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Given four vectors, $\vec{a},\vec{b},\vec{c},\vec{d}$, if you know $\vec{a}\cdot\vec{b}=0$ and $\vec{c}\cdot\vec{d}=0$ what can you say about $\vec{c}\cdot\vec{b}-\vec{a}\cdot\vec{d}$ if $|a|=|b|$ and $|c|=|d|$

There is nothing that I could attempt, or simplify. I guess, this reduces to zero. but I can't prove it.

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What are $p,q$? – Christopher A. Wong Feb 28 '13 at 22:40
sorry, that was typo. – user45099 Feb 28 '13 at 22:43
up vote 3 down vote accepted

I'm afraid there is nothing you can say.

Take $a,b$ orthogonal such that $\|a\|=\|b\|$.

Then set $$ c:=a-b\qquad d:=a+b. $$ You can check that $c$ and $d$ are orthogonal and $\|c\|=\|d\|$. Also $$ (a,d)=\|a\|^2=\|b\|^2=-(c,b). $$

Now set $$c:=a+b\qquad d:=a-b. $$ You have $c$ and $d$ are orthogonal, $\|c\|=\|d\|$, and $$ (a,d)=\|a\|^2=\|b\|^2=(c,b). $$

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It works in $R^2$ if the two ordered bases have opposite orientation. – Will Jagy Feb 28 '13 at 23:27
@WillJagy Yes, sure, thanks for the note. – 1015 Feb 28 '13 at 23:28

EDDITTT: In the first paragraph I write (d,c), in the second paragraph (c,d) which are different. These are often called ordered or oriented bases.


Even in $\mathbb R^2$ this does not have a single answer. Given the way you chose the letters, the order is critical: if $(d,c)$ is a rotation and scaling of $(a,b),$ then your expression is indeed $0.$

However, if $(c,d)$ is a rotation and scaling of $(a,b),$ you get anything between certain bounds. For example, let $a,d$ point in the same direction, so that $d$ is a positive multiple of $a.$ Then $c$ is a negative multiple of $b.$ In symbols, some $\lambda > 0,$ you get $d = \lambda a, \; \; c = - \lambda b.$ So $$ b \cdot c - a \cdot d = - \lambda b^2 - \lambda a^2 = -2 \lambda a^2, $$ where $a^2 = b^2$ is the squared length of either vector.

Take the previous paragraph and simply negate $c,d,$ now you get $2 \lambda a^2$

The key word here is ORIENTATION.

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I think you need "is not a rotation" in one of the clauses. – Rahul Feb 28 '13 at 23:29
@RahulNarain, no, in the first one I wrote (d,c), in the second (c,d). – Will Jagy Feb 28 '13 at 23:31
Sorry, I hadn't noticed that. Nevertheless, strictly speaking, scalings don't count as rotations... – Rahul Mar 4 '13 at 9:38

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