INSTRUCTOR: Find p cross q for p equal to the vector 5 comma 1 comma 2 and q equal to the vector negative 2 comma 0 comma 1. Express the answer using standard unit vectors. Now, there are two ways to calculate this. I'm going to do this in both ways.

But the first is directly from theorem 2.9, and it says that if we label these two vectors as u1, u2, u3-- that's my p vector-- and q, if we label that as v1, v2, and v3 of those components, then defines u cross v as equal to u2 multiplied by v3 minus u3 multiplied by v2 closed parentheses i minus parentheses u1 v3 minus u3 v1 close parentheses j plus parentheses u1 v2 minus u2 v1 close parentheses k.

That is the definition of calculating the cross product. So I'm going to just fit my values to this. So u2 is 1 multiplied by v3-- that is also a 1-- minus u3, which is 2, multiplied by v2, which is 0, close parentheses i minus open parentheses u1, which is 5, v3, which is 1.

Multiply those two minus u3, which is 2, multiplied by v1, which is negative 2 close parentheses j plus u1, which is 5, multiplied by v2, which is 0, minus u2, which is 1, multiplied by v1, which is negative 2, close parentheses k.

So this would be equal to-- just double-checking my arithmetic here. OK. That would be equal to 1i-- so I have i-- minus 9j plus 2k. Going straight from the definition. Now, if you're familiar with matrices at all, there is another method to use, and I will use this later.

Going you write the cross product using a determinant of a matrix. My first row in my matrix will be i, j, and k respectively. The next row will be the values from u. So this will be 5, 1, 2. And my third row will be the values from q, negative 2, 0, and 1, respectively.

Now, strictly speaking, you can't actually find a determinant of a matrix like this, but we are going to relax our standards a little bit. And the way that a determinant works with a 3 by 3 matrix is you take the first component in the first row, and then you find the determine of the minor matrix. That is, neglecting that column, the first column, I'm going to find the determinant of 1, 2, 0, 1, that submatrix.

So that matrix, the determinant of it-- I'll just go ahead and write it here. It's going to be 1, 2, 0, 1 multiplied by i minus-- for the j component, I'm going to find the minor matrix, ignoring the j column. So if I take out the second column, my submatrix is 5, 2, negative 2, 1, plus, for the k component, I will find the minor matrix, eliminating the k column as the third column.

So this would be 5, 1, negative 2, 0. So we just turned a 3 by 3 matrix into three 2 by 2 matrices. Now, the determinant of each one of those matrices is ad minus bc. I'm just going to write this over here. If I have the matrix a, b, c, d, the determinant of that is a times d minus b times c.

So we multiply the diagonal, and we subtract multiplying the other two diagonals, or the two elements in the other off diagonal. So that means I'm going to have 1 times 1 minus 2 times 0, which is 0-- and that will be the value for my i component-- minus 5 times 1, which is 5, minus 2 times negative 2. So I have a minus a negative 4.

So 5 minus negative 4j. Again, it was a minus in front of that, so don't forget that either. Plus 5 times 0, which is 0, minus 1 times negative 2 is negative 2. So 0 minus negative 2k. 1 minus 0 is 1, so I end up with i minus-- that would be 5 minus negative 4 is 9, so minus 9j.

And 0 minus negative 2 is 2, so I have plus 2k. End up in the same place. However, if you're familiar with determinants at all, this is way quicker. It also ties back to something else that you know without having to memorize which subscripts go with which, so hopefully that is helpful.