### Video Transcript

Which of the following is the parametric form of the equation of the plane that contains the line ๐ฅ plus one over negative two is equal to ๐ฆ minus two over two is equal to ๐ง minus five over four and the vector ๐ is equal to one, three, one. Is it option (A) ๐ฅ is equal to negative one minus two ๐ก sub one plus ๐ก sub two, ๐ฆ is equal to two plus two ๐ก sub one plus three ๐ก sub two, and ๐ง is equal to five plus four ๐ก sub one plus ๐ก sub two? Is it option (B) ๐ฅ is equal to negative one minus two ๐ก sub one plus ๐ก sub two, ๐ฆ is equal to two plus two ๐ก sub one plus three ๐ก sub two, and ๐ง is equal to negative five plus four ๐ก sub one plus ๐ก sub two. Is it option (C) ๐ฅ is equal to negative one minus two ๐ก sub one plus ๐ก sub two, ๐ฆ is equal to negative two plus two ๐ก sub one plus three ๐ก sub two, and ๐ง is equal to five plus four ๐ก sub one plus ๐ก sub two. Option (D) ๐ฅ is equal to negative one plus two ๐ก sub one plus five ๐ก sub two, ๐ฆ is equal to negative two plus two ๐ก sub one plus three ๐ก sub two, and ๐ง is equal to five plus four ๐ก sub one plus ๐ก sub two. Or is it option (E) ๐ฅ is equal to negative one plus two ๐ก sub one plus five ๐ก sub two, ๐ฆ is equal to two plus two ๐ก sub one plus three ๐ก sub two, and ๐ง is equal to negative five plus four ๐ก sub one plus ๐ก sub two?

In this question, weโre 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, weโre given two pieces of information about the plane. First, weโre told that the plane contains a line given in Cartesian form. Weโre also told that the plane contains the direction vector ๐ is equal to one, three, one. And itโs 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, weโll start by recalling that a plane passing through the point ๐ with coordinates ๐ฅ sub ๐, ๐ฆ sub ๐, ๐ง sub ๐ parallel to two distinct vectors ๐ฎ is equal to ๐ข sub ๐ฅ, ๐ข sub ๐ฆ, ๐ข sub ๐ง and ๐ฏ is equal to ๐ฃ sub ๐ฅ, ๐ฃ sub ๐ฆ, ๐ฃ sub ๐ง will have parametric equations ๐ฅ is equal to ๐ฅ sub ๐ plus ๐ข sub ๐ฅ ๐ก sub one plus ๐ฃ sub ๐ฅ ๐ก sub two. ๐ฆ is equal to ๐ฆ sub ๐ plus ๐ข sub ๐ฆ ๐ก sub one plus ๐ฃ sub ๐ฆ ๐ก sub two. And ๐ง is equal to ๐ง sub ๐ plus ๐ข sub ๐ง ๐ก sub one plus ๐ฃ sub ๐ง ๐ก sub two, where ๐ก sub one and ๐ก sub two can take any scalar value.

And itโs worth noting we need vectors ๐ฎ and ๐ฏ to be noncollinear. And we can apply this definition to the plane given to us in the question. Since the plane is parallel to ๐, weโll set our vector ๐ฏ equal to ๐, the vector one, three, one. Weโre 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 weโll set ๐ฎ 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 ๐ฎ by one-half. However, as weโll see, itโs 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 ๐ has coordinates negative one, two, five. We can now substitute the coordinates of ๐ and the components of vectors ๐ฎ and ๐ฏ into our parametric equations of the plane. Once weโve 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 ๐ฅ is equal to negative one minus two ๐ก sub one plus ๐ก sub two, ๐ฆ is equal to two plus two ๐ก sub one plus three ๐ก sub two, and ๐ง is equal to five plus four ๐ก sub one plus ๐ก sub two.