# I have three vectors: vec(u) , vec(v) and vec(w) and this equality: vec(u) * vec(v) =vec(u) * vec(w). Then the question is if vec(v) =vec(w).

Nathanial Levine 2022-10-01 Answered
I have three vectors: $\stackrel{\to }{u}$ ,$\stackrel{\to }{v}$ and $\stackrel{\to }{w}$ and this equality: $\stackrel{\to }{u}·\stackrel{\to }{v}=\stackrel{\to }{u}·\stackrel{\to }{w}$. Then the question is if $\stackrel{\to }{v}=\stackrel{\to }{w}$. At first i tried to write the equality representing components:
$\stackrel{\to }{u}·\stackrel{\to }{v}={u}_{1}·{v}_{1}+\dots +{u}_{n}·{v}_{n}$
$\stackrel{\to }{u}·\stackrel{\to }{w}={u}_{1}·{w}_{1}+\dots +{u}_{n}·{w}_{n}$
Then:
${u}_{1}·{v}_{1}+\dots +{u}_{n}·{v}_{n}={u}_{1}·{w}_{1}+\dots +{u}_{n}·{w}_{n}.$
But I got stuck right there; I thought if ${c}_{i},i\in \left[1,n\right]$ could be canceled from both terms, but don't know if that is possible somehow.
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Frida King
Let $\stackrel{\to }{u}=\stackrel{\to }{v}=\stackrel{\to }{0}$and $\stackrel{\to }{w}\ne \stackrel{\to }{0}$ then $\stackrel{\to }{u}\cdot \stackrel{\to }{v}=\stackrel{\to }{u}\cdot \stackrel{\to }{w}=\stackrel{\to }{0}$
More generally $\stackrel{\to }{u}\cdot \stackrel{\to }{v}=\stackrel{\to }{u}\cdot \stackrel{\to }{w}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\stackrel{\to }{u}\cdot \left(\stackrel{\to }{v}-\stackrel{\to }{w}\right)=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\stackrel{\to }{u}$ perpendicular to $\left(\stackrel{\to }{v}-\stackrel{\to }{w}\right)$
And the above example is true as $\stackrel{\to }{0}$ is perpendicular to every vector.
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st3he1d0t
You can rearrange and use associativity of the dot product:
$\stackrel{\to }{u}\cdot \stackrel{\to }{v}=\stackrel{\to }{u}\cdot \stackrel{\to }{w}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\stackrel{\to }{u}\cdot \left(\stackrel{\to }{v}-\stackrel{\to }{w}\right)=0$
Which implies that:
The vector (of nonzero norm) formed by the difference $\stackrel{\to }{v}-\stackrel{\to }{w}$ is orthogonal to $\stackrel{\to }{u}$ (also of nonzero norm)
Or
$\stackrel{\to }{u}$ is a null vector
AND/OR
$\stackrel{\to }{v}-\stackrel{\to }{w}$ is a null vector.
The final condition implies $\stackrel{\to }{v}=\stackrel{\to }{w}$, but it is clearly not the only possibility.