### Video Transcript

If the angle between the two planes
ππ₯ minus four π¦ plus five π§ equals negative five and π« dot seven, negative one,
zero equals four is 60 degrees, find the value of the positive constant π.

Okay, so here weβre given the
equations of two planes. Hereβs one equation, and here is
the second. For the first plane, we can say
that a vector normal to its surface has components π, negative four, and positive
five. Letβs call this vector π§ one. And again, itβs normal or
perpendicular to the first planeβs surface.

Our second plane is given to us in
a form called vector form. Written this way, on the left-hand
side, we have a dot product of a vector to an arbitrary point in the plane with a
vector that is normal to the plane. We can say then that a vector
normal to our second plane, weβll call it π§ two, has components seven, negative
one, and zero.

Letβs recall now that the general
equation for the cosine of the angle between two planes with normal vectors π§ one
and π§ two is given by this expression. In this example, weβre actually
told the value of what weβve called π. Itβs 60 degrees. What we donβt yet know and what we
want to solve for is this value π multiplying π₯ in the equation of our first
plane. So if we substitute everything we
know into this equation for the cosine of the angle between our planes, we can now
begin simplifying the right- and the left-hand sides with the goal of solving for
π, which weβre told is a positive constant value.

First, carrying out the dot product
in our numerator, we get a result of seven π plus four plus zero. And squaring the components of our
normal vectors in our denominator, we have the square root of π squared plus 16
plus 25 times the square root of 49 plus one plus zero. This equals the magnitude of seven
π plus four divided by the square root of π squared plus 41 times the square root
of 50.

Now, at this point, since weβre
told that π is a positive value, we know that seven times π must be positive. And then when we add four to it,
that will be positive two. This means we can remove the
absolute value bars from our numerator since that number will be positive
anyway. And now, looking at the left-hand
side of this expression, letβs recall that the cos of 60 degrees is equal to exactly
one-half.

Bringing it all together, we know
that one-half is equal to this right-hand side. And again, itβs the value of π
that we want to solve for. To begin doing that, letβs multiply
both sides of our equation by the entire denominator on the right. We then have that the square root
of 50 over two times the square root of π squared plus 41 equals seven π plus
four. And if we now square both sides of
the equation, we have on our left side 50 over four times the quantity π squared
plus 41 is equal on the right to 49π squared plus 56π β thatβs two times four
times seven π β plus 16.

On the left-hand side, if we
multiply through by this factor of 50 over four and then subtract both 50π squared
over four and 50 times 41 over four from both sides of this equation, then that
makes the value on the left zero. And we have an equation that is
quadratic in π. 49 minus 50 over four simplifies to
73 over two, while 16 minus 50 times 41 over four simplifies to negative 993 divided
by two.

Our equation overall then is that
zero equals seventy-three halves times π squared minus 56π minus 993 over two. In terms then of an equation that
lets us solve for the roots of a quadratic equation, where we solve for the roots of
π₯ using these constant values π, π, and π, in our particular quadratic equation,
we would say that 73 over two is π, negative 56 is π, and negative 993 over two is
π.

We can use these labels so long as
weβre careful not to be confused that this value for π is not the same as the π
that weβre trying to solve for. That π corresponds to π₯ in this
equation for the roots of a quadratic equation. Knowing that, the π that we do
want to solve for is given by this equation. There are two solutions here
because of this plus and minus sign in our numerator. When we calculate all this out, as
expected, we get two results. When we use the minus sign in the
numerator, we get negative 331 over 73. When we use the plus sign, our
result is simply three.

At this point, letβs recall that
our problem statement tells us that π is a positive constant. Since itβs positive, that rules out
our negative 331 over 73 result. As our final answer then, we say
that π is equal to positive three.