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Is (A vector +B vector) always perpendicular to (A vector - B vector)?

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    Have you tried it with $B = \vec 0$? – 2017-01-01

4 Answers 4

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No, of course not. Why do you think it would? Try almost any two random vectors in the basic spaces $\;\Bbb R^n\;$ , for example, with the usual inner product:

$$a:=\binom10,\,\,b:=\binom11\implies a+b=\binom21\;,\;\;a-b=\binom0{-1}$$

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Try to get resultant from the parallelogram law, and you will see that $(\vec A + \vec B)$ and $(\vec A - \vec B)$ are perpendicular only when their magnitude is equal.

Resultant of $(\vec A + \vec B)=\sqrt{a^2+b^2+2AB\cos\theta}$ and resultant of $(\vec A - \vec B)=\sqrt{a^2+b^2-2AB\cos\theta}$ where $\theta$ is the angle between $\vec A $ and $\vec B$.

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No.

Let $a, b$ be two vectors in a vector space equipped with an inner product, the necessary condition would be: $\left\|a\right\| = \left\|b\right\|$.

The derivation is

\begin{align} (a + b)(a - b) = a*a - b*b = \left\|a\right\| - \left\|b\right\| = 0 \end{align}

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They should only satisfy the following formula: $$(\vec A + \vec B) \cdot (\vec A-\vec B)=0$$

$$A^2-B^2 = 0 \Rightarrow A=\pm B$$ where $V$ is the magnitude of the vector $\vec v$.

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    You should make it clear that you are using $A$ to denote the magnitude of $\vec A$. I think the norm and absolute value notations are much more common. Plus, the fact that you include $\pm$ with no discussion of $A$ and $B$ being necessarily nonnegative, serves to further obscure this intent. – 2017-01-01