### Video Transcript

Given that the plane πΎπ§ plus two
π₯ plus three π¦ equals negative four is parallel to the plane πΏπ¦ minus two π₯
minus two π§ equals three, find the values of πΎ and πΏ.

Okay, so in this exercise we have
these two planes given by these equations. If we write them in standard form,
the first one is two π₯ plus three π¦ plus πΎπ§, where πΎ is some constant value, is
all equal to negative four. And the second is negative two π₯
plus πΏπ¦, where πΏ is some unknown constant, minus two times π§ is equal to
three. Because these equations are now
written in standard form, we can pick out the components of the normal vectors to
each plane. The normal vector of the first
plane will have an π₯-component of two, a π¦-component of three, and a π§-component
of πΎ. Weβll call this vector π§ one and
write it out this way. And then for the second plane, its
normal vector will have an π₯-component of negative two, a π¦-component of πΏ, and a
π§-component of negative two. And weβll call this vector π§
two.

Our problem statement tells us that
these two planes are parallel to one another. This fact can point us to recalling
the mathematical condition for two planes to be parallel. Two parallel planes with normal
vectors π§ one and π§ two have their components related by a constant; weβve called
it πΆ. This constant could be positive or
negative, but whatever its sign, it means that there is a consistent ratio between
the components of these two normal vectors. Knowing this, letβs look at our two
normal vectors π§ one and π§ two and see what this constant of proportionality πΆ
might be.

Comparing the π₯-components of
these vectors, we see the π₯-component of π§ one is two and that of π§ two is
negative two. We know that two is equal to
negative one times negative two. And in fact, this is the only value
by which we can multiply negative two to yield positive two which tells us that πΆ,
in this case, is negative one. And this fact is the key for
solving for the values of πΎ and πΏ in our normal vector equations because just as
we can write this equation out for the π₯-components of our normal vectors, so we
can write it out for the π¦- and π§-components.

If we substitute negative one in
for πΆ in both of these equations, then we see that three is equal to negative one
times πΏ, telling us that πΏ is equal to negative three. And since πΎ is equal to negative
one times negative two, that means πΎ is equal to two. So weβve used the fact that our two
planes are parallel to solve for the unknown parts of their normal vectors. πΏ is equal to negative three, and
πΎ is equal to positive two.