### Video Transcript

Find, in parametric form, the
equation of the plane that passes through the point π΄ one, two, one and the two
vectors π one equals one, negative one, two and π two equals two, negative one,
one.

Okay, we have here all these
possible answers for the parametric form of the equation of our plane. And as weβve seen, this plane
passes through this point π΄ and contains these two vectors π one and π two. Visually then, our plane might look
like this. In writing the equation of this
plane, the idea is that we start at our known point and then we move out from that
point in multiples of the vectors π one and π two. We could say then that a vector
describing our planeβs entire surface β weβll call it π« β is equal to a vector to
our known point on the plane plus one parameter that varies across all possible
numbers multiplied by our one vector π one added to another parameter that also
varies across all possible numbers multiplied by the vector π two.

Even though it may seem weβre
looking at just one equation here, actually, there are three involved: one for the
π₯-dimension, one for the π¦, and one for the π§. To see that, we can replace this
vector π« with its components. And now we see that, for example,
π₯ is equal to one plus π‘ one times one plus π‘ two times two. Then, likewise, π¦ is equal to two
plus π‘ one times negative one plus π‘ two times negative one. And then, lastly, we also have an
equation for the π§-component of our vector. π§ equals one plus two times π‘ one
plus π‘ two. This is the parametric form of the
equation of our plane. And if we look through our answer
options, we see it matches up with option (A). The equation of our plane in
parametric form is π₯ equals one plus π‘ one plus two times π‘ two, π¦ equals two
minus π‘ one minus π‘ two, π§ equals one plus two π‘ one plus π‘ two.