 So we saw that given any single vector, we can find a vector orthogonal to it by solving the equation, making the dot product equal to zero. What if we want to find a vector orthogonal to two given vectors? Or what if we want to find a vector orthogonal to three given vectors? And the easy answer to that is, well, this is mathematics. If in doubt, set up and solve an equation. So let's try to find a non-zero vector, we'll call it x, y, z. It is orthogonal to both 1, 4, negative 1, and 3, 1, 4. So let's think about this a moment. The vector x, y, z is going to be orthogonal to 1, 4, negative 1 if the dot product is zero. Well, we know how to find the dot product, and that gives us an equation. We also want x, y, z to be orthogonal to 3, 1, 4. And again, this is orthogonal if the dot product is zero, and that gives us a second equation. And so whatever the values x, y, and z are, they have to satisfy a system of two equations and three unknowns. Now there's a number of ways of solving systems of equations, but the most efficient is by addition. And what that requires is getting the coefficient of a variable in one equation equal but opposite to the coefficient of the variable in another equation. And so in this case, we can solve this by multiplying the first equation by minus three and adding it to the second. So multiplying our equation by minus three, and we add our two equations. And we no longer need all three equations. We can drop one of them, and let's get rid of this one. So talking about systems of equations is something you'll spend a lot more time on in linear algebra. But here's a preview. If you have more equations than variables, you might be able to choose values for some of the variables and solve for the rest. So here we have two equations and three unknowns. We might be able to choose a value for some of the variables and find the others. So the other strategy that you'll see in linear algebra is it helps to work backwards. So our last variable is z. We can choose a value for z and solve for y. So let's choose z equal, how about zero? If z is equal to zero, our last equation becomes nv solve. Then since z is zero and y is zero, our first equation becomes, and so our vector x, y, zero has x equals zero, y equals zero, z equals zero. Except this is a trivial solution and we don't want that. We want a non-zero vector. So let's choose something else. And here's something that'll work. We'll choose z equals 11, y11. Well, notice that our coefficient of y is negative 11. And that means at some point we're going to be dividing by 11 to solve for y. Choosing z equals 11 helps us avoid fractions. So if z equals 11, we can solve the first equation for y. But now we know z is equal to 11, y is equal to 7. So we can solve our first equation for x and that gives us a value of x, y, and z. And consequently, we have our orthogonal vector. So let's try and find a vector perpendicular to 3, negative 1, 3, and 2, 5, negative 4. So again, we want the dot product to be zero, so that gives us the equations. And in this system of equations, notice that if we multiply the first equation by 5, then our y coefficients will be equal but opposite to the y coefficient in the second equation. So if we add the second equation, now since we got this third equation by combining the first two equations, we don't actually need both of the first two equations. We can get rid of one of them and let's get rid of the second one this time. So again, we'll choose a value for z and notice that we'll be dividing by 17 and so to avoid fractions, we'll let z equals 17, the coefficient of x, and this gives us negative 11. And if x equals negative 11, z equals 17, we find y. And that gives us our orthogonal vector. Or let's find a vector perpendicular to both negative 5, 3, 8, and 3, 2, negative 9. So again, we want the dot product of our vector x, y, z with the two vectors to be zero, and so this gives us the equations. We'd like to eliminate one variable from one equation, so let's get rid of the x and make our x coefficients equal but opposite. And we could do that by multiplying the first equation by 3 and the second by 5. And if we add the two equations, and again since this new equation combines the first and second equation, we can get rid of one of those first two equations, so let's get rid of the second one. Now for variety, we'll pick y this time. And if we look at our equation, we see that if we're going to solve for z, we're going to need to be dividing by 21. So since we're going to be dividing by 21 divided z, we'll let y equal 21 and solving for z. And with y equal 21, z equals 19, we can solve for x and we get our solution, the vector 43, 21, 19. And we'll make the same observation we did with the dot product. Notice we're always doing the same thing, forming an equation where one dot product is zero, forming a second equation where the other dot product is zero, adding the two equations to eliminate a variable, and then choosing a value for a variable and finding the values of the other two. And remember, because we're always doing the same thing, what we can do is we can write down a formula to do all the steps at once. And fair warning, in contrast to the dot product where there was an advantage to using the dot product to find the angle between two vectors, our formula is a little bit more opaque. We'll take a look at that next.