 We can use vectors to write equations for lines and planes. Here's a useful idea. A vector is normal to a glurable if it is orthogonal to the glurable. What's a glurable? You ask? Well, if you watch the video that I posted on Star Day 2437.6, I... Wait, I haven't posted that video yet. Actually, this idea of a glurable is completely generic. A glurable is anything a vector can be orthogonal to. So we can say that a vector is normal to another vector, to a plane, to a line, to a curve, to anything at all. So for example, we can talk about finding a vector normal to the vector 154. We don't typically use this language. You just say find O and that's orthogonal. But we could. The vector is orthogonal if the dot product is zero, and so we've required the dot product of whatever our vector is. Let's be creative and call it x, y, z, and the given vector is equal to zero. We know that dot product, and since we have one equation with three unknowns, we can choose values for two of the unknowns and find the remainder. So you could choose your own values, but I'm going to let y equals 1 and z equals 3 because I like the numbers 1 and 3. Anyway, if we do that, we find x is negative 17, and so our normal vector will be x equals negative 17, y equals 1, z equals 3. Now, when you're talking about geometric objects, it helps to talk a little bit about geometry. So we've been looking at these things from an algebraic perspective, writing down equations and solving them, but a little geometry is helpful. A vector in a plane can point in any direction, but three component vectors that are orthogonal to a plane all have to point in the same direction or in the opposite direction. And we can make use of this to write the equation of a plane. So for example, let's try to find the equation of the plane through these three given points. So the important idea to begin with is that if a vector is orthogonal to the plane, it will be orthogonal to all the vectors in the plane. So we need to find some vectors in the plane, and since we know points in the plane and a vector is a set of directions for getting from a point to another point, we can find vectors by going from one of these points to another point. For example, since 2, 1, 5 and 1, 4, negative 3 are in the plane, the vector from 2, 1, 5 to 1, 4, negative 3 is going to be in the plane. And we find this vector will be, similarly, I might go from 2, 1, 5 to 1, 2, negative 4, and the vector from 2, 1, 5 to 1, 2, negative 4 is in the plane, and this vector will be, and that gives me two vectors in the plane. So now we want to find a normal vector, because that's the one thing we can rely on is that the normal vector will be orthogonal to all vectors in the plane. To find a vector orthogonal to both, we can find the cross product or solve a system of equations. Since solving a system of equations is easier, we'll find the cross product. That's because you should be able to set up and solve the system of equations on your own without any additional help. The cross product takes a little bit of managing. So our cross product will be the determinant of the matrix whose first row is I, J, K, and whose second and third rows are the components of our two vectors. We'll use the rule of Saint-Rue to find the determinant, we'll double the matrix, and then take our down right diagonals, then subtract the down left diagonals. And that gives us our orthogonal vector, negative 19, negative 1, 2. Now let's think about this. Suppose I have a point x, y, z in the plane. Then the vector from any point in the plane to x, y, z will be orthogonal to this normal vector, negative 19, negative 1, 2. For example, the vector from 2, 1, 5 to x, y, z is x minus 2, y minus 1, c minus 5. And since this is orthogonal to negative 19, negative 1, 2, then the dot product of the two vectors will be 0. Well, I know how to set up the dot product, how to find the dot product, and we can do a little algebra. Now, while we don't have to do anything else, it's traditional to have the constant over on the right-hand side. And it's aesthetically pleasing not to have a negative leading coefficient, so let's multiply through by negative 1, and that gives us our equation. The important thing to understand is everything after this dot product equal to 0 is really just an algebraic simplification to make our final answer look nice. So remember that any time we're doing the same sequence of steps, we can combine them all to produce a formula. And so we can combine all these steps to produce a formula for the equation of a plane, but we won't. The thing to remember is that formulas are a fast path to failure. Understand the concept. If you understand the underlying concept, then the formula is irrelevant and often more trouble than it's worth. In this case, finding the equation for a plane relied on a couple of important concepts. First, a vector is a set of directions for getting from some point to another point. And this problem, that meant that given the point in the plane, we could also find vectors that were in the plane. The second was this geometric concept that a vector normal to a plane is perpendicular to every vector in the plane. The third concept here is really a matter of convenience, which is to remind us that the cross product does give us an orthogonal vector. And lastly, the dot product gives us the angle between vectors.