Video Transcript
Write the vector equation of the
straight line that passes through the point six, negative nine with direction vector
nine, negative two.
We begin by recalling the general
form for the vector equation of a straight line. We have 𝐫 is equal to 𝐫 sub zero
plus 𝑘 multiplied by 𝐝, where 𝐫 sub zero is the position vector of a point that
lies on the line, 𝐝 is the direction vector of the line, and 𝑘 is a scalar
quantity. And we note we can use any letter
to represent this. We are told in this question that
our straight line passes through the point with coordinates six, negative nine. And we know that any point 𝑝 with
coordinates 𝑥 sub zero, 𝑦 sub zero will have position vector 𝐫 sub zero equal to
𝑥 sub zero, 𝑦 sub zero. The position vector of our point is
therefore equal to six, negative nine. We are also told that the direction
vector of the line is nine, negative two.
Substituting both of these vectors
into the general form, we have 𝐫 is equal to six, negative nine plus 𝑘 multiplied
by nine, negative two. Whilst we could simplify the
right-hand side, this is a perfectly valid way of leaving our answer. The vector equation of the straight
line that passes through the point six, negative nine with direction vector nine,
negative two is 𝐫 is equal to six, negative nine plus 𝑘 multiplied by nine,
negative two.
