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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra2 Marks
Question
Prove that $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$, if and only if $\vec{a}, \vec{b}$ are perpendicular, given $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$.
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Answer

Given: $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$
To prove: vectors $\vec{a}$ and $\vec b$ are mutually perpendicular to each other.
$\Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=0$
$\because(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$
$\Rightarrow(\vec{\mathrm{a}} \cdot \vec{\mathrm{a}}+\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{a}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{b}})=|\vec{\mathrm{a}}|^{2}+|\vec{\mathrm{b}}|^{2}$
$(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a})$ scalar product is commutative.
$\Rightarrow\left(|\vec{\mathrm{a}}|^{2}+|\vec{\mathrm{b}}|^{2}+2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}\right)=|\vec{\mathrm{a}}|^{2}+|\vec{\mathrm{b}}|^{2}$
$\Rightarrow(2 . \vec{\mathrm{a}} . \vec{\mathrm{b}})=0$
$\Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=0$
Hence, $\vec a$ and $\vec b$ are mutually perpendicular to each other. as $\vec{a} \neq 0$ and $\overrightarrow{\mathrm{b}} \neq 0$ is given.

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