Prove that $\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c}$ and $\vec{a}\times\vec{b} = \vec{a}\times\vec{c} \Rightarrow \vec{b} = \vec{c}$

When I manipulate the first expression I get $$\vec{a}\cdot(\vec{b}-\vec{c})=0$$ implying that $\vec{b}-\vec{c}$ is perpendicular to $\vec{a}$.

For the second expression, I have been told I should "left multiply (in the vector sense) by $\vec{a}$."

Eventually I should be left with an expression that implies that $\vec{b}-\vec{c}$ is parallel to $\vec{a}$, so that $\vec{b}-\vec{c}$ has to be the zero vector. I'm not sure how to manipulate the second expression after the multiplication to achieve this though?

• Do you know the vector identity $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u}\cdot\vec{w})\vec{v} - (\vec{u}\cdot\vec{v})\vec{w}$? What is $\vec{a} \times (\vec{a} \times \vec{x})$? Feb 28 '13 at 11:25
• A related question. Mar 1 '13 at 1:22

You can use the BAC-CAB rule. Since the vectors you're given are themselves $a, b, c$, I'll write the identity in terms of three vectors $f, g, h$ to avoid confusion:
$$f \times (g \times h) = g (f \cdot h) - h (f \cdot g)$$
Applying this identity to $a \times (a \times b)$ yields $a (a \cdot b) - b |a|^2$. You should be able to take it from here by using the same identity on $a, c$ instead.
• Yeah, basically. The way I looked at it, you can reduce the expression down to $|a|^2 b = |a|^2 c$ (since you know the dot products were equal, they can be eliminated), but it's all basically the same idea. Feb 28 '13 at 11:32