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# Vectors maths question help! watch

1. See attached question..
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2. (Original post by BeatricePrior)
See attached question..
look at which way the arrows go- vectors have direction. To get from p to r, follow the directions needed to be taken. The first route is opposite to the arrow direction so you have -ve the corresponding vector given. The second arrow is in the direction we're going so we add on that vector- we therefore subtract the first from the second and whatever is left is our answer
3. (Original post by EnglishMuon)
look at which way the arrows go- vectors have direction. To get from p to r, follow the directions needed to be taken. The first route is opposite to the arrow direction so you have -ve the corresponding vector given. The second arrow is in the direction we're going so we add on that vector- we therefore subtract the first from the second and whatever is left is our answer
That's what I though but If you go -PS+SR no components have been given relating to SR so how would you work that out?
4. (Original post by Mystery.)
That's what I though but If you go -PS+SR no components have been given relating to SR so how would you work that out?
Parallel.
5. (Original post by Zacken)
Parallel.
It's parallel to PQ but there's no info for that either?
6. (Original post by Mystery.)
That's what I though but If you go -PS+SR no components have been given relating to SR so how would you work that out?
You can find what pt is then subtract the components that give it its 'height' to get the vector from p to the midpoint of the base diagonal. Then just double this
7. (Original post by BeatricePrior)
See attached question..
Maybe I'm being dimwitted, but I'm not sure that there's enough information to solve the problem. Even if P lay at the origin and S was on the x-axis (so PS was along the x-axis), then we could locate T's z-displacement above the x-y plane, but that doesn't seem to preclude, say, PQ lying at any arbitrary angle to the y-axis.

I think we need some info about the angle between the PTS plane and the QTR plane.

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