Prove that ( a + b ) ( a + b )= | a |^2+ | b |^2, if and only if a , b are perpendicular, given a 0 , b 0 .

( a + b ) ( a + b )= | a |^2+ | b |^2 a a + a b + b a + b b = | a |^2+ | b |^2 [Distributivity of scalar products over addition] | a |^2+2 a b + | b |^2= | a |^2+ | b |^2 [ a b = b a (Scalar product is commutative) ] 2 a b =0 a b =0 a and b are perpendicular. [ a 0 , b 0 (Given) ]

Prove that ( a + b ) ( a + b )= | a |^2+ | b |^2, if and only if …

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Question

Prove that $\big(\vec{\text{a}}+\vec{\text{b}}\big)\cdot\big(\vec{\text{a}}+\vec{\text{b}}\big)=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2,$ if and only if $\vec{\text{a}},\vec{\text{b}}$ are perpendicular, given $\vec{\text{a}}\neq\vec{\text{0}},\vec{\text{b}}\neq\vec{\text{0}}.$

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Answer

$\Big(\vec{\text{a}}+\vec{\text{b}}\Big)\cdot\Big(\vec{\text{a}}+\vec{\text{b}}\Big)=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2$
$\Leftrightarrow\vec{\text{a}}\cdot\vec{\text{a}}+\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{a}}+\vec{\text{b}}\cdot\vec{\text{b}}=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2$ [Distributivity of scalar products over addition]
$\Leftrightarrow\big|\vec{\text{a}}\big|^2+2\vec{\text{a}}\cdot\vec{\text{b}}+\Big|\vec{\text{b}}\Big|^2=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2\ \ \ $ $\Big[\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{b}}\cdot\vec{\text{a}}\ \text{(Scalar product is commutative)}\Big]$
$\Leftrightarrow2\vec{\text{a}}\cdot\vec{\text{b}}=0$
$\Leftrightarrow\vec{\text{a}}\cdot\vec{\text{b}}=0$
$\therefore\vec{\text{a}}\ \text{and}\ \vec{\text{b}} \ \text{are perpendicular.}$ $\ \ \ \Big[\vec{\text{a}}\neq\vec{\text{0}},\ \vec{\text{b}}\neq\vec{\text{0}}\ \text{(Given)}\Big]$