SalmonPanna2022
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Show that: |A⃗ + B⃗ |² - |A⃗ - B⃗ |² = 4 A⃗.B⃗ .​

Show That A B A B 4 AB class=

Sagot :

SalmonWorldTopper

Given Question:-

Prove that :

[tex] \sf \: { |\vec{A} + \vec{B}| }^{2} - { |\vec{A} - \vec{B}| }^{2} = 4 \: \vec{A} \: . \: \vec{B}[/tex]

[tex] \green{\large\underline{\sf{Solution-}}}[/tex]

Consider, LHS

[tex]\rm :\longmapsto\: { |\vec{A} + \vec{B}| }^{2} - { |\vec{A} - \vec{B}| }^{2} [/tex]

We know,

[tex]\rm :\longmapsto\:\boxed{\tt{ |\vec{x}| ^{2} = \vec{x}.\vec{x}}}[/tex]

So, using this, we get

[tex]\rm \:  =  \: (\vec{A} + \vec{B}).(\vec{A} + \vec{B}) - (\vec{A} - \vec{B}).(\vec{A} - \vec{B})[/tex]

[tex]\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}][/tex]

[tex]\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}][/tex]

[tex]\red{ \bigg\{  \sf \: \because \: \vec{A}.\vec{B} = \vec{B}.\vec{A} \bigg\}}[/tex]

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

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

[tex]\rm \:  =  \: 4 \: \vec{A}.\vec{B}[/tex]

Hence,

[tex] \sf \:\boxed{\tt{ \: \: { |\vec{A} + \vec{B}| }^{2} - { |\vec{A} - \vec{B}| }^{2} = 4 \: \vec{A} \: . \: \vec{B} \: \: }}[/tex]

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Additional Information

[tex]\boxed{\tt{ \vec{A}.\vec{B} = \vec{B}.\vec{A}}}[/tex]

[tex]\boxed{\tt{ \vec{A}.\vec{A} = { |\vec{A}| }^{2} }}[/tex]

[tex]\boxed{\tt{ \vec{A} \times \vec{B} = - \vec{B} \times \vec{A}}}[/tex]

[tex]\boxed{\tt{ \vec{A} \times \vec{A} = 0}}[/tex]

[tex]\boxed{\tt{ \vec{A}.\vec{B} = 0 \: \rm\implies \:\vec{A} \: \perp \: \vec{B}}}[/tex]

[tex]\boxed{\tt{ \vec{A} \times \vec{B} = 0 \: \rm\implies \:\vec{A} \: \parallel \: \vec{B}}}[/tex]

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