cdj8498
Answered

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

Sagot :

Answer:

Taking LHS:-

[tex] { | \overrightarrow{A} + \overrightarrow{B} | }^{2} - { | \overrightarrow{A} - \overrightarrow{B} | }^{2}[/tex]

[tex] \boxed{ \mathsf{using \: \overrightarrow{a} . \overrightarrow{a} = { |a| }^{2} }}[/tex]

[tex] \implies { | \overrightarrow{A} + \overrightarrow{B} | }^{2} = (\overrightarrow{A} + \overrightarrow{B})(\overrightarrow{A} + \overrightarrow{B})[/tex]

[tex] \mathsf{using \: distributive \: property \: of \: vector \: multiplication}[/tex]

eqn 1:-

[tex] \implies \mathsf{ \overrightarrow{A} \overrightarrow{A}+\overrightarrow{A} \overrightarrow{B} + \overrightarrow{B}\overrightarrow{A} + \overrightarrow{B}\overrightarrow{B}}[/tex]

same with the other one

[tex] \boxed{ \mathsf{using \: \overrightarrow{a} . \overrightarrow{a} = { |a| }^{2} }}[/tex]

[tex] { | \overrightarrow{A} - \overrightarrow{B} | }^{2} = (\overrightarrow{A} - \overrightarrow{B})(\overrightarrow{A} - \overrightarrow{B})[/tex]

[tex] \mathsf{using \: distributive \: property \: of \: vector \: multiplication}[/tex]

eqn. 2:-

[tex] \implies \mathsf{ \overrightarrow{A} \overrightarrow{A} - \overrightarrow{A} \overrightarrow{B} - \overrightarrow{B}\overrightarrow{A} + \overrightarrow{B}\overrightarrow{B}}[/tex]

eqn. 1 - eqn. 2 :-

[tex] \implies \mathsf{ \overrightarrow{A} \overrightarrow{A}+\overrightarrow{A} \overrightarrow{B} + \overrightarrow{B}\overrightarrow{A} + \overrightarrow{B}\overrightarrow{B} - ( \overrightarrow{A} \overrightarrow{A} - \overrightarrow{A} \overrightarrow{B} - \overrightarrow{B}\overrightarrow{A} + \overrightarrow{B}\overrightarrow{B})}[/tex]

after distributing the minus sign inside the braces AA and BB get canceled, while the other two add up.

[tex] \implies \mathsf{2 \overrightarrow{A}\overrightarrow{B} + 2\overrightarrow{B}\overrightarrow{A}}[/tex]

[tex] \boxed{ \mathsf{\overrightarrow{A} \overrightarrow{B}= \overrightarrow{B}\overrightarrow{A}} }[/tex]

[tex] \implies \mathsf{ 4\overrightarrow{A}\overrightarrow{B} }[/tex]

LHS = RHS

Hence, Proved! =D

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