Find the direction cosines of the normal to the plane four 𝑥 plus eight 𝑦 minus three 𝑧 equals 28.
Starting with the equation of this plane, we can first rearrange it so that it’s in what’s called general form. This involves shifting quantities around so that zero appears on one side of the equation. For the plane given to us, subtracting 28 from both sides lets us write it in general form. And expressed this way, we can recognize that the factors by which we multiply 𝑥, 𝑦, and 𝑧 represent the components of a vector that’s normal to this plane. That is, we can say that a normal vector 𝐧 has components four, eight, and negative three.
This is helpful because the direction cosines of the normal to the plane can be thought of as the components of a unit vector in the same direction as 𝐧. In other words, if 𝐧 had a magnitude of one, then its components would be the direction cosines of the normal to the plane.
To find the direction cosines then, we’ll want to divide each of these components by the magnitude of the vector 𝐧. To do that, let’s recall that, given a three-dimensional vector, here we’ve called it 𝐯, the magnitude of that vector equals the square root of the sum of the squares of its 𝑥-, 𝑦-, and 𝑧-components. So then the magnitude of 𝐧 equals the square root of four squared plus eight squared plus negative three quantity squared. This equals the square root of 16 plus 64 plus nine, or the square root of 89.
We’re now ready to write out our direction cosines. Like we mentioned, these are the components of our normal vector divided by the magnitude of that vector. That gives us four divided by the square root of 89, eight divided by the square root of 89, and negative three divided by that same square root.
Before we report this as our final answer though, let’s rationalize our denominator. If we multiply both numerator and denominator by the square root of 89, then we get this equivalent form. These are the direction cosines of the normal to our plane.