Question Video: Finding the Vector Form of the Equation of the Plane given Its Normal Vector Equation Mathematics

Find the vector form of the equation of the plane that has normal vector 𝐧 = 𝐢 + 𝐣 + 𝐤 and contains the point (2, 6, 6).


Video Transcript

Find the vector form of the equation of the plane that has normal vector 𝐧 equals 𝐢 hat plus 𝐣 hat plus 𝐤 hat and contains the point two, six, six.

Okay, so in this example, we have a plane. We’ll say this is it. And we’re told that relative to some set of coordinate axes, there’s a vector normal or perpendicular to the plane. And if we write that vector in vector form, we see that it has components of one, one, and one in the 𝑥-, 𝑦-, and 𝑧-directions, respectively. Along with this, we’re told that the plane contains a point, we’ll call it 𝑃 zero, with coordinates two, six, six. And knowing all this, we want to solve for the vector form of the plane’s equation.

To write the plane’s equation that way, we’ll want to define a vector that lies in the plane so that if we take the dot product of that vector and the normal vector 𝐧, we get zero. Here’s how we can go about doing that. First, let’s define a vector from our origin to the point 𝑃 zero. We’ll call the vector 𝐫 zero. And since it comes from the origin, it must have components two, six, six. And next, let’s do this. Let’s pick a point at random in our plane, we’ll call it 𝑃, and we’ll say this point has coordinates 𝑥, 𝑦, 𝑧.

We don’t specify what these values are, but nonetheless this point will help us because now we can draw a vector from our origin to this point 𝑃, call that vector 𝐫, which we see has components 𝑥, 𝑦, 𝑧. And then if we subtract the vector 𝐫 zero from 𝐫, and this was the whole purpose of defining 𝐫 in the first place, then we get a vector shown here, which lies in the plane, which means that this vector is indeed perpendicular to the normal vector 𝐧. And that means if we take the dot product of 𝐧 and 𝐫 minus 𝐫 zero, the result we’ll get is zero.

Another way to write this is 𝐧 dot 𝐫 is equal to 𝐧 dot 𝐫 zero. And we can now substitute in the known values for the normal vector 𝐧 and the vector 𝐫 zero. We’ll do that down here, where we see that the vector one, one, one dotted with 𝐫 is equal to one, one, one dotted with two, six, six. We see the vector 𝐫 is made up of the components 𝑥, 𝑦, 𝑧. But since we don’t know those more specifically, we’ll leave this left-hand side as it is. On the right-hand side, though, we can compute this dot product by multiplying the respective components together, giving us two plus six plus six.

And since these numbers add up to 14, we can write a simplified form of our equation like this. This is the vector form of the equation of our plane.

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