-- it'll be minus 12 for our j component. And finally, 18 for our k amounts. We have 0i minus 12j plus 18k. Now, because our area is the magnitude of that resulting cross product, the area is going to be equal to the square root of 0 squared plus negative 12 squared plus 18 squared.

If we sum those two values after we square them, or those three values, we'll get the square root of 468. That actually will simplify as 6 square root of 13. So our area is 6 square root of 13 units squared. That is the area of our parallelogram.