Show that B|A|+ A|B| and A|B|−B|A| are orthogonal Where A and B are vectors, I personally am not really sure what this question is even really asking, maybe I'm interpreting it wrong. I know how to take dot products and cross products but is that even involved here?
 A: Yeah you should use the dot product here to calculate $$\left(B|A|+A|B|\right)\cdot\left(A|B|-B|A|\right)$$ and you want to show that this is equal to zero. Remember that you can distribute this dot-product and pull out scalars, and that $A\cdot A=|A|^2$.
A: Geometrical proof of the orthogonality of these vectors :
$\|A\|B$ and $\|B\|A$ have the same norm ($\|A\|.\|B\|$).
Thus, their sum $\|A\|B +\|B\|A$ is the diagonal of a rhombus, which is orthogonal to the other diagonal, which can be expressed as the difference $ \|B\|A - \|A\|B$.
In other words, we deal with the internal and external line bissectors of the axes defined by $A$ and $B$ resp. 

Remark : we would have had a similar result by adding and subtracting the norm-one vectors $\frac{A}{\|A\|} $ and $\frac{B}{\|B\|}.$
A: Two vectors are orthogal if their inner product is zero
$(A\|B\| + \|A\|B)\cdot (A\|B\| - \|A\|B) = A\cdot A \|B\|^2 + \|A\|\|B\|A\cdot B - \|A\|\|B\| A\cdot B -\|A\|^2B\cdot B$
$\|A\|^2 = A\cdot A$ 
And as you see everything cancels.
Geometrically $(A\|B\| + \|A\|B)$ and $(A\|B\| - \|A\|B)$ are the diagonals of a rhombus.
