### Video Transcript

If vector π is equal to β, β plus
two and vector π is equal to three β, four β minus one, then one of the values of β
that makes vector π parallel to vector π is what. Option (A) seven, option (B) five,
option (C) negative five, or option (D) negative seven.

In this question, weβre given two
vectors π and π. Their components include this
unknown value of β. We need to find one of the values
of β such that vector π and π will be parallel. So, letβs begin by recalling what
it means to have two parallel vectors. In this case, as we have vector π
and π, we would say that theyβre parallel if we can write that vector π is equal
to π times vector π for some scalar quantity π, where π is not equal to
zero. So, letβs set our two vectors as
parallel vectors by saying that vector π is equal to π times vector π. We can then fill in the values of
the vectors that weβre given. Vector π is β, β plus two, and
vector π is three β, four β minus one.

On the right-hand side, we can
multiply π by each of the π₯- and π¦-components. We can then evaluate the
π₯-components. So we would have β is equal to
three πβ. Dividing both sides by β, we have
that one is equal to three π. Dividing three by three, we have
one-third is equal to π or π is equal to one-third. Next, letβs consider the
π¦-components. We can write that β plus two is
equal to π times four β minus one. We have already established that π
is equal to one-third, so letβs substitute this into the equation. We then have that β plus two is
equal to one-third times four β minus one. Now letβs expand the parentheses on
the right-hand side. One-third multiplied by four β will
give us four-thirds β, and one-third multiplied by negative one gives us negative
one-third.

We can then subtract β from both
sides, remembering that if we have four-thirds β and we take away β, that will leave
us with one-third β. We can then add one-third to both
sides, remembering that two and one-third is equivalent to seven-thirds. Finally, we can multiply by three
on both sides of this equation, which leaves us with seven is equal to one β or
simply seven is equal to β. Therefore, one of the values of β
that makes π parallel to π is seven. So, the answer is that given in
option (A) seven. We can check this by plugging the
value of seven into vectors π and π. When β is equal to seven, vector π
will be seven, and seven plus two is nine. For vector π, when β is equal to
seven, three times seven is 21 and four times seven is 28; subtract one will give us
27.

So, are vectors π and π scalar
multiples? Yes, they are because we could
write that vector π is equal to one-third of vector π. So, weβve confirmed that when β is
seven, vectors π and π are parallel. But letβs check if any of the
options given in (B), (C), or (D) for the value of β will also make the vectors π
and π parallel. This time, weβll take the value
given in the answer option for the value of β and then check if the vector π is
parallel to vector π.

So, for option (B), weβre checking
the value of β is equal to five. And vector π will therefore be
five, and five plus two is seven. And vector π will be three times
five, which is 15, and four times five is 20 minus one is 19. So, is there a scalar value π for
which the vector five, seven is equal to π times the vector 15, 19? Well, no, thereβs not. If we look at the π₯-components,
weβll have five is equal to 15π. That would mean that π is equal to
one-third. However, π is equal to one-third
would not fit the equation seven is equal to π times 19. So, that means when β is equal to
five, these two vectors five, seven and 15, 19 are not parallel. So, we can eliminate option
(B).

Letβs check option (C). This time weβre checking the value
of β is equal to negative five. So, vector π will be negative
five, negative three, and vector π will be negative 15, negative 21. Once again, there is no value of π
for which vector π is equal to π times vector π. So, we can eliminate option (C)
because when β is equal to negative five, these two vectors are not parallel. Finally then, in option (D), weβre
checking the value of β is equal to negative seven. When we plug this value into the
vectors π and π, once again, there is no value of π which makes negative seven,
negative five and negative 21, negative 29 parallel. We have therefore eliminated
options (B), (C), and (D), leaving us with the value of seven.