Question

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

Answer 1

Taking LHS:-

`{ | \overrightarrow{A} + \overrightarrow{B} | }^(2) - { | \overrightarrow{A} - \overrightarrow{B} | }^(2)`

`\boxed{ \mathsf{using \: \overrightarrow{a} . \overrightarrow{a} = ^(2) }}`

`\implies { | \overrightarrow{A} + \overrightarrow{B} | }^(2) = (\overrightarrow{A} + \overrightarrow{B})(\overrightarrow{A} + \overrightarrow{B})`

`\mathsf{using \: distributive \: property \: of \: vector \: multiplication}`

eqn 1:-

`\implies \mathsf{ \overrightarrow{A} \overrightarrow{A}+\overrightarrow{A} \overrightarrow{B} + \overrightarrow{B}\overrightarrow{A} + \overrightarrow{B}\overrightarrow{B}}`

same with the other one

`\boxed{ \mathsf{using \: \overrightarrow{a} . \overrightarrow{a} = a^(2) }}`

`{ | \overrightarrow{A} - \overrightarrow{B} | }^(2) = (\overrightarrow{A} - \overrightarrow{B})(\overrightarrow{A} - \overrightarrow{B})`

`\mathsf{using \: distributive \: property \: of \: vector \: multiplication}`

eqn. 2:-

`\implies \mathsf{ \overrightarrow{A} \overrightarrow{A} - \overrightarrow{A} \overrightarrow{B} - \overrightarrow{B}\overrightarrow{A} + \overrightarrow{B}\overrightarrow{B}}`

eqn. 1 - eqn. 2 :-

`\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})}`

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

`\implies \mathsf{2 \overrightarrow{A}\overrightarrow{B} + 2\overrightarrow{B}\overrightarrow{A}}`

`\boxed{ \mathsf{\overrightarrow{A} \overrightarrow{B}= \overrightarrow{B}\overrightarrow{A}} }`

`\implies \mathsf{ 4\overrightarrow{A}\overrightarrow{B} }`

LHS = RHS

Hence, Proved! =D