When Are Two Vectors Orthogonal in Vector Algebra?

[terryds]

Homework Statement

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

Homework Equations

u⋅v = |u||v| cos θ

The Attempt at a Solution

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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
Thanks in advance

 

[blue_leaf77]

It seems the answer is C

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.

 

[terryds]

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.

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

 

[terryds]

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.

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

 

[blue_leaf77]

In both

u ⋅ (u + v) = x ⋅ (u + v)

and

u ⋅ (u - x) = - v (u - x)

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.