# Question Video: Identifying the Vector Equation of a Line Mathematics

Which of the following is a vector equation for the straight line passing through the points π΄(2, β1) and π΅(3, 5)? [A] π« = β©2, β1βͺ + πβ©β1, 6βͺ [B] π« = β©2, β1βͺ + πβ©2, 12βͺ [C] π« = β©2, 12βͺ + πβ©2, β1βͺ [D] π« = β©3, 5βͺ + πβ©β1, 6βͺ [E] π« = β©2, 12βͺ + πβ©3, 5βͺ

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### 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, letβs start by recalling what is meant by a vector equation of a line. Itβs 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, letβs 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 pointβs 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, letβs 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.