### Video Transcript

Which of the following is a vector
equation for the straight line passing through the points π΄ two, negative one and
π΅ three, five? Option (A) π« equals two, negative
one plus π times negative one, six. Option (B) π« equals two, negative
one plus π times two, 12. Option (C) π« equals two, 12 plus
π times two, negative one. Option (D) π« equals three, five
plus π times negative one, six. Or is it option (E) π« equals two,
12 plus π times three, five?

In this question, we are asked to
determine which of five given options is a correct vector equation of a line passing
through two given points π΄ and π΅. To do this, letβs start by
recalling what is meant by a vector equation of a line. Itβs an equation of the form π«
equals π« sub zero plus π times π.

In the vector equation of a line,
the vector π« is the position vector of any point on the line depending on the
parameter π and our choices of π« sub zero and π. The vector π« sub zero is the
position vector of a single point on the line. π is a parameter; it can take any
real values. And π is a direction vector of the
line; it is any nonzero vector parallel to the line.

This means we can construct vector
equations of lines by using two vectors: a position vector of a point on the line
and a direction vector of the line. We could try eliminating options at
this stage. And while this is possible, it is
quite difficult to do this. Instead, letβs find the direction
vector of the line passing through π΄ and π΅. We can recall that the vector from
π΄ to π΅ will be a direction vector of this line, since it is parallel to the line,
and that this is given by the position vector of π΅ minus the position vector of
π΄.

We can then recall that the
position vector of a point has components equal to the pointβs coordinates. So we substitute the coordinates of
these points into the vectors to obtain the vector three, five minus the vector two,
negative one. To subtract two vectors, we
subtract their corresponding components. So we get the vector three minus
two, five minus negative one.

We can then evaluate this to obtain
the vector one, six. This is one possible direction
vector of the line. It is important to note that this
is not the only possible direction vector of this line. We can take any nonzero scalar
multiple of this vector to find a different direction vector of this line. And they are all valid choices of
π.

We can distribute the scalar
multiplication by π‘ over the vector by multiplying each component by π‘ to get the
vector π, six π. All direction vectors of this line
are given by this vector for some nonzero value of π. We can use this to eliminate
options.

One way of doing this is to note
that since π‘ is nonzero and the coefficients of π‘ have the same signs, the
components of π must have the same signs. This allows us to eliminate any
options with direction vectors with the opposite signs in their components. We see that options (A), (C), and
(D) cannot be vector equations of this line since their direction vectors are not
parallel to the vector from π΄ to π΅.

In a similar way, we can eliminate
option (E). For the first component of the
direction vector π to be three, we need π‘ equals three. However, for the second component
to be five, we need π‘ equals five over six. This means that the vector three,
five is not parallel to the vector from π΄ to π΅. And so it is not a possible
direction vector of the line. This only leaves option (B).

And for due diligence, letβs make
sure that this is a valid vector equation of the line. We first note that the vector two,
negative one is the position vector of point π΄. So this is a position vector of a
point on the line. We can also note that if we set π‘
equals two, then we see that the vector two, 12 is parallel to the vector from π΄ to
π΅. This means that it is a valid
direction vector of the line. This confirms that π« equals two,
negative one plus π times two, 12 is a vector equation of the line between the
points two, negative one and three, five.