# Orthogonal vector condition

1. May 2, 2016

### terryds

1. The problem statement, all variables and given/known data

Vector u, v, and x are not zero. Vector u + v will be perpendicular (orthogonal) to u-x if

A. |u+v| = |u-v|
B. |v| = |x|
C. u ⋅ u = v ⋅ v, v = -x
D. u ⋅ u = v ⋅ v, v = x
E. u ⋅ u = v ⋅ v

2. Relevant equations
u⋅v = |u||v| cos θ

3. The attempt at a solution

Two vectors are orthogonal to each other if the dot product is zero.

(u + v) ⋅ (u - x) = 0
(u ⋅ u) - (u ⋅ x) + (v ⋅ u) - (v ⋅ x) = 0
u ⋅ (u + v) - x ⋅ (u + v) = 0
u ⋅ (u + v) = x ⋅ (u + v)

u = x

or

u ⋅ (u - x) + v ⋅ ( u - x) = 0
u ⋅ (u - x) = - v (u - x)

so, u = -v

x = -v
v = -x

It seems the answer is C
But, how to get the condition u⋅u = v⋅v

2. May 2, 2016

### blue_leaf77

I don't think so.
Just do it by inspection on the equation (u ⋅ u) - (u ⋅ x) + (v ⋅ u) - (v ⋅ x) = 0. Try the answer choices one by one to find which one satisfies that equation.

3. May 2, 2016

### terryds

Okay.. Just by inspection.. I get D as the answer..
I thought too far and too much hahahaha...
Thank you

4. May 2, 2016

### terryds

Anyway, can you show me the error in my calculation before?
I'm curious where I get wrong

5. May 2, 2016

### blue_leaf77

In both
and
you are concluding that if $(a,b) = (c,b)$ for some vector(s) $b$, then $a=c$ - this conclusion is incorrect. It would have been true had the vector $b$ stands for any vectors in the space.