3u vector question (1 Viewer)

[catha230]

I have trouble solving part b of question 20. I tried to use dot product as the question instructs but it led me to nowhere. Any help would be greatly appreciated.

 

[TheOnePheeph]

In a square, the diagonals are perpendicular and the sides are perpendicular, so simply compute the dot product of OP and OR, and set it equal to 0, then do the same thing for OQ and PR to get two simultaneous equations. You will have the term a.b present in both, so simply eliminate that as you normally would in a simultaneous equation, and you should end up with the desired result after some manipulation.

 

[Drongoski]

Don't know how to do the vector notations. So unable to post solution. Also, there's too much to type out.

But vector PR = vector OR - vector OP = (2a - 3b) - (a+b) = a - 4b.

You can do as suggested by TheOnePheeph or solve simultaneously any pair of related results, like (2a-3b).(a+b) = 0
and (a+b).(a+b) = (2a-3b).(2a-3b)

 

[Paradoxica]

In a square, the square of the diagonal length is twice the square of the side length.

2(a+b)·(a+b)=(3a-2b)·(3a-2b)

2|a|²+2|b|²+4(a·b)=9|a|²+4|b|²-12(a·b)

⇔7|a|²+2|b|²=16(a·b)

I got PR is a-4b, so correct me if I am wrong.

Diagonals of a square are perpendicular. Use the perpendicular property of the dot product.

(a-4b)·(3a-2b)=0

3|a|²+8|b|²-14(a·b)=0

⇔3|a|²+8|b|²=14(a·b)

Multiply the first result by 7 and the second result by 8.

49|a|²+14|b|²=112(a·b)

24|a|²+64|b|²=112(a·b)

Subtract.

25|a|²-50|b|²=0

⇔|a|²=2|b|²