The question I need help addressing has to do with part iv)
The solution is as follows:
This is the vector equation of a line. From what I gather, the solution has the following logical forethought:
- The vector equation of a line allows us to find the position vector of any point on a given line, which is usually specified by the difference between two points. For example, in $r = a + t(b-a)$, due to the way adding vectors works, when we impose vector $\vec a$ and $\vec {b-a}$ and create the resultant vector, it will still be a position vector on that line. This means the scalar t can be used to find us the position of any point on that line.
- In the case of (iv), we need to find the midpoint of line $\overline {PQ}$. We can find any point on line $\overline {PQ}$ by finding $\vec P – \vec Q$ and applying $r_x = r_1 + t(\vec P – \vec Q)$ to find any given point on the line given the scalar. If we set constant $t$ to $\frac {1}{2}$, shouldn't that then give us $X$ in the graphic?
However, I take issue with the solution, primarily the graphic. My qualms with this solution:
- The two vectors both have the same $x$ coordinate, but seemed to be separated in the graphic. In fact, given their coordinates, their positions in this graphic don't make sense to me.
- I don't understand why $r_1, r_2$ and $ a$ in the graphic are halved in size.
- If I want to find a point on $\overline {PQ}$ that is 3 units from X, what value of $t$ do I set?
Please do let me know if any of my logical forethoughts were incorrect, and why, and if you have any answers to my qualms stated above, especially the "3 units away from X" one.
EDIT: It also seems I need to use $r_x = r_1 + t(\vec Q – \vec P)$ instead of $r_x = r_1 + t(\vec P – \vec Q)$ to get the right answer? Why is that? Why is it $r = a + t(b-a)$ instead of $r = a + t(a-b)$?
Best Answer
A line described by $(1-t)P+tQ$, where the parameter $t$ is taken from the interval $[0,1]$ gives at $t=\frac{1}{2}$ the middle point you had as the answer.