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Answer 1

Given Question

Prove that :

$\sf \: { |\vec{A} + \vec{B}| }^{2} - { |\vec{A} - \vec{B}| }^{2} = 4 \: \vec{A} \: . \: \vec{B}$

$\green{\large\underline{\sf{Solution-}}}$

Consider, LHS

$\rm :\longmapsto\: { |\vec{A} + \vec{B}| }^{2} - { |\vec{A} - \vec{B}| }^{2}$

We know,

$\rm :\longmapsto\:\boxed{\tt{ |\vec{x}| ^{2} = \vec{x}.\vec{x}}}$

So, using this, we get

$\rm \:  =  \: (\vec{A} + \vec{B}).(\vec{A} + \vec{B}) - (\vec{A} - \vec{B}).(\vec{A} - \vec{B})$

$\rm \:  =  \:[ \vec{A}.\vec{A} + \vec{A}.\vec{B} + \vec{B}.\vec{A} + \vec{B}.\vec{B}] - [\vec{A}.\vec{A} - \vec{A}.\vec{B} - \vec{B}.\vec{A} + \vec{B}.\vec{B}]$

$\rm \:  =  \: [ { |\vec{A}| }^{2} + \vec{A}.\vec{B} + \vec{A}.\vec{B} + { |\vec{B}| }^{2}] - [ { |\vec{A}| }^{2} - \vec{A}.\vec{B} - \vec{A}.\vec{B} + { |\vec{B}| }^{2}]$

$\red{ \bigg\{  \sf \: \because \: \vec{A}.\vec{B} = \vec{B}.\vec{A} \bigg\}}$

$\rm \:  =  \: [ { |\vec{A}| }^{2} + 2\vec{A}.\vec{B} + { |\vec{B}| }^{2}] - [ { |\vec{A}| }^{2} -2 \vec{A}.\vec{B} + { |\vec{B}| }^{2}]$

$\rm \:  =  \: { |\vec{A}| }^{2} + 2\vec{A}.\vec{B} + { |\vec{B}| }^{2}- [{ |\vec{A}| }^{2} + 2 \vec{A}.\vec{B} - { |\vec{B}| }^{2}$

$\rm \:  =  \: 4 \: \vec{A}.\vec{B}$

Hence,

$\sf \:\boxed{\tt{ \: \: { |\vec{A} + \vec{B}| }^{2} - { |\vec{A} - \vec{B}| }^{2} = 4 \: \vec{A} \: . \: \vec{B} \: \: }}$

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Additional Information

$\boxed{\tt{ \vec{A}.\vec{B} = \vec{B}.\vec{A}}}$

$\boxed{\tt{ \vec{A}.\vec{A} = { |\vec{A}| }^{2} }}$

$\boxed{\tt{ \vec{A} \times \vec{B} = - \vec{B} \times \vec{A}}}$

$\boxed{\tt{ \vec{A} \times \vec{A} = 0}}$

$\boxed{\tt{ \vec{A}.\vec{B} = 0 \: \rm\implies \:\vec{A} \: \perp \: \vec{B}}}$

$\boxed{\tt{ \vec{A} \times \vec{B} = 0 \: \rm\implies \:\vec{A} \: \parallel \: \vec{B}}}$