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  • Need solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 19

Need solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 19

Answers (1)

Answer:

Option (a)  2 \sin \frac{\theta}{2}

Hint:

Use the identity (a-b)^{2}=a^{2}-2 a b+b^{2}

Given:

\vec{a} \text { and } \vec{b} are unit vectors inclined at an angle \theta

Solution:

\begin{aligned} &|\vec{a}|=|\vec{b}|=1 \text { [unit vectors }] \\\\ &\vec{a} \cdot \vec{b}=|\vec{a}| \vec{b} \mid \cos \theta \\\\ &\vec{a} \cdot \vec{b}=\cos \theta \end{aligned}

Now,

\begin{aligned} &|\vec{a}-\vec{b}|^{2}=(\vec{a})^{2}+(\vec{b})^{2}-2 \vec{a} \cdot \vec{b} \\\\ &|\vec{a}-\vec{b}|^{2}=1-2 \cos \theta+1 \\\\ &|\vec{a}-\vec{b}|^{2}=2-2 \cos \theta \quad\left[\because \cos \theta=1-2 \sin ^{2} \frac{\theta}{2}\right] \end{aligned}

\begin{aligned} &|\vec{a}-\vec{b}|^{2}=2(1-\cos \theta) \\\ &|\vec{a}-\vec{b}|^{2}=2\left(2 \sin ^{2} \frac{\theta}{2}\right) \\\ &|\vec{a}-\vec{b}|^{2}=4 \sin ^{2} \frac{\theta}{2} \\\ &|\vec{a}-\vec{b}|=2 \sin \frac{\theta}{2} \end{aligned}

Hence, option (a) is correct

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