vector multiplication 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.
 A: 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
$$
(a,d)=\|a\|^2=\|b\|^2=(c,b).
$$
A: 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.
ORIGINAL: 
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.
