Video Transcript
Using point four, nine, negative
nine and direction vector negative five, one, negative six, give the position vector
𝑟 of the point on this line corresponding to parameter 𝑡 equals nine.
Let’s begin by identifying exactly
what this question is asking us. We’re being asked to find a
position vector 𝑟 of a point on a line defined by the point four, nine, negative
nine and a direction vector negative five, one, negative six. Let’s begin by recalling the vector
equation of a line. It’s 𝑟 equals 𝑟 naught plus
𝑡𝑑. In this equation, 𝑟 naught
describes the vector that takes us from the origin to a point on this line. 𝑑 is the direction vector. And this is a little bit like the
gradient or the slope in the formula for the Cartesian equation of a straight
line. 𝑡 is just a real constant. So let’s begin by substituting what
we know about our line into this formula.
The vector that gets us from the
origin to a point on the line is given by four, nine, negative nine and the
direction vector is stated in the question as negative five, one, negative six. And this means that the vector
equation of our line is four, nine, negative nine plus 𝑡 times negative five, one,
negative six. We’re looking to find the position
vector of the point on a line that corresponds to the parameter 𝑡 equals nine. And so we substitute nine into our
equation. And we obtain 𝑟 to be equal to
four, nine, negative nine plus nine times negative five, one, negative six.
We can multiply this constant by
the second vector by multiplying nine by each element; that’s by negative five, one,
and negative six. And when we do, we obtain that the
position vector 𝑟 is given by four, nine, and negative nine plus negative 45, nine,
negative 54. We’re simply going to add each
element in our vector. Four plus negative 45 is negative
41. Nine plus nine is 18. And negative nine plus negative 54
is negative 63. And we see that the position vector
𝑟 of the point on our line corresponding to parameter 𝑡 equals nine is negative
41, 18, negative 63.
