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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 $\vec{a} \neq \overrightarrow{0}, \vec{b} \neq \overrightarrow{0}$.
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Verified Answer
$$
\begin{aligned}
&(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=\vec{a} \cdot(\vec{a}+\vec{b})+\vec{b} \cdot(\vec{a}+\vec{b}) \\
&=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b} \\
&=|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b} \quad[\because \vec{b} \cdot \vec{a}=\vec{a} \cdot \vec{b}]
\end{aligned}
$$
(i) let $\vec{a} \perp \vec{b}=\vec{a} \cdot \vec{b}=0$
$$
\begin{aligned}
&\therefore(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b}) \\
&=|\vec{a}|^2+|\vec{b}|^2 \quad[\because \vec{a} \cdot \vec{b}=0]
\end{aligned}
$$
(ii) Again if $\vec{a}$ is not perpendicular to $\vec{b}$
$$
\Rightarrow \quad \vec{a} \cdot \vec{b} \neq 0
$$
$\operatorname{Now}(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}$
$\neq|\vec{a}|^2+|\vec{b}|^2$ unless $\vec{a} \cdot \vec{b}=0$,
Hence $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2$
when $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0 \quad$ or vector $\overrightarrow{\mathrm{a}}$ is $\perp^{\mathrm{r}}$ to $\overrightarrow{\mathrm{b}}$.
\begin{aligned}
&(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=\vec{a} \cdot(\vec{a}+\vec{b})+\vec{b} \cdot(\vec{a}+\vec{b}) \\
&=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b} \\
&=|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b} \quad[\because \vec{b} \cdot \vec{a}=\vec{a} \cdot \vec{b}]
\end{aligned}
$$
(i) let $\vec{a} \perp \vec{b}=\vec{a} \cdot \vec{b}=0$
$$
\begin{aligned}
&\therefore(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b}) \\
&=|\vec{a}|^2+|\vec{b}|^2 \quad[\because \vec{a} \cdot \vec{b}=0]
\end{aligned}
$$
(ii) Again if $\vec{a}$ is not perpendicular to $\vec{b}$
$$
\Rightarrow \quad \vec{a} \cdot \vec{b} \neq 0
$$
$\operatorname{Now}(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}$
$\neq|\vec{a}|^2+|\vec{b}|^2$ unless $\vec{a} \cdot \vec{b}=0$,
Hence $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2$
when $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0 \quad$ or vector $\overrightarrow{\mathrm{a}}$ is $\perp^{\mathrm{r}}$ to $\overrightarrow{\mathrm{b}}$.
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