# Is this a Valid proof? [Vectors]

• Nov 8th 2009, 08:31 AM
simpleas123
Is this a Valid proof? [Vectors]
Suppose that a, b ∈ ℝ3 and a × b = 0 Prove that there exist scalars λ, μ at least one of
which is non-zero, such that $\displaystyle \lambda a= \mu b.$

Suppose we have λ, μ such that μ does not equal zero. And that a and b are vectors.

Using λa = μb we can re-arrange to get (λ/μ)a = b

Plugging this into a x b = 0, we get;

a x ( a(λ/μ)) = 0, re-arranging we get.

(a x a) (λ/μ) = 0

((a x a)/μ) = 0

Am i right so far? / Can any 1 show me the proof ?
• Nov 8th 2009, 08:50 AM
Plato
Quote:

Originally Posted by simpleas123
Suppose that a, b ∈ ℝ3 and a × b = 0 Prove that there exist scalars λ, μ at least one of
which is non-zero, such that $\displaystyle \lambda a= \mu b.$

I don't see where you are going with that.

If either $\displaystyle a \text{ or }b$ is the zero vector you are done.
So suppose neither is the zero vector.
If $\displaystyle \phi$ is the angle between them then $\displaystyle \sin (\phi ) = \frac{{\left\| {a \times b} \right\|}}{{\left\| a \right\|\left\| b \right\|}}$.
From the given, what does that tell you about $\displaystyle \phi?$
• Nov 8th 2009, 09:27 AM
simpleas123
Quote:

Originally Posted by Plato
I don't see where you are going with that.

If either $\displaystyle a \text{ or }b$ is the zero vector you are done.
So suppose neither is the zero vector.
If $\displaystyle \phi$ is the angle between them then $\displaystyle \sin (\phi ) = \frac{{\left\| {a \times b} \right\|}}{{\left\| a \right\|\left\| b \right\|}}$.
From the given, what does that tell you about $\displaystyle \phi?$

But dont we have to prove that either Lander or Mew is zero?
And it does it mean Theta = 180?
• Nov 8th 2009, 09:44 AM
Plato
Quote:

Originally Posted by simpleas123
But dont we have to prove that either Lander or Mew is zero?
And it does it mean Theta = 180?

Actually it means that $\displaystyle \phi=0\text{ or }\phi=\pi$.
In turn that means that $\displaystyle a~||~b$ or one is a multiple of the other.
Just recall that neither is the zero vector.
• Nov 8th 2009, 11:14 AM
simpleas123
Quote:

Originally Posted by Plato
Actually it means that $\displaystyle \phi=0\text{ or }\phi=\pi$.
In turn that means that $\displaystyle a~||~b$ or one is a multiple of the other.
Just recall that neither is the zero vector.

Right i sort of understand it, so how would i write out the full proof?