Question

Given two non-zero vectors a and b such that [a+b]=[a-b], find the values of a*b.
(Where "[" inidcates magnitudes and "*" is the dot product)

Answer 1

looks a*b must be 0

Answer 2

That's what I was thinking by playing on geogebra, the magnitudes are equal when the two vector are perpendicular. But I'm not sure how to prove or confirm that.

Answer 3

[a+b] = sqrt{(a1+b1)^2+(a2+b2)^2+(a3+b3)^2}
[a+b] = sqrt{(a1)^2+(a2)^2+(a3)^2+2(a1b1+a2b2+a3b3)+(b1)^2+(b2)^2+(b3)^2}
[a+b] = sqrt([a]^2+[b]^2+2a*b)
[a-b] = sqrt{(a1-b1)^2+(a2-b2)^2+(a3-b3)^2}
[a-b]= sqrt{(a1)^2+(a2)^2+(a3)^2+2(a1b1+a2b2+a3b3)+(b1)^2+(b2)^2+(b3)^2}
= sqrt([a]^2+[b]^2-2a*b)
because [a+b] = [a-b], so
sqrt([a]^2+[b]^2+2a*b) = sqrt([a]^2+[b]^2-2a*b)

Answer 4

square to both sides, gives
[a]^2+[b]^2+2a*b = [a]^2+[b]^2-2a*b
solve for ab

Answer 5

4ab = 0
ab = 0/4 = 0

Answer 6

I don't get what you are doing in the second line of you first comment...

Answer 7

Forget it, I see you already expanded and simplified the expressions...

Answer 8

u just neeed to expand (a1+b1)^2 = ....

Answer 9

yea... ^^^

Answer 10

also (a1-b1)^2 = ....

Answer 11

ops... miss index of a and b

Answer 12

(a+b)^2 and (a-b)^2 i meant

Answer 13

But in the third line, how do you simplify it to [a]^2?

Answer 14

just remember that define of
[a] = sqrt{(a1)^2+(a2)^2+(a3)^2}, square to both sides gives
[a]^2 = (a1)^2+(a2)^2+(a3)^2

Answer 15

also
[b] = sqrt{(b1)^2+(b2)^2+(b3)^2}, square to both sides gives
[b]^2 = (b1)^2+(b2)^2+(b3)^2

Answer 16

and a*b = a1b1 + a2b2 + a3b3

Answer 17

does this make sense ?

Answer 18

just compare this
sqrt(4) = 2
square both sides, be
4 = 4

Answer 19

Yeah thanks! I had done the same in two dimensions, but was missing the jump from a1b1+a2b2=a*b-0. Anyway thanks for all the help.

Answer 20

Sorry, meant to say: a1b1+a2b2=a*b=0

Answer 21

yeah... np
very welcome