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  • Explain solution RD Sharma class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer question 8

Explain solution RD Sharma class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer question 8

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

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

HINTS:Use algebraic identities and formulas of cosθ

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

\begin{aligned} & &\Rightarrow 2 \vec{a} \cdot \vec{b}=2|\vec{a}||\vec{b}| \\ & \end{aligned}

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

Now we know that  \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}

\begin{aligned} &\Rightarrow \cos \theta=\frac{|\vec{a}||\vec{b}|}{|\vec{a}||\vec{b}|} \qquad[\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|] \\ & \end{aligned}

\Rightarrow \cos \theta=1 \quad \Rightarrow \theta=0 \qquad[\because \cos \theta=1]

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

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