Prove these 2 vectors are perpendicular

[fred2028]

I guess this is the most appropriate forum since no vectors forum exists. Anyways, a test question was:

Given |a+b|=|a-b|, and the fact that a and b are both unit vectors, show that a and b are perpendicular using the dot product only.

Here's what I did.

|a+b|=|a-b|
|a+b|^2=|a-b|^2
(a+b).(a+b)=(a-b).(a-b)
a.a+2a.b+b.b=a.a-2a.b+b.b
2a.b=-2a.b
a.b=-a.b

Only 0 works for the above since no # is equal to its negative.

a.b=0
|a||b|cos(theta) = 0

Since a and b are unit vectors, neither equal 0.

cos(theta)=0
theta = 90 degrees, or perpendicular.

Would you say that this is a good answer? I asked my dad and he said that instead of saying a.b=-a.b, I should've continued and did

a.b=-a.b
2a.b=0
a.b=0

Do you think I'd lose a mark there? Thanks!

 

[royhaas]

Either way you have proved it.