If vector $a = b$ and $a = c$, then $b = c$? 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 $\vec{v} =  \vec{w}$.
At first i tried to write the equality representing components:
$$\vec{u} · \vec{v} = u_1·v_1+\ldots+u_n·v_n$$
$$\vec{u} · \vec{w} = u_1·w_1+\ldots+u_n·w_n$$
Then:
$$u_1·v_1+\ldots+u_n·v_n = u_1·w_1+\ldots+u_n·w_n.$$
But I got stuck right there; I thought if $c_i, i \in [1,n]$ could be canceled from both terms, but don't know if that is possible somehow.
 A: Let $\vec u=\vec v = \vec0$ and $\vec w\neq \vec0$ then $\vec u \cdot\vec v=\vec u\cdot\vec w =\vec0$.
More generally $\vec u \cdot\vec v=\vec u\cdot\vec w \implies \vec u\cdot(\vec v-\vec w)=0\implies\vec u$ perpendicular to $(\vec v-\vec w)$.
And the above example is true as $\vec0$ is perpendicular to every vector.
A: The answer is no if $\vec{u} \cdot (\vec{v}-\vec{w})=0$ which is possible even if $\vec{u}\neq 0$ that means let the $\vec{d}=\vec{v}-\vec{w}$ and assume angle between $\vec{u}$ and $\vec{d}$ 90°.So, it not necessary that $\vec{d}=\vec{0}$
Example let, $\vec{u}=\vec{i}+\vec{j}+\vec{k}$ and let $\vec{d}=-2\vec{i}+\vec{j}+\vec{k}$ Now there are many pairs with $a-b=-2$ ,$c-d=1$, $e-f=1$ where we set $\vec{v}=a\vec{i}+c\vec{j}+e\vec{k}$ and $\vec{u}=b\vec{i}+d\vec{j}+f\vec{k}$
Done.
A: You can rearrange and use associativity of the dot product:
$\vec u \cdot \vec v = \vec u \cdot \vec w \implies \vec u \cdot (\vec v - \vec w) = 0$
Which implies that:

*

*The vector (of nonzero norm) formed by the difference $\vec v - \vec w$ is orthogonal to $\vec u$ (also of nonzero norm)

OR


*$\vec u$ is a null vector

AND/OR


*$\vec v - \vec w$ is a null vector.

The final condition implies $\vec v = \vec w$, but it is clearly not the only possibility.
A: $$\vec u\cdot v=\vec u\cdot w\implies \vec u\cdot(\vec v-\vec w)=0.$$
This expresses that $\vec v-\vec w$ is orthogonal to $\vec u$, and not necessarily $\vec0$.
