### Video Transcript

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.