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  • Provide solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 22

Provide solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 22

Answers (1)

best_answer

Answer:

Option (d) \frac{2 \pi}{3}<\theta<\pi

Hint:

You must know the formula of magnitude of vector =\sqrt{a^{2}+b^{2}+c^{2}}

Given:

\vec{a} \text { and } \vec{b} are unit vectors inclined at an angle \theta such that |\vec{a}+\vec{b}|<1

Solution:

\begin{aligned} &|\vec{a}+\vec{b}|<1 \\\\ &|\vec{a}+\vec{b}|^{2}<1 \end{aligned}

\begin{aligned} &|\vec{a}|^{2}+|\vec{b}|^{2}+2|\vec{a}| \cdot|\vec{b}| \cos \theta<1 \\\\ &1+1+2 \cos \theta<1 \\\\ &2(1+\cos \theta)<1 \end{aligned}

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

\left|\cos \frac{\theta}{2}\right|<\frac{1}{2}

We know \theta is always lies between \left [ -\pi ,\pi \right ]

\frac{2 \pi}{3}<\theta<\pi

Hence, option (d) is correct

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