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  • Please Solve RD Sharma Class 12 Chapter 23 Scalar or Dot Product Exercise Very Short Answer Question 41 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 23 Scalar or Dot Product Exercise Very Short Answer Question 41 Maths Textbook Solution.

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Answer:  \frac{2\pi}{3}

Given:  |\vec{a}|=|\vec{b}|=1 \text { and }|\vec{a}+\vec{b}|=1

Hint: Apply the identity  (a+b)^{2}

Solution:  |\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2 \cdot \vec{a} \cdot \vec{b}

\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]

\Rightarrow(1)^{2}=(1)^{2}+(1)^{2}+2 \cdot \vec{a} \cdot \vec{b}

[\because|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=1,|\overrightarrow{\mathrm{a}}|=1 \&|\overrightarrow{\mathrm{b}}|=1]

\begin{aligned} &\Rightarrow 1=2+2 \vec{a} \cdot \vec{b} \\ & \end{aligned}

\Rightarrow-1=2 \vec{a} \cdot \vec{b} \\

\Rightarrow \vec{a} \cdot \vec{b}=\frac{-1}{2}

Now, we know   \operatorname{Cos} \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}

\begin{aligned} &\Rightarrow|\vec{a}||\vec{b}| \operatorname{Cos} \theta=\frac{-1}{2}\qquad\left[\because \vec{a} \cdot \vec{b}=\frac{-1}{2}\right] \\ & \end{aligned}

\Rightarrow \operatorname{Cos} \theta=\frac{-1}{2}\qquad[\because|\vec{a}|=|\vec{b}|=1] \\

\Rightarrow \theta=\frac{2 \pi}{3} \qquad\left[\because \operatorname{Cos} \frac{2 \pi}{3}=\frac{-1}{2}\right]

 

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