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  • Explain solution RD sharma class 12 chapter 24 Vector or cross product exercise Fill in the blanks question 11

Explain solution RD sharma class 12 chapter 24 Vector or cross product exercise Fill in the blanks question 11

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best_answer

\therefore \theta=\frac{\pi}{6}

Hint: Use formula of dot product &cross product & then compare

Given:

\begin{aligned} &|\vec{a} \times \vec{b}|=1 \\ &|\vec{a} \cdot \vec{b}|=\sqrt{3} \end{aligned}

If

\begin{aligned} &|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta \\ &1=|\vec{a}||\vec{b}| \sin \theta \\ &\frac{1}{\sin \theta}=|\vec{a}||\vec{b}| \ldots \ldots .(1) \\ \end{aligned}

\begin{aligned} &|\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}| \cos \theta \\ &\sqrt{3}=|\vec{a}||\vec{b}| \cos \theta \\ &\frac{\sqrt{3}}{\cos \theta}=|\vec{a}||\vec{b}| \ldots \ldots . .(2) \end{aligned}

From (1) &(2)

\begin{aligned} &\frac{1}{\sin \theta}=\frac{\sqrt{3}}{\cos \theta} \\ &\frac{1}{\sqrt{3}}=\frac{\sin \theta}{\cos \theta} \\ &\frac{1}{\sqrt{3}}=\tan \theta \end{aligned}

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