Prove 2 vectors are perpendicular

[RM5152]

Hello. I’m stuck on this question (6). I know that for 2 vectors to be perpendicular their product must equal zero and the angle between them is 90 degrees. But how can I prove this without having values and only lengths? Can anyone help? Thanks ?

 

[Deleted member 4993]

I know that for 2 vectors to be perpendicular their dot-product must equal zero

A small but important addition

Hint: Dot product can be distributed over addition or subtraction.

 

[topsquark]

Hello. I’m stuck on this question (6). I know that for 2 vectors to be perpendicular their product must equal zero and the angle between them is 90 degrees. But how can I prove this without having values and only lengths? Can anyone help? Thanks ? View attachment 33176View attachment 33177

Hint: If [imath]|| \textbf{x} + \textbf{y} || = || \textbf{x} - \textbf{y} ||[/imath] then [imath]\sqrt{ (\textbf{x} + \textbf{y}) \cdot (\textbf{x} + \textbf{y}) } = \sqrt{ (\textbf{x} - \textbf{y}) \cdot (\textbf{x} - \textbf{y}) }[/imath].

-Dan

 

[pka]

Do you understand(know) that [imath]\bf\|A\|^2=A\cdot A[/imath]
From which it follows that [imath]\bf (A+B)\cdot(A+B)=(A-B)\cdot(A-B)[/imath]
Can you see that [imath]\bf 4 A\cdot B=0[/imath]. How & Why?
What does that imply?
[imath][/imath]

 

[Steven G]

Have you drawn the two vectors, x+y and x-y, to see what is going on?
Doing what has been suggested above is fine, but if you don't know what is going on, then draw the two vectors!