Question

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

Answer 1

Answer:

Taking LHS:-

${ | \overrightarrow{A} + \overrightarrow{B} | }^{2} - { | \overrightarrow{A} - \overrightarrow{B} | }^{2}$

$\boxed{ \mathsf{using \: \overrightarrow{a} . \overrightarrow{a} = { |a| }^{2} }}$

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

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

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