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ExtremeBDS [4]
3 years ago
7

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

Mathematics
1 answer:
Inessa05 [86]3 years ago
4 0
<h3><u>Given </u><u>Question:</u><u>-</u></h3>

Prove that :

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

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

<u>Consider, LHS</u>

\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}

<u>Hence, </u>

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h3><u>Additional Information</u></h3>

\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}}}

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