# Vectors problem

• November 19th 2011, 09:16 AM
kandyfloss
Vectors problem
a and b are two position vectors of two distinct points A and B , neither of which is the origin. show that if (modulus) a+b = (modulus) a-b , then a is perpendicular to b using :
a. a vector algebraic method
b. a geometric argument

I don't quite understand what they mean by these^
I've drawn a quick diagram using word, see image below.
http://img14.imageshack.us/img14/1426/vectors.png
• November 19th 2011, 09:32 AM
Plato
Re: Vectors problem
Quote:

Originally Posted by kandyfloss
a and b are two position vectors of two distinct points A and B , neither of which is the origin. show that if (modulus) a+b = (modulus) a-b , then a is perpendicular to b using :
a. a vector algebraic method
b. a geometric argument

For part a), recall that $\|a\pm b\|^2=(a\pm b)\cdot (a\pm b)$.

For part b), what type of quadrilateral diagonals of equal lengths?
• November 19th 2011, 09:36 AM
kandyfloss
Re: Vectors problem
part a) oh yeah, i agree on that rule. but how does it help me explain a is perpendicular to b?
part b) parallelogram ?
• November 19th 2011, 09:47 AM
Plato
Re: Vectors problem
Quote:

Originally Posted by kandyfloss
part a) oh yeah, i agree on that rule. but how does it help me explain a is perpendicular to b?
part b) parallelogram ?

You are given that $\|a+b\|=\|a-b\|$ squaring we get
$(a+b)\cdot(a+b)=(a-b)\cdot(a-b)$.
Expand to see what happens.
• November 19th 2011, 09:50 AM
kandyfloss
Re: Vectors problem
Ok, thanks. and about the part b) since it's a parallelogram the diagonals intersect at 90 deg and adjacent sides are perpendicular to each other. right?