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  • Explain solution RD Sharma class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer question 9

Explain solution RD Sharma class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer question 9

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ANSWER:  \vec{a} \& \vec{b}  are perpendicular

GIVEN:  |\vec{a}+\vec{b}|=|\vec{a}|-|\vec{b}|

HINTS: Use the identity of  (a+b)^{2} \&(a-b)^{2}

SOLUTION:

\begin{aligned} &|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \\ & \end{aligned}

(|\vec{a}+\vec{b}|)^{2}=(|\vec{a}-\vec{b}|)^{2}\qquad[\text { Squaring both sides }]

\begin{aligned} &\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b}\qquad\left[\because(\mathrm{a}+\mathrm{b})^{2}=\mathrm{a}^{2}+b^{2}-2 a b \&(\mathrm{a}-\mathrm{b})^{2}=\mathrm{a}^{2}+b^{2}-2 a b\right] \\ & \end{aligned}

\Rightarrow 2 \vec{a} \cdot \vec{b}=-2 \vec{a} \cdot \vec{b} \\

\Rightarrow \vec{a} \cdot \vec{b}=-\vec{a} \cdot \vec{b} \\

\Rightarrow \vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{b}=0 \Rightarrow 2 \vec{a} \cdot \vec{b}=0

\therefore \vec{a} is perpendicular to  \vec{b}

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