Find the general equation of the plane that passes through the point eight, negative nine, negative nine and cuts off equal intercepts on the three coordinate axes.
Let’s start on our solution by plotting in our coordinate frame and then noting the second point about our plane that it cuts off equal intercepts on the three coordinate axes. This tells us that the distance between the origin and the 𝑥-axis intercept of our plane, whatever that is, is equal to the distance between the origin and the 𝑦-axis intercept to the plane and the 𝑧-axis intercept to the plane. That is, these three line segments in pink all have the same length. Our plane then would look something like this with respect to this coordinate frame.
Our goal is to solve for the general equation of this plane. We can do that given two pieces of information about it. If we have, first, a point that lies in the plane and, second, a vector that is normal or perpendicular to it, then we can solve for the vector form of the plane’s equation and convert that to the general form. Notice that we are given a point that lies in our plane. That means all we’re lacking is a normal vector. How can we figure out the components of such a vector? We can do it by thinking a bit about the slope or the gradient of our plane.
Thinking of it as a two-dimensional surface, we know that that surface is stretched equally, we could say, in the 𝑥-, 𝑦-, and 𝑧- directions. We know this because it intercepts these axes the same distance from the origin. This means that just as the 𝑥-, 𝑦-, and 𝑧-intercepts are equally balanced, we could say, so will the components be of a vector normal or perpendicular to the plane. That is, if one of the components of our normal vector is, say, 𝑎, then the other two components must share that same value. Now, 𝑎 could be any nonzero number. After all, there are infinitely many vectors that are normal to our plane. For simplicity’s sake, we can let 𝑎 equal positive one. Our normal vector 𝐧 then has components one, one, and one.
And now that we have a vector normal to our plane and a point that lies in it, let’s recall what’s known as the vector form of a plane’s equation. This equation tells us that if we take the dot product of a vector normal to our plane with a vector to an arbitrary point in the plane, then that’s equal to the dot product of our normal vector and a vector to a known point in the plane. In our case, our known point, we can call it 𝑃 zero, is eight, negative nine, negative nine. And this means if we sketch in a vector from our origin to 𝑃 zero, then that vector, we’ve called it 𝐫 zero, will have the same components as the coordinates of 𝑃 zero. Knowing both the components of 𝐫 zero and those of a normal vector, we can now substitute into the vector form of a plane’s equation.
On the left-hand side, we have the dot product of one, one, one and a vector to a general point in our plane. We leave it with coordinates 𝑥, 𝑦, and 𝑧. On the right, we have a dot product between our normal vector and a vector to our known point. We can now start carrying out these dot products by multiplying together their respective components. On the left, we get 𝑥 plus 𝑦 plus 𝑧; on the right, eight minus nine minus nine. This equals negative 10. And if, as a last step, we add positive 10 to both sides of our equation, we get this result, which is the general form of our plane’s equation. Written this way, our plane has the equation 𝑥 plus 𝑦 plus 𝑧 plus 10 equals zero.