 In this video we'll provide the solution for question number 15 from the practice midterm exam number two for math 2270. We're given three vectors u, v, and w which will all live inside of R5 and we say that w is the span of these three vectors so it's a subspace of R5. We need to find a basis for the orthogonal complement of w. Now the trick to this one here is to remember that the row space of a matrix if you take it to orthogonal complement this is equal to the null space. So what we're going to do is we're going to set up a matrix so that w equals the row space of a. So this would look something like the following. A looks like the matrix. The first row will just be u, 1, 7, negative 2, 98, and 0. The second row will be v, 3, 0, 1, negative 14, and 168. And then row 3 will be w, 6, 1, negative 1, 0, and 224, like so. And so notice that the way we set this up that the row space of a here it's going to equal the span of its rows which would be u, v, and w and therefore that's equal to capital W. So we've created a matrix whose row space is equal to w. The next thing we're going to do is we're then going to compute the row reduced echelon form of this matrix for which the RREF would look like the following. Again, I'm going to skip the details of this but we're going to get 1, 0, 0, negative 3, 43, and then we get 0, 1, 0, 13, and 5, and then 0, 0, 1, negative 5, and 39. Now the instructions do not say I have to show all of my row operations so therefore I think I can jump to the echelon form without any further calculation there. You'll notice that we have pivots in the first, second, and third columns. So now what we have here is we want to compute the null space of a because that'll equal the orthogonal complement. If w is the row space of a, w-perp is then going to equal the null space of a. So we need to find a basis for the null space for which this is something we're well practiced with. So we see that w-perp is going to equal the span of well since we have two free variables in this in this linear system here looking at this equation or this matrix right here and thinking of the associated homogeneous system of equations we're going to have we're going to have two spanners that'll correspond to our free variables. So what you can see me doing right here is that I am just recording down I'm just recognizing that okay I have a one in the fourth position I have a one in the fifth position these are my free variables and so then looking at the columns associated to those free variables I'm going to write down the opposite of those numbers in which case you're then going to get three negative 13 and positive five and then for the last column we end up with negative 43 negative 5 and negative 39 and so this then gives me a basis for the this gives me a basis for the null space of a which coincides with the orthogonal complement. So in summary this is what we do if you need to find the basis of the orthogonal complement create a matrix whose rows coincide with the spanning set of set vector space then row reduce that matrix find a basis for the null space of set matrix using the usual techniques and then the basis of the null space will give us a basis for the orthogonal complement.