### Video Transcript

In this video, our topic is
equations of parallel and perpendicular planes. Weβll see how planes,
two-dimensional surfaces in three-dimensional space, can be parallel or
perpendicular to one another and how to recognize these conditions
mathematically.

As we get started, letβs recall a
bit about just how a plane is defined. The two pieces of information we
often have about a plane that let us uniquely define it in three-dimensional space
are, first, a point that lies on the plane and, second, a vector β weβve called it
π§ β that is normal or perpendicular to the plane. In general, given a vector thatβs
normal to a plane where that vector has components π, π, and π and a point π
zero on the plane with coordinates π₯ zero, π¦ zero, and π§ zero, we can say that
our normal vector dotted with the vector that goes to a general point on the plane
with coordinates π₯, π¦, π§ is equal to that same normal vector dotted with a vector
going to our point π zero.

And we can say that this line here
represents the vector form or the vector equation of our plane, while if we multiply
out the dot products, then we get this expression here. This is called the scalar form or
scalar equation of a plane. And then going a bit further, if we
group the terms on the right-hand side together, calling them π, then we arrive at
an equation for our plane thatβs sometimes called the general or normal form. So anyway, given a vector thatβs
normal to the surface of a plane and a point that is in the plane, weβre able to
generate various expressions for the equation of the plane. And it turns out, not only does the
normal vector to a plane help us define that planes equation, it also helps us
figure out what other mathematical planes this one might be parallel or
perpendicular to.

For example, say that we have one
plane with a normal vector like this and another plane with this normal vector. If we call these vectors π§ one and
π§ two and we assume that these two planes are parallel to each other, then we can
say something about the way that normal vector π§ one relates to normal vector π§
two. If the two planes are parallel,
then these normal vectors must also be parallel or possibly antiparallel. Either way, we could write the one
normal vector in terms of a constant, whether positive or negative, multiplied by
the other normal vector.

So just for example, letβs say that
our normal vector π§ one has components one, two, three, while our normal vector π§
two has these components. Comparing these vectors, we see
that we could multiply the π₯-component of π§ one by negative two to get the
π₯-component of π§ two. And likewise, we could multiply the
π¦-component of π§ one by that same factor negative two to get the π¦-component of
π§ two. And the same is true for the
π§-components of these vectors. In this case then, we would say
that our constant πΆ is equal to negative two. And because there is such a
constant by which we can multiply π§ two so that it equals π§ one, we say that these
two planes are parallel.

This expression telling us that the
normal vector of one plane equals a constant times the normal vector of a second
plane is the condition for two planes being parallel. If two planes have normal vectors
that satisfy this condition for any constant value πΆ, then we can safely say that
the planes are parallel. But now letβs imagine two planes
that are oriented in a different way. Say that our two planes are now
perpendicular to one another. Well, since their normal vectors
are perpendicular to the surface of each plane, respectively, we can expect that if
the planes are perpendicular, then π§ one and π§ two are perpendicular as well. And in general, if two vectors are
perpendicular, that means if we take the dot product of those vectors, the result
weβll find is zero.

Geometrically, this means that
thereβs no overlap between the two vectors weβre dotting together. So, given the normal vectors of two
planes, if those normal vectors have a dot product of zero, then the planes must be
perpendicular to one another. And these two equations, one for
parallel planes and one for perpendicular, give us helpful tests to apply once we
know the normal vectors of two planes to see whether they are parallel or
perpendicular. The best way to really learn all
this is through some practice. So letβs look now at an example
exercise.

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.

Letβs look now at a second parallel
plane example.

Find the equation of the plane
passing through the point π, π, π and parallel to the plane π₯ plus π¦ plus π§
equals zero.

All right, so we want to solve for
the equation of a plane, and we know that this plane passes through a point with
coordinates π, π, π. So if we draw a sketch, letβs say
that this is the plane whose equation we want to solve for and we know that
somewhere on this plane is this point, π, π, π. Along with this, weβre told that
this plane is parallel to another plane whose equation is given here. And note that this planeβs equation
is given in such a form that weβre able to use it to solve for the components of the
planeβs normal vector. Note that all three of the
variables π₯, π¦, and π§ are effectively being multiplied by one. These multiplying factors, these
ones, give us components of a vector thatβs normal to this plane.

If we call that vector π§, its
π₯-component could be given by one, its π¦-component by one, and its π§-component by
the same value. And itβs helpful to know this
normal vector because weβre told that the plane whose equation we want to solve for
is parallel to the plane whose equation is given here. And therefore, the normal vector of
this unknown plane, weβll call this vector π§ sub π, must be parallel or
antiparallel to π§. And in fact, we could set π§ sub π
equal to π§ in general. And we do this based on our
knowledge of the fact that the two planes weβre considering are parallel.

At this point, note that we have a
vector thatβs normal to the plane whose equation we want to solve for. And we also have the coordinates of
a point on that plane. Taken together, these two pieces of
information are enough to let us solve for this planeβs equation. We can recall that given a normal
vector to a plane as well as a vector to a random point on that plane, the dot
product of those two vectors equals the dot product of the normal vector with a
vector to a known point on the plane. To apply this relationship to our
scenario, weβll say that our normal vector π§ sub π dotted with a vector that goes
to any point on our plane with the general components π₯, π¦, and π§ is equal to our
normal vector π§ sub π dotted with the vector to our known point π, π, π.

We can now substitute in the value
for our normal vector π§ sub π. That gives us this equation
here. And if we then carry out both of
these dot product operations, we find that π₯ plus π¦ plus π§ is equal to π plus π
plus π. And this result is the equation of
the plane that passes through the point π, π, π and is parallel to the plane π₯
plus π¦ plus π§ equals zero.

Now letβs look at an example
involving two planes that are perpendicular to one another.

Given that the plane three π₯ minus
three π¦ minus three π§ equals one is perpendicular to the plane ππ₯ minus two π¦
minus π§ equals four, find the value of π.

Okay, in this exercise, weβre given
equations of two planes and told that these planes are perpendicular to one
another. And we can say that if two planes
are perpendicular like these are, then it must also be the case that their normal
vectors are perpendicular. And this points us to the
mathematical condition that is satisfied for two perpendicular planes. The dot product of the normal
vectors of these planes must be zero. This suggests that we might solve
for the normal vectors of our two given planes and then apply this condition to see
if it helps us solve for π. Since both of these equations are
given in standard form, we can identify the components of each planeβs normal
vector.

For the first plane, its normal
vector has an π₯-component of three, a π¦-component of negative three, and a
π§-component also of negative three. We can write that like this,
calling this normal vector π§ one. For our second plane, its normal
vector has an π₯-component of π, a π¦-component of negative two, and a π§-component
of negative one. And weβll call this normal vector
π§ two. Next, knowing that these two planes
are perpendicular, weβll apply this condition. Weβll say that π§ one, this vector
here, dotted with π§ two, this vector here, equals zero. If we then carry out this dot
product operation, we find that three times π plus six plus three equals zero. And this means that three times the
unknown value π is equal to negative nine. Dividing both sides by three, we
find that π is equal to negative three. This is the value of π.

Letβs now look at another
perpendicular planes example.

Find the general equation of the
plane which passes through the point two, eight, one and is perpendicular to the two
planes negative six π₯ negative four π¦ plus six π§ equals negative five and five π₯
plus three π¦ minus six π§ equals three.

Okay, so in this exercise, weβre
told about three planes, two of which we have the equations for. The third plane whose general
equation we want to solve for is perpendicular to these two and it also passes
through this given point. Now, in general, we can write the
equation for a plane if we have two bits of information about it. First, if we know a vector that is
normal to the surface of the plane and, second, if we know a point on the plane. In this scenario, we see that,
indeed, we are given a point which lies on this unknown plane. But we donβt yet know the
components of a vector normal to it.

However, notice this. We do know that this plane,
whatever its equation, is perpendicular to these two planes given by these
equations. This implies that the normal vector
π§ we want to solve for is also perpendicular to the normal vectors of these two
planes, meaning if we can solve for the normal vectors of these planes and then find
a vector perpendicular to them both, weβll have an expression for π§. We can begin doing this by
recalling that when the equation of a plane is given in whatβs called standard form,
like we see it here, then the components of vectors normal to those planes are given
by the factors by which we multiply the variables π₯, π¦, and π§.

For example, if we call a vector
normal to our first plane π§ one, then this may have components negative six,
negative four, six. And if we call a vector normal to
our second plane π§ two, this will have components five, three, negative six. And now that we know π§ one and π§
two, recall that π§ is perpendicular to both of these vectors. So hereβs what we can do. We can take the cross product of π§
one and π§ two. And if we recall that the cross
product of two vectors β weβve called them π and π, which are three-dimensional β
is equal to the determinant of this three-by-three matrix, then we can apply this
relationship to our scenario, where the second and third rows of our matrix are
populated by the respective components of our vectors π§ one and π§ two.

We see that the π’-component of π§
one is negative six, the π£-component is negative four, and the π€-component is
positive six. Likewise, the π’-component of π§
two is five, its π£-component is three, and its π€-component negative six. Weβre now ready to calculate this
cross product. And weβll do it component by
component, starting with π’. The π’-component of the vector
resulting from crossing π§ one and __ two is equal to the determinant of this
two-by-two matrix. Thatβs a negative four times
negative six or 24 minus six times three or 18. We then move on to the
π£-component, which is negative the determinant of this two-by-two matrix. Negative six times negative six is
positive 36. And we subtract from that six times
five or 30.

And lastly then, we compute the
π€-component equal to the determinant of this two-by-two matrix. Negative six times three is
negative 18 minus negative four times five or negative 20. Adding these three components
together, we get the cross product of π§ one and π§ two. Thatβs equal to six π’ minus six π£
plus two π€ or, written in vector form, six, negative six, two. Now that we know this cross
product, letβs recall that we were calculating it because we knew that π§ is
perpendicular to π§ one and π§ two. And since this vector resulting
from crossing π§ one and π§ two is perpendicular to them both by the rule of the
cross product, we can say that itβs equal to the normal vector π§.

Now that we know this and we also
know a point on our plane π, we can make quick progress in finding the general
equation of the plane. We can write the equation of our
plane as the normal vector π§ dotted with a vector to a general point on the plane
with coordinates π₯, π¦, and π§ being equal to that same normal vector dotted with
the vector to the known point on the plane with coordinates two, eight, one. And since we know the vector form
of π§, we could substitute that into this equation. And having done that, if we then
calculate both of these dot products, we find that six π₯ minus six π¦ plus two π§
equals 12 minus 48 plus two. This is equal to negative 34.

So if we add 34 to both sides of
our equation, we can write that six π₯ minus six π¦ plus two π§ plus 34 equals
zero. And noticing that all the terms on
the left are divisible by two, we find that three π₯ minus three π¦ plus π§ plus 17
equals zero. And this is the general equation of
the plane which passes through the point two, eight, one and is perpendicular to the
two planes given.

Letβs finish up now by summarizing
a few key points. In this lesson, we saw that two
planes are parallel when their normal vectors π§ one and π§ two satisfy this
condition, that they are a constant multiple of one another. We also learned that two planes are
perpendicular when their normal vectors π§ one and π§ two satisfy the condition that
their dot product equals zero. And lastly, we saw that if two
planes are parallel or perpendicular, so are their corresponding normal vectors.