Question Video: Finding the Parametric Form of the Equation of a Plane That Passes through Three Points Mathematics

Find the parametric form of the equation of the plane that passes through the points 𝐴(1, 5, 1), 𝐡(3, 4, 3), and 𝐢(2, 3, 4).

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Video Transcript

Find the parametric form of the equation of the plane that passes through the points 𝐴 one, five, one; 𝐡 three, four, three; and 𝐢 two, three, four.

Okay, so we have this plane that contains these three points 𝐴, 𝐡, and 𝐢. And we want to figure out which of these five options gives us the parametric form of the equation of the plane. We can recall that to define the parametric form of a plane’s equation, we need to know one point on the plane and two vectors that lie in it. Here we have three points. And we can actually use these points to define two coplanar vectors. For example, if we subtract point 𝐴 from point 𝐡, we get this vector in pink. And likewise, if we subtract 𝐴 from 𝐢, we get this vector.

Substituting in the coordinates of these points, for 𝐡 minus 𝐴, three minus one is two, four minus five is negative one, and three minus one is two. And then, for 𝐢 minus 𝐴, two minus one is one, three minus five is negative two, and four minus one is three. And what we’re saying is that these coordinates are actually the components of two vectors in our plane. We’ll call them 𝐯 one and 𝐯 two.

And now note that we have a point β€” we’ve picked point 𝐴 β€” as well as two vectors that lie in this plane. We can now recall the most general way to write the parametric form of a plane’s equation. It’s in terms of a point in the plane that we have a vector going to and two vectors, 𝐯 one and 𝐯 two, that lie in the plane, each of which is multiplied by its own parameter.

Applying this equation to our scenario, we can write that the vector 𝐫 with components π‘₯, 𝑦, 𝑧 is equal to a vector to our known point, one, five, one, plus a parameter 𝑑 one times our first vector 𝐯 one plus a second parameter 𝑑 two times 𝐯 two. Note that from this expression, we can get equations for π‘₯ and 𝑦 and 𝑧. For example, π‘₯ is equal to one plus two times 𝑑 one plus one times 𝑑 two. 𝑦 is equal to five minus one times 𝑑 one minus two times 𝑑 two. And then 𝑧 equals one plus two times 𝑑 one plus three times 𝑑 two. And these equations altogether are the parametric form of the equation of our plane. Looking over our answer options, we see a match with choice (E).

Now, it’s true that there are many equivalent ways to express the parametric form of a plane’s equation. For example, we might’ve picked a different point, say 𝐡 or 𝐢, rather than point 𝐴. And we might’ve solved for different coplanar vectors than these two here. Nonetheless, we still wouldn’t have chosen options (A), (B), (C), or (D) because none of those options use point 𝐴, 𝐡, or 𝐢 as the point that lies in the plane.

Our final answer then is choice (E). π‘₯ equals one plus two times 𝑑 one plus 𝑑 two, 𝑦 equals five minus 𝑑 one minus two 𝑑 two, and 𝑧 equals one plus two 𝑑 one plus three 𝑑 two.

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