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hi .
please help to solve : Vector Dot Problem
if we know that $\displaystyle \vec{A}.\vec{B_1} = \vec{A}.\vec{B_2}$ can we infer from that : $\displaystyle \vec{B_1} = \vec{B_2}$ ?
thanks in advance ...
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hi .
please help to solve : Vector Dot Problem
if we know that $\displaystyle \vec{A}.\vec{B_1} = \vec{A}.\vec{B_2}$ can we infer from that : $\displaystyle \vec{B_1} = \vec{B_2}$ ?
thanks in advance ...
No, I don't think so...
$\displaystyle \left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right). \left(\begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right) = \left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right) . \left(\begin{array}{c} 4 \\ 0 \\ 2 \end{array} \right)$
But
$\displaystyle \left(\begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right) \neq \left(\begin{array}{c} 4 \\ 0 \\ 2 \end{array} \right)$
EDIT: I'm not sure of that... yet.
EDIT2: Confirmed!
No, assuming $\displaystyle \vec{A} \neq \vec{0}$.
Let $\displaystyle \vec{A} = <a_1,a_2>$, $\displaystyle \vec{B_1} = <u_1,u_2>$, and $\displaystyle \vec{B_2} = <v_1,v_2>$. Then, $\displaystyle \vec{A} \cdot \vec{B_1} = a_1u_1 + a_2u_2$ and $\displaystyle \vec{A} \cdot \vec{B_2} = a_1v_1 + a_2v_2$.
If $\displaystyle \vec{A} \cdot \vec{B_1} = \vec{A} \cdot \vec{B_2}$, then $\displaystyle a_1u_1 + a_2u_2 = a_1v_1 + a_2v_2$. The preceding equation cannot be solved; therefore, the answer is no.
I found a disproof on Wikipedia: Dot product - Wikipedia, the free encyclopedia
No, you are correct. My reasoning was flawed. Here is a disproof on Wikipedia: Dot product - Wikipedia, the free encyclopedia