Question Video: Finding the General Equation of a Plane Mathematics

Video Transcript

Find the general equation of the plane 𝑥 equals four plus seven 𝑡 one plus four 𝑡 two, 𝑦 equals negative three minus four 𝑡 two, 𝑧 equals one plus three 𝑡 one.

Alright, so in this exercise, we’re given the equation of a plane in parametric form. That means we have three equations for 𝑥, 𝑦, and 𝑧. And they’re written in terms of two parameters, 𝑡 one and 𝑡 two. We want to convert from the parametric form to the general form of the equation of this plane. And we can remember that the general or Cartesian form of a plane’s equation is given by 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equaling zero. So the question is, how do we express these equations this way?

We can start by remembering that the parametric form of a plane’s equation involves one point in the plane and two vectors that also lie in it. We can write that mathematically like this. This equation means that we can get to any point in our plane by starting at a known point in the plane and then moving out from that in these two different directions defined by these vectors 𝐯 one and 𝐯 two, where these vectors critically are varied by these parameters 𝑡 one and 𝑡 two.

Now, these three given equations for 𝑥, 𝑦, and 𝑧 fit into this general parametric form. What we mean is that, for example, 𝑥 is equal to four — that would be 𝑥 zero in this form — plus seven times 𝑡 one — so 𝐯 one 𝑥 would be seven — plus four times 𝑡 two — so 𝐯 two 𝑥 would be four. Carrying this out for 𝑥, 𝑦, and 𝑧 altogether, we get this result, which tells us that according to these parametric equations, our plane contains the point four, negative three, one and the vectors seven, zero, three and four, negative four, zero.

Now, at this point, let’s recall that the general form of a plane is based on defining a plane in terms of a vector normal to it and one point on it, while on the other hand the parametric form comes from defining a plane in terms of a point on it and two vectors that lie in it. To go from parametric to general form then, we need to take these two coplanar vectors and combine them in some way to get a vector normal to the plane.

We can do exactly this by taking the cross product of our two coplanar vectors. We’ve called them 𝐯 one and 𝐯 two. Note that the components of 𝐯 one are seven, zero, three and those of 𝐯 two are four, negative four, zero. And so those are the values we use in the second two rows of our three-by-three matrix. Calculating this determinant, we get 𝐢 times zero minus negative 12 minus 𝐣 times zero minus 12 plus 𝐤 times negative 28 minus zero. That’s equal to 12𝐢 plus 12𝐣 minus 28𝐤 or, written another way, 12, 12, negative 28.

Notice that if we divide all the components by four in this vector, then we will get a vector that is reduced but that is still normal to the plane of interest. For simplicity’s sake then, we’ll say that this is our normal vector 𝐧.

Okay, so now we’ve got a vector that’s normal to our plane. And we also have a point that lies in it: four, negative three, one. We can use this information to write the equation of our plane as our normal vector dotted with a vector to a general point in the plane being equal to our normal vector dotted with a vector to a point we know to lie in the plane. Carrying out those dot products gives us three 𝑥 plus three 𝑦 minus seven 𝑧 being equal to 12 minus nine minus seven, which is equal to negative four.

With our equation written this way, we can see we’re very close to the general form of a plane. As a last step, let’s add positive four to both sides of this equation, leaving us with this result. So the general form of the equation of our plane is three 𝑥 plus three 𝑦 minus seven 𝑧 plus four equals zero.