 So let's go back to that problem of finding the angle between two vectors, and remember we can find the angle between two vectors u and v using the law of cosines. So again, if I imagine our two vectors originating from the same point, vector u and vector v, the third side of this triangle is going to correspond to the vector we might write as u minus v, and the law of cosines gives us a relationship between the angle theta and the magnitude of the three sides. Now we can rewrite this using the dot product. So remember the magnitude of any vector is the square root of the dot product of the vector with itself. And so this square of the magnitude of u, well that's just the dot product of u with itself, the square of the magnitude of v, well that's the dot product of v with itself, and the square of the magnitude of u minus v, well that's the dot product of u minus v with itself. And we'll leave the rest of the expression the same for now. Now since u minus v is itself a vector, then we have this distributive property for vectors that gives us u dot u minus v minus v dot u minus v. And we can apply the distributive property a second time, and remember that the dot product is commutative, so u dot v and v dot u are the same, and so we can simplify our dot product. Now notice both sides have a u dot u and a v dot v, and since these are just real numbers we can remove them from both sides and solve for the cosine of the angle between the two vectors, and this leads to the following theorem. Let u and v be two vectors with real components, the angle theta between the two vectors satisfies the following, the cosine of theta is the dot product divided by the product of the magnitudes. So let's try and find the angle between our two vectors, so we need to find the dot product, we need to find the magnitude of the two individual vectors, and our theorem says the cosine of the angle is the dot product divided by the product of the magnitudes, so we'll replace with the values we've found, and we can use that to find the angle itself. Now, we don't really care that much about the angle between two vectors. Well, okay, if you're a physicist or an engineer or an astronomer or a navigator or anybody else who uses vectors for something practical, you do actually care about the angle, but mathematicians don't really care that much about the angle, except there are sometimes when it's useful to have vectors that are perpendicular or because we want to generalize this idea, we use the term orthogonal. And so the question you want to ask yourself is, self, how can we tell if two vectors are orthogonal? And here's where the dot product is useful. If the two vectors are orthogonal, then the angle between them is theta equal to 90 degrees, but the cosine of 90 degrees is zero, and so our formula that gives us the angle between two vectors reduces down to the dot product equal to zero, and so we say that two vectors are perpendicular. Again, we also use the term orthogonal if their dot product is zero. And this leads to an important problem. Let's find a vector that's orthogonal to 158. And so we know that the vector will have components x, y, z, and our dot product should be zero. Well, we know how to compute the dot product. And the thing to observe here is that we have one equation with three unknowns, and we know that there is no unique solution. However, because we have one equation with three unknowns, if we pick values for two of the variables, we can solve for the value of the third. So let's choose, well, how about x equals zero, y equals zero? If we choose these as our values, we get z equal to zero, and so our vector is x, y, z, all zero. And before we box in this as our answer, let's think about this, a vector tells us how to get to a point from a point. The vector zero, zero, zero tells us to go nowhere. So it tells us how to get from where we are to where we are. Well, that doesn't seem very useful. And we say that it is a trivial solution to the problem where a trivial solution is a solution that doesn't really tell you anything. Can we find a non-trivial solution? Sure, let's choose a different set of values for x and y. So we don't want them all to be zero, so what if we choose x equals zero, y equals three? In that case, we get, and so we have our vector components, zero, three, negative 15 eighths, and so this vector will be orthogonal. And in fact, we could find other vectors by other choices of two of the variables. For example, if we let x equals three and y equals one, we get another vector that's orthogonal. And we don't have to choose values for x and y and find z. We could choose values for x and z and find y or y and z and find x. And so we could find as many orthogonal vectors as we want. So how many do we want? We'll take a look at that next.