Multi-Variable Calculus: Cancellation of dot products

[Dembadon]

Homework Statement

In real-number multiplication, if uv₁ = uv₂ and u ≠ 0, then we can cancel the u and conclude that v₁ = v₂. Does the same rule hold for the dot product: If u • v₁ = u • v₂ and u ≠ 0, can you conclude that v₁ = v₂? Give reasons for your answer.

Homework Equations

The Attempt at a Solution

If we let u = k<u₁, u₂> with u₂ = 0 and scalar k, then the dot product of u with any other vector v = k<v₁, v₂> will simply be the component kv₁ because u₂ will make the ku₂kv₂ product always zero regardless of its value. Thus, v can be infinitely many different vectors and still have the same dot product with u.

 

[Pyrrhus]

The hint is can you compute an inverse of \vec{u} ? and thus multiply both by that inverse in order that both vectors v1 and v2 are equal.

 

[lanedance]

also a counter example might be good here, how about considering when
u • v1 = u • v2 = 0

what does this mean geometrically? using that it should be easy to find a counter example in 3D space

 

[vela]

If we let u = k<u₁, u₂> with u₂ = 0 and scalar k, then the dot product of u with any other vector v = k<v₁, v₂> will simply be the component kv₁ because u₂ will make the ku₂kv₂ product always zero regardless of its value. Thus, v can be infinitely many different vectors and still have the same dot product with u.

Just out of curiosity, why do you introduce the scalar k? Doesn't your argument work without the k's?

 

[Dembadon]

The hint is can you compute an inverse of \vec{u} ? and thus multiply both by that inverse in order that both vectors v1 and v2 are equal.

Interesting! I hadn't thought of that. I'll have to play around with it.

also a counter example might be good here, how about considering when
u • v1 = u • v2 = 0

what does this mean geometrically? using that it should be easy to find a counter example in 3D space

It would mean that both v₁ and v₂ are orthogonal to u, or that u is orthogonal to v₁ and v₂, right?

Just out of curiosity, why do you introduce the scalar k? Doesn't your argument work without the k's?

You're absolutely right. Thank you for pointing that out.

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Thanks for the input, everyone. I appreciate your time.