Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

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

share|cite|improve this answer
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. $$(x\cdot z)(y\cdot z)=x^Tzy^Tz \,.$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.