Question Video: Finding the Parametric Equation of a Plane Mathematics

Video Transcript

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

In this question, we’re given five parametric equations. And we need to determine which of these is the correct parametric form of the equation of a plane which contains two given lines in Cartesian form. To answer this question, let’s start by recalling what we mean by the parametric form of the equation of a plane. We recall a plane passing through the point 𝑃 with coordinates 𝑥 sub 𝑃, 𝑦 sub 𝑃, 𝑧 sub 𝑃 which contains two noncollinear vectors 𝐮 with components 𝑢 sub 𝑥, 𝑢 sub 𝑦, 𝑢 sub 𝑧 and 𝐯 with components 𝑣 sub 𝑥, 𝑣 sub 𝑦, 𝑣 sub 𝑧 will have parametric equations given by the following set of three equations.

We can see the equation for 𝑥 is entirely given in terms of the 𝑥-coordinate for 𝑃 and the 𝑥-components of vectors 𝐮 and 𝐯. And something very similar is true for our equations for 𝑦 and 𝑧. We can use this definition to determine the parametric form of the equation of this plane. To do this, we start by noting we’re given two lines which are contained within the plane. And both of these lines are given in Cartesian form. And we can recall we could find the direction vectors of lines given in Cartesian form by considering the denominators of each part of the question. The direction vector of each term will have components equal to the denominators of these fractions. So, we can set our vector 𝐮 equal to the vector negative two, negative one, three.

And now we might be tempted to apply the exact same reasoning to set vector 𝐯 equal to the vector negative four, negative two, six. However, there’s one problem with this. We’re told that we need to choose noncollinear vectors 𝐮 and 𝐯. And we can notice all of the components of vector 𝐯 share a factor of two. In fact, vector 𝐯 is just equal to two times vector 𝐮. So, they’re scalar multiples of each other; in other words, these vectors are collinear. So, we need to determine another vector contained within the plane. To do this, let’s sketch what we’re given.

We have two lines contained within the plane, and we’ve shown that their direction vectors are collinear. So, the lines must be parallel. And we can also note that the lines are not coincident because the equations are not scalar multiples of each other. So, we have two distinct lines in our plane. And we can also take our vector 𝐮 to be one of the direction vectors of the line. So, we can find the vector contained within the plane by finding a point on each line. Then, the vector between these two points will be contained within the plane. We can find a point on each line by setting each equation equal to zero. So, let’s clear some space and calculate our vector 𝐯.

We start by finding the position vector of the first point on the line. We set each equation equal to zero. We get the vector one, negative one, one. We then do the exact same with the equation of the second line. It contains the point with position vector zero, two, negative one. Then, we can set 𝐯 equal to the difference between these two vectors. And then we can evaluate the subtraction by subtracting the vectors component-wise. We get 𝐯 is the vector negative one, three, negative two. And at this point, it is worth reiterating we can choose any scalar multiples of these two vectors to be our vectors 𝐮 and 𝐯.

And now we’re almost ready to substitute this information into our parametric equations. However, we might want to choose our point 𝑃 first. However, there’s many different points we could choose for 𝑃. For example, we’ve already found the point on each line. And we can construct even more points on the line or even more points on the plane. And all of these would give valid parametric equations of the line. So, instead, we’re just going to substitute our vectors 𝐮 and 𝐯 into the equation and then determine the necessary coordinates of point 𝑃 from the options given. Substituting the components of 𝐮 and 𝐯 into the parametric equations of the line gives us the following three equations.

We can then notice this only matches one of our options, option (B). Since the coefficients of 𝑡 sub one and 𝑡 sub two match for all three of these equations, this will be a valid parametric equation of the plane if the point with coordinates one, negative one, one lies on the plane. And we’ve already shown that this point does lie on the plane because we’ve already shown that it lies on the first line. So, we choose 𝑃 to be the point with coordinates one, negative one, one and then substitute the coordinates at this point into our parametric equations, which we can then see exactly matches option (B).

Therefore, we were able to show that the parametric form of the equation of the plane which contains the two lines 𝑥 minus one over negative two is equal to 𝑦 plus one over negative one is equal to 𝑧 minus one over three and 𝑥 over negative four is equal to 𝑦 minus two over negative two is equal to 𝑧 plus one over six is option (B). 𝑥 is equal to one minus two 𝑡 sub one minus 𝑡 sub two, 𝑦 is equal to negative one minus 𝑡 sub one plus three 𝑡 sub two, and 𝑧 is equal to one plus three 𝑡 sub one minus two 𝑡 sub two.