### Video Transcript

Consider the diagram. π΄π΅π·, π΄ππΆ, and π΅ππΆ are
straight lines, where π΅ is the midpoint of π΄π· and π is the midpoint of π΅πΆ. The vector πΆπ΄ is equal to nine
π, the vector πΆπ΅ is equal to six π, and the vector πΆπ is equal to ππ, where
π is a constant. πππ· is a straight line. Find the value of π.

Letβs begin by adding the
information weβre given about the vectors πΆπ΄, πΆπ΅, and πΆπ onto the diagram. So we have πΆπ΄ is equal to nine
π, πΆπ is equal to ππ, and πΆπ΅ is equal to six π. Weβre also told in the question
that πππ· is a straight line. This means that the vectors ππ·
and ππ must be parallel or phrased another way ππ· is a scalar multiple of
ππ.

Letβs see if we can find
expressions for these two vectors, starting with ππ. We donβt have a vector for going
directly from π to π, but we can go via the point πΆ. The vector ππ is, therefore,
equal to ππΆ plus πΆπ.

Now, we havenβt been given the
vector ππΆ, but we have been given the vector πΆπ. And as weβre travelling in the
opposite direction, this means that we change the sign. So we have that ππ is equal to
negative πΆπ plus πΆπ.

Weβre also told in the question
that π is the midpoint of π΅πΆ. This means that the vectors πΆπ
and ππ΅ are each half of the vector πΆπ΅. As the vector πΆπ΅ is equal to six
π, this means that the vectors πΆπ and ππ΅ are each equal to three π. Substituting the vectors for πΆπ
and πΆπ into our expression for ππ, we have that ππ is equal to negative ππ
plus three π.

Now, letβs consider the vector
ππ·. We donβt have a vector that
describes how to go directly from π to π·. But we can go via the point π΅. The vector ππ· is equal to ππ΅
plus π΅π·. Now, we already have an expression
for the vector ππ΅, itβs three π. So we just need to find the vector
π΅π·.

As π΅ is the midpoint of the line
π΄π·, the vector π΅π· will be equal to the vector π΄π΅. We can find the vector π΄π΅ by
going via the point πΆ. The vector π΄π΅ is equal to π΄πΆ
plus πΆπ΅, both of which we have expressions for. The vector πΆπ΅ is six π. And as weβre travelling in the
opposite direction from π΄ to πΆ, the vector π΄πΆ is negative nine π. So this gives an expression for
π΄π΅. And as π΄π΅ is equal to π΅π·, we
also have an expression for π΅π·.

Substituting the vectors for ππ΅
and π΅π· into our expression for ππ·, we have three π minus nine π plus six
π. This simplifies to negative nine π
plus nine π. And we have our vector for
ππ·.

Now, remember we said that ππ· is
a scalar multiple of ππ, which means that the vector ππ· is equal to π
multiplied by the vector ππ for some number π. Substituting the expressions we
found for ππ· and ππ gives negative nine π plus nine π is equal to π
multiplied by negative ππ plus three π. We can expand the bracket on the
right side of this equation to give negative nine π plus nine π is equal to
negative πππ plus three ππ.

Now, remember weβre looking to find
the value of π. And to do so, we need to equate the
coefficients of π and π on the two sides of this equation. Equating the coefficients of π, we
have negative nine is equal to negative ππ. And equating the coefficients of
π, we have that nine is equal to three π.

We can solve the second equation to
find the value of π by dividing both sides by three. This tells us that π is equal to
three.

Finally, we can substitute this
value into our first equation, which tells us that negative nine is equal to
negative three π. The negatives on each side of the
equation will cancel out. And to find the value of π, we
need to divide both sides by three. This gives three is equal to
π.

The value of π is three.