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I am asked to prove when if vectors $u \cdot v = u \cdot w$ then it must follow that $v = w$

If I cannot prove it, I am asked to bring a counter example.

If $u \cdot v = u \cdot w$ then that implies that

$ \{u_1 v_1 + u_2 v_2 + ... + u_n v_n \} = \{v_1 w_1 + ... + v_n w_n \}$ and I can't think of any number that when multiplied with another number, gives two solutions..

Have I sufficiently proven that for $u \cdot v =u \cdot w \therefore v = w$ ?

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  • $\begingroup$ (1,0,0).(0,1,0) = (1,0,0).(0,0,1) = 0 , but (0,1,0)$\ne$(0,0,1) $\endgroup$
    – Peter
    Jan 22 '14 at 20:08
  • $\begingroup$ Hint: The dot product is also equal to $\lvert u \rvert \lvert v \rvert\cos(\theta)$, where $\theta$ is the angle between $u$ and $v$. $\endgroup$ Jan 22 '14 at 20:09
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If $\vec{u} = \vec{0}$, then $\vec{v}$ doesn't necessarily equal $\vec{w}$.

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Hint: what can you say about the case where $\vec{u}\neq 0$, $\vec{v}$ is a non-zero vector orthogonal to $\vec{u}$, and $\vec{w}=-\vec{v}$?

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For example, let $v=(1,0)$ and $w=(0,1)$. Let $u=(1,1)$.

Or else, more trivially, let $v$ and $w$ be arbitrary, and let $u$ be the $0$ vector.

If we are looking for a non-zero $u$, then for any vectors $v$ and $w$ such that neither is a multiple of the other, we can find a non-zero $u$ such that $u\cdot v=u\cdot w$.

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  • $\begingroup$ Well it just states in the reals, so I guess it does NOT follow that v = w. Thank you. $\endgroup$
    – Paze
    Jan 22 '14 at 20:18
  • $\begingroup$ You are welcome. Definitely, from $u\cdot v=u\cdot w$ for a particular $u$, we cannot conclude that $v=w$. $\endgroup$ Jan 22 '14 at 20:31

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