# Algebraic vectors

arildno
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Note that if S-P=R-Q, then you have R-S=Q-P.
See how you may translate this idea into something useful. Hootenanny
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Another little hint, think of how you can get to R using the vectors you have been given (this involves adding and subtracting as you say) Last edited:
masterofthewave124 said:
arunbg: what order are you talking about?
The order that I am talking of is the order in which you take the vertices.
Suppose you are given three non collinear points on a piece of paper and you are asked to find out the fourth point such that the figure formed by joining the points is a parallelogram . If the order in which you join the points (say P to Q, Q to R ,R to S and S to Q) is not specified, then there are 3C1 = 3 possible locations for R. Do this on a paper and you can see why .In this question you have assumed PQRS to be the order.

In my earlier post, I was referring to the property that diagionals of a parallelogram bisect each other or the midpoints of the two diagonals coincide . Remember section formula ?

Or of course, for vector method, what is the pallelogram law of vector addition ?

ok its been a while but i'm only coming back to this question now.

taking the vertices in order, i found the coordinates of R to be (1,5,-8)

for b), would the perimeter be equal to 2(|PQ| + |QR|)?

for c), how do i find the height? would it be |RS x RQ|?

Last edited:
b), would the perimeter be equal to 2(|PQ| + |QR|)?
Yes that wouls be correct .

c), how do i find the height? would it be |RS x RQ|?
No, the height would be
$$\vert\vec{RS}\times\widehat{RQ}\vert$$

where $\widehat{RQ}$ is a unit vector along RQ .
Do you see why ?
As a side note, the area of the parallelogram can be directly found using
|RS x RQ| .