1. ## Vector Dot Problem

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hi .

if we know that $\vec{A}.\vec{B_1} = \vec{A}.\vec{B_2}$ can we infer from that : $\vec{B_1} = \vec{B_2}$ ?

2. No, I don't think so...

$\left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right). \left(\begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right) = \left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right) . \left(\begin{array}{c} 4 \\ 0 \\ 2 \end{array} \right)$

But

$\left(\begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right) \neq \left(\begin{array}{c} 4 \\ 0 \\ 2 \end{array} \right)$

EDIT: I'm not sure of that... yet.

EDIT2: Confirmed!

3. No, assuming $\vec{A} \neq \vec{0}$.

Let $\vec{A} = $, $\vec{B_1} = $, and $\vec{B_2} = $. Then, $\vec{A} \cdot \vec{B_1} = a_1u_1 + a_2u_2$ and $\vec{A} \cdot \vec{B_2} = a_1v_1 + a_2v_2$.

If $\vec{A} \cdot \vec{B_1} = \vec{A} \cdot \vec{B_2}$, then $a_1u_1 + a_2u_2 = a_1v_1 + a_2v_2$. The preceding equation cannot be solved; therefore, the answer is no.

I found a disproof on Wikipedia: Dot product - Wikipedia, the free encyclopedia

4. Hm... maybe I'm wrong... or is my example an exception?

5. No, you are correct. My reasoning was flawed. Here is a disproof on Wikipedia: Dot product - Wikipedia, the free encyclopedia