### Video Transcript

If the angle between the plane
three π₯ plus ππ¦ minus five π§ is equal to two and the straight line π₯ plus two
is equal to three minus π¦ over two is equal to π§ minus one over two is 45 degrees,
find the value of the positive constant π given that π is greater than zero and
less than 10.

In this question, weβre given the
equation of a plane and the equation of a straight line. And our equation of a plane
involves an unknown value of π. We need to determine the value of
π by using the fact itβs a positive constant which is less than 10 and the angle
between the line and the plane is 45 degrees. And to find this value of π, letβs
start by recalling how we find the angle between a line and a plane. We know if the vector π is a
direction vector of a line πΏ and the vector π§ is a normal vector to a plane π,
then the acute angle π between the line and the plane will satisfy the equation sin
of π is equal to the absolute value of the dot product of π§ and π divided by the
magnitude of vector π§ multiplied by the magnitude of vector π.

And there is one small thing worth
noting. Since weβre taking the dot product
of vector π§ and π, we are working in three dimensions for both our line and our
plane. Weβre told in the question the
angle between the given line and plane is 45 degrees, so our value of π is 45. We can find vectors π and π§ by
looking at the equations of the line in the plane. Letβs start by finding the
direction vector of the line. To find the direction vector of
this line, we recall a line in the form π₯ minus π₯ one over π is equal to π¦ minus
π¦ one over π is equal to π§ minus π§ one over π will have a direction vector of
π, π, π. And our line is almost given in
this form. Thereβs just a few changes we need
to make.

First, on the leftmost part of this
equation, weβre going to divide through by one. Next, in the numerator of the
middle part of this equation, we donβt have π¦ in the numerator; we have negative
π¦. So weβre going to need to multiply
the numerator and denominator by negative one. If we distributed the parentheses,
this would now be in the required form, and we can find the direction vector of this
line. The direction vector π is the
vector one, negative two, two.

Letβs now find the normal vector to
the plane. We can do this by recalling the
plane ππ₯ plus ππ¦ plus ππ§ is equal to zero will have a normal vector of π, π,
π. In other words, the components of
the normal vector to the plane are the coefficients of the variables and is the
vector three, π, negative five. We could now substitute our value
for π and our vectors π§ and π into our equation. However, itβs easier to evaluate
the numerator and denominator of the right-hand side separately.

So letβs start by clearing some
space and then calculating the dot product between vectors π§ and π. We can do this by recalling to
evaluate the dot product of two vectors of equal dimension, we just need to find the
sum of the products of the corresponding components of the vectors. Applying this, we get one times
three plus negative two multiplied by π plus two times negative five, which if we
evaluate gives us negative two π minus seven.

Letβs now determine the magnitudes
of the two vectors. Letβs start with the magnitude of
vector π§. We can find this by recalling the
magnitude of a vector is the square root of the sum of the squares of its
components. So we get the square root of three
squared plus π squared plus negative five squared, which if we evaluate gives us
the square root of 34 plus π squared. We can do the same to find the
magnitude of vector π. Itβs the square root of one squared
plus negative two squared plus two squared, which we can evaluate is the square root
of nine, which is just equal to three.

Letβs now clear some space and
substitute all of this information into our equation involving the acute angle
between the line and plane. This then gives us that the sin of
45 degrees must be equal to the absolute value of negative two π minus seven
divided by three times the square root of 34 plus π squared. We now want to solve this equation
for π. We can do this by first noting the
sin of 45 degrees is equal to root two over two. And we can also multiply both sides
of the equation through by three root 34 plus π squared. This then gives us root two over
two multiplied by three times the square root of 34 plus π squared is equal to the
absolute value of negative two π minus seven.

And remember, weβre told in the
question that the value of π is positive. So negative two multiplied by π is
a negative multiplied by a positive. Itβs a negative number. We then subtract seven, so this is
negative. And when we take the absolute value
of a negative number, we just multiply it by negative one. So the right-hand side of this
equation simplifies to give us two π plus seven.

We can simplify the left-hand side
of this equation. We have root two multiplied by root
34 π squared. We can simplify this as the square
root of two times 34 plus π squared. This gives us the square root of 68
plus two π squared. We need to multiply this by three
over two. However, weβll just multiply this
by three and then multiply both sides of our equation through by two. This gives us that three root 68
plus two π squared is equal to two times two π plus seven. We can distribute the two over the
parentheses on the right-hand side of our equation. This gives us four π plus 14.

Now, to solve our equation for π,
weβre going to need to square both sides of our equation. On the left-hand side, three
squared is nine. And then to square a square root,
we can just remove the square root symbol. We get nine times 68 plus two π
squared. And on the right-hand side of this
equation, we have the square of a binomial. We can solve this by using the FOIL
method or binomial expansion. Either way, we get 16π squared
plus 112π plus 196.

Now, weβre going to distribute nine
over our parentheses and then rearrange to get a quadratic in π. Doing this, we get two π squared
minus 112π plus 416 is equal to zero. And we can simplify this slightly
by noting all three of our terms share a factor of two. So we can divide our equation
through by two to get π squared minus 56π plus 280 is equal to zero. We can then solve this by using the
quadratic formula, or we can look for two numbers which multiply to give 208 and add
to give negative 56.

Using either method, we can find
two solutions. Either our value of π is equal to
four or our value of π is 52. However, weβre told in the question
our value of π is positive and less than 10. So only one of these is a valid
value for π. Therefore, we were able to show
there was only one possible value of the positive constant π between zero and
10. And that value of π was four.