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Explain solution RD Sharma class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 4 maths

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Answer:

Option (c) write what is given in option c from book pdf

Hint:

If  \vec{a} \text { and } \vec{b} are unit vectors then |\vec{a}|=1 \text { and }|\vec{b}|=1

Given:

\vec{a} \text { and } \vec{b} are two unit vectors and \alpha is an angle between them

Solution:

According to the given question,

|\vec{a}|+|\vec{b}|=1

Squaring both side, we get

|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{b}|^{2}=1 \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]

As both are unit vectors

1+2|\vec{a}| \cdot|\vec{b}| \cos \alpha+1=1 \quad[\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta]

\begin{aligned} &2 \cos \alpha=1-2 \\\\ &\cos \alpha=\frac{-1}{2} \quad\left[\because \cos \left(\frac{-\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)\right] \end{aligned}

\alpha=\frac{2 \pi}{3}

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