How to show that-

Question

$[|\vec{A} × \vec{B}|^2+|\vec{A} \bullet \vec{B}|^2]=[|\vec{A}|^2 |\vec{B}|^2]$ I have tried to do this by putting components of both vectors but couldn't able to.

Answer 1

$|u\times v|=|u||v|\sin\angle (u,v) $

$|u. v|=|u||v||\cos\angle (u,v)| $

Answer 2

Hint

$$a\cdot b=|a|\cdot |b|\cdot \cos\theta$$also prove that $$|a\times b|=|a|\cdot |b|\cdot |\sin\theta|$$

Answer 3

. It's actually easy one I just considered it too taough. All we have to do is just to put components.

Answer 4

Here is a different approach which you could also concider to use the following identity:

$$ \bbox[5px,border:2px solid #000000]{(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d}) = (\vec a\cdot\vec c)(\vec b\cdot\vec d) - (\vec a\cdot\vec d)(\vec b\cdot\vec c). } $$

From this point it is straightforward:

$$\begin{align} \left|\vec a\times\vec b\right|^2+\left(\vec a\cdot\vec b\right)^2 &= \left(\vec a\times\vec b\right)\cdot\left(\vec a\times\vec b\right) + \left(\vec a\cdot\vec b\right)^2\\ &= (\vec a\cdot\vec a)(\vec b\cdot\vec b) - (\vec a\cdot\vec b)(\vec a\cdot\vec b)+ \left(\vec a\cdot\vec b\right)^2\\ &= \left|\vec a\right|^2\;\left|\vec b\right|^2 \end{align}$$

When it boils down to vector identities, avoid as much as possible to write down its components. Generally, you do not need them.