# Question Video: Finding the Parametric Equation of a Plane Mathematics

Which of the following is the parametric form of the equation of the plane that contains the line (𝑥 + 1)/−2 = (𝑦 − 2)/2 = (𝑧 − 5)/4 and the vector 𝐝 = ⟨1, 3, 1⟩? [A] 𝑥 = −1 − 2𝑡₁ + 𝑡₂, 𝑦 = 2 + 2𝑡₁ + 3𝑡₂, 𝑧 = 5 + 4𝑡₁ + 𝑡₂ [B] 𝑥 = −1 − 2𝑡₁ + 𝑡₂, 𝑦 = 2 + 2𝑡₁ + 3𝑡₂, 𝑧 = −5 + 4𝑡₁ + 𝑡₂ [C] 𝑥 = −1 − 2𝑡₁ + 𝑡₂, 𝑦 = −2 + 2𝑡₁ + 3𝑡₂, 𝑧 = 5 + 4𝑡₁ + 𝑡₂ [D] 𝑥 = −1 + 2𝑡₁ + 5𝑡₂, 𝑦 = −2 + 2𝑡₁ + 3𝑡₂, 𝑧 = 5 + 4𝑡₁ + 𝑡₂ [E] 𝑥 = −1 + 2𝑡₁ + 5𝑡₂, 𝑦 = 2 + 2𝑡₁ + 3𝑡₂, 𝑧 = −5 + 4𝑡₁ + 𝑡₂

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### Video Transcript

Which of the following is the parametric form of the equation of the plane that contains the line 𝑥 plus one over negative two is equal to 𝑦 minus two over two is equal to 𝑧 minus five over four and the vector 𝐝 is equal to one, three, one. Is it option (A) 𝑥 is equal to negative one minus two 𝑡 sub one plus 𝑡 sub two, 𝑦 is equal to two plus two 𝑡 sub one plus three 𝑡 sub two, and 𝑧 is equal to five plus four 𝑡 sub one plus 𝑡 sub two? Is it option (B) 𝑥 is equal to negative one minus two 𝑡 sub one plus 𝑡 sub two, 𝑦 is equal to two plus two 𝑡 sub one plus three 𝑡 sub two, and 𝑧 is equal to negative five plus four 𝑡 sub one plus 𝑡 sub two. Is it option (C) 𝑥 is equal to negative one minus two 𝑡 sub one plus 𝑡 sub two, 𝑦 is equal to negative two plus two 𝑡 sub one plus three 𝑡 sub two, and 𝑧 is equal to five plus four 𝑡 sub one plus 𝑡 sub two. Option (D) 𝑥 is equal to negative one plus two 𝑡 sub one plus five 𝑡 sub two, 𝑦 is equal to negative two plus two 𝑡 sub one plus three 𝑡 sub two, and 𝑧 is equal to five plus four 𝑡 sub one plus 𝑡 sub two. Or is it option (E) 𝑥 is equal to negative one plus two 𝑡 sub one plus five 𝑡 sub two, 𝑦 is equal to two plus two 𝑡 sub one plus three 𝑡 sub two, and 𝑧 is equal to negative five plus four 𝑡 sub one plus 𝑡 sub two?

In this question, we’re given five possible parametric equations for a plane. We need to determine which of these five options is a correct form for the parametric equation of the plane. And to do this, we’re given two pieces of information about the plane. First, we’re told that the plane contains a line given in Cartesian form. We’re also told that the plane contains the direction vector 𝐝 is equal to one, three, one. And it’s worth noting when we say that a plane contains a vector, this just means the vector runs parallel to the plane. And in fact, something very similar is true about our line. If the line is contained within the plane, then this means that our line runs parallel to the plane. So the direction vector of this line is parallel to the plane. This means we have two vectors parallel to the plane.

This then gives us two different ways of answering this question. We could now recall exactly what is meant by the parametric equation of a plane. We could then consider the five given options and determine which one is in the correct form for the parametric equation of a plane. And we can do this by eliminating options. However, this method requires us to be given the five options.

So instead, we’ll start by recalling that a plane passing through the point 𝑝 with coordinates 𝑥 sub 𝑝, 𝑦 sub 𝑝, 𝑧 sub 𝑝 parallel to two distinct vectors 𝐮 is equal to 𝑢 sub 𝑥, 𝑢 sub 𝑦, 𝑢 sub 𝑧 and 𝐯 is equal to 𝑣 sub 𝑥, 𝑣 sub 𝑦, 𝑣 sub 𝑧 will have parametric equations 𝑥 is equal to 𝑥 sub 𝑝 plus 𝑢 sub 𝑥 𝑡 sub one plus 𝑣 sub 𝑥 𝑡 sub two. 𝑦 is equal to 𝑦 sub 𝑝 plus 𝑢 sub 𝑦 𝑡 sub one plus 𝑣 sub 𝑦 𝑡 sub two. And 𝑧 is equal to 𝑧 sub 𝑝 plus 𝑢 sub 𝑧 𝑡 sub one plus 𝑣 sub 𝑧 𝑡 sub two, where 𝑡 sub one and 𝑡 sub two can take any scalar value.

And it’s worth noting we need vectors 𝐮 and 𝐯 to be noncollinear. And we can apply this definition to the plane given to us in the question. Since the plane is parallel to 𝐝, we’ll set our vector 𝐯 equal to 𝐝, the vector one, three, one. We’re also told that the plane contains the line given in Cartesian form. And we know the denominators of these fractions will give us a direction vector of this line. So we’ll set 𝐮 to be the direction vector of this line, the vector negative two, two, four. And it is worth noting we can take any scalar multiple of these vectors if we prefer. For example, we could multiply vector 𝐮 by one-half. However, as we’ll see, it’s not necessary in this question.

Now, all we need to do is find a point which lies on the plane. And since the plane contains the line, it contains all the points on the line. So we just need to find a single point on this line. And one way of doing this is to solve each part of the equation equal to zero. We get that 𝑝 has coordinates negative one, two, five. We can now substitute the coordinates of 𝑝 and the components of vectors 𝐮 and 𝐯 into our parametric equations of the plane. Once we’ve substituted these values in, we can simplify. And when we do, we get the following set of parametric equations, which we can see is equal to option (A).

Therefore, we were able to show the parametric form of the equations of the plane given to us in the question is 𝑥 is equal to negative one minus two 𝑡 sub one plus 𝑡 sub two, 𝑦 is equal to two plus two 𝑡 sub one plus three 𝑡 sub two, and 𝑧 is equal to five plus four 𝑡 sub one plus 𝑡 sub two.

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