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Given that $\vec{x}$, $\vec{y}$ & $\vec{z}$ are three arbitrary vectors in $\mathbb{R}^3$, is there a concise way to express $ \left(\vec{x}\cdot\vec{z}\right)\left(\vec{y}\cdot\vec{z}\right) $ in terms of $\vec{z}$, $\left(\vec{x}\times\vec{y}\right)$ and maybe $\left(\vec{x}\cdot\vec{y}\right)$ only?

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What you are asking for is impossible in general. There is no way at all, never mind concise. –  Andrey Sokolov Feb 27 '13 at 13:21

2 Answers 2

$$ (y\cdot z)(x\cdot z) = (x\cdot y)\,||z||^2 -\ \left( (x\times z) \cdot (y\times z) \right) $$ This is obtained by expansion of the third term.

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Correct, but not the answer I was looking for -- it is in terms of $x \times z$ & $y \times z$, not $x times y$. –  Avijit Feb 27 '13 at 14:45

$(x\times y)\times z=(y\cdot z)x-(x\cdot z)y\\ x\times(y\times z)=(x\cdot y)z-(x\cdot z)y$

Subtracting these we get $(y\cdot z)x-(x\cdot y)z$, and scalar multiplying by $z$, as $z\perp (x\times y)\times z$, it yields $$(y\cdot z)(x\cdot z) = -\ \left( (x\times (y\times z))\,\cdot z\right) \ + (x\cdot y)\,||z||^2$$

Well, this is not really nice. Probably with matrix multiplication, it is more useful, as $(x\cdot y)=x^Ty$ where vectors are regarded as culomn matrices. $$(x\cdot z)(y\cdot z)=x^Tzy^Tz \,.$$

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Hi @Berci -- I was looking for an answer in terms of $\vec{x} \times \vec{y}$ ... what you've said is correct, but it gives the result in terms of $\vec{y} \times \vec{z}$ ... –  Avijit Feb 27 '13 at 12:05
    
I guess, there is no such an expression that uses only $z$ and $x\times y$ and $x\cdot y$. –  Berci Feb 27 '13 at 16:53

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