# Simple problem on vector

1. Sep 5, 2008

### spandan

If |vector A + vector B| (magnitude of sum of vectors A and B) = |vector A + vector B| (magnitude of difference between vectors A and B), how can we show that theta is equal to pi/2?

I know that vector A - vector B can be equal to the sum of vectors A and B if vector B is a null vector, but with magnitudes, angles, I'm all confused.

Last edited: Sep 5, 2008
2. Sep 5, 2008

### jackiefrost

If you are allowed to use dot products you can use the fact the magnitude of a vector (such as the resultant vectors for the sum and difference of A and B) equals the square root of the dot product.

Provide the vectors with components:

A = <a1, a2>
B = <b1, b2>

So the magnitudes of the sum and diff resultant vectors would be

mag_sum = Sqrt[(a1+b1)^2 + (a2+b2)^2]
mag_diff = Sqrt[(a1-b1)^2 + (a2-b2)^2]

You can then use the fact that mag_sum = mag_diff and expand the squared terms, etc... and arrive at a very simple equation of the form,

X+Y = -(X+Y)

From there, [assuming you recognize (X+Y) ...] and using the definition of "dot product" as |A||B|*Cos(theta), it's easy to see what Cos(theta), and therefor theta itself, must be for the equality to be true.

Probably a little more involved than what's necessary... but hey - it works :rofl:

jf

3. Sep 5, 2008

### tiny-tim

Welcome to PF!

Hi spandan! Welcome to PF!

Yes it does! … but the joy of vectors is that you can often prove things without using coordinates …

in this case, |A + B|2 = (A + B).(A + B) (by definition of || ),

and |A - B|2 = (A - B).(A - B) …

so |A + B| = |A - B| if … ?

4. Sep 7, 2008

### spandan

No, how do we showw that theta is equal to pi/2?

Of course, pi = 3.14159..........

5. Sep 8, 2008

### tiny-tim

........265358979........

π/2 = 90º