Video Transcript
Find the general equation of the plane which passes through the point three, negative eight, negative seven and contains the 𝑥-axis.
As we get started, let’s visualize what this plane might look like. If we draw in an 𝑥-axis, we’re told that the plane contains this axis, which means that it might look, say, like this. But then we realize that it’s also possible to rotate this plane and still have it contain the 𝑥-axis. In fact, there are actually infinitely many planes that contain the 𝑥-axis. However, only one of those infinitely many contains this given point. So whichever of these rotated planes contains the points three, negative eight, negative seven, that’s the plane whose general equation we want to write.
And we can recall that the general equation of a plane is written in this form. 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero. Importantly, the factors by which we multiply 𝑥, 𝑦, and 𝑧, respectively, are known to be the components of a vector that is normal or perpendicular to the surface of the plane. And in general, if we know a vector that’s normal to the plane’s surface, as well as a point in the plane, we can define it precisely. Since we already know a point that lies in our plane, our task now is to solve for a vector normal to it.
Now, let’s say that out of all the planes that do contain the 𝑥-axis, this is the one that also contains our point three, negative eight, negative seven. Let’s call that point 𝑃 zero. Now, since our plane also contains the 𝑥-axis, we can name another point in it. Since the point zero, zero, zero lies along the 𝑥-axis, it must also lie in our plane. The reason we’ve named this second point is that now, since we have two points in the plane, we can connect them by a vector that will itself lie in our plane. We can call this vector 𝐫 zero. And since it travels from our origin to the known point 𝑃 zero, the components of 𝐫 zero are simply the coordinates of 𝑃 zero.
Now, we have one vector that we know to lie in our plane. If we can find a second vector that also lies in this plane and is not parallel to 𝐫 zero, then we can calculate the cross product of those two vectors and wind up with a vector normal to our plane. That’s our objective. So now let’s consider another vector that must lie in our plane. Well, again, since the 𝑥-axis lies in our plane, so must the point one, zero, zero. Drawing a vector from the origin to this new point, we can call the vector 𝐫 one. Its components are one, zero, zero. And we can see that 𝐫 one and 𝐫 zero are not parallel. In other words, it’s impossible to multiply either one of these vectors by a constant and get the other vector.
This is good because it means if we take the cross product of these two vectors, say 𝐫 zero cross 𝐫 one, then we really will get a vector that’s normal to our plane of interest. 𝐫 zero cross 𝐫 one is equal to the determinant of this three-by-three matrix. Here, we filled in the first row with our 𝐢, 𝐣, and 𝐤 unit vectors and the second and third rows with the corresponding components of our two vectors, 𝐫 zero and 𝐫 one. Computing this determinant component by component, we have 𝐢 times the determinant of this matrix, that’s zero, minus 𝐣 times the determinant of this two-by-two matrix, that’s seven, plus 𝐤 times the determinant of this two-by-two matrix, that’s eight. Writing it out in simplified form then, this gives us this vector. And we’ll call this vector 𝐧 because, as we said, it’s normal to our plane.
We’re now very close to being able to write the general equation of our plane. Now that we have a point in the plane as well as a vector normal to it, we can clear a bit of space and remind ourselves of the vector equation of a plane. This equation tells us that if we have a normal vector as well as a point in the plane, then that normal vector dotted with the vector to a general point in the plane equals 𝐧 dot 𝐫 zero, the vector to our known point. Applying this to our scenario, we have our normal vector zero, negative seven, eight dotted with a vector to a general point in our plane with components 𝑥, 𝑦, and 𝑧 being equal to our normal vector dotted with a vector to our known point.
If we now calculate these two dot products, on the left-hand side, we get negative seven plus eight 𝑧. And on the right, we get 56 minus 56, or zero. And now we have the general equation of our plane. It’s negative seven 𝑦 plus eight 𝑧 equals zero. And notice that this equation doesn’t depend on the 𝑥-values in our plane. This is consistent with the fact that the plane contains the entire 𝑥-axis.
