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  • Provide solution for RD Sharma Maths Class 12 Chapter 24 Vector or Cross Product exercise Very Short Answer Question, question 11.

Provide solution for RD Sharma Maths Class 12 Chapter 24 Vector or Cross Product exercise Very Short Answer Question, question 11.

Answers (1)

ANSWER:

  \theta = \frac{\pi}{3}

HINT:

Just apply the formula of cross product and dot product.

GIVEN:

  \begin{gathered} \vec{a} \times \vec{b}=\sqrt{3} \\ \vec{a} \cdot \vec{b}=1 \end{gathered}

SOLUTION:    

\begin{aligned} \vec{a} \times \vec{b} &=\sqrt{3} \\ \vec{a} \cdot \vec{b} &=1 \\ \text { Using }|\vec{a} \times \vec{b}| &=|\vec{a}||\vec{b}| \sin \theta \\ \sqrt{3} &=|\vec{a}||\vec{b}| \operatorname{sin} \theta \\ \therefore \frac{\sqrt{3}}{\operatorname{sin} \theta} &=|\vec{a}||\vec{b}| \ldots \ldots .....(1) \end{aligned}

\begin{aligned} \text { Using } \vec{a} \cdot \vec{b} &=|\vec{a}||\vec{b}| \cos \theta \\ 1 &=|\vec{a}||\vec{b}| \cos \theta \\ \therefore \frac{1}{\operatorname{cos} \theta} &=|\vec{a}||\vec{b}| \ldots \ldots ......(2) \end{aligned}

From (1) and (2)

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

\therefore \theta =\frac{\pi}{3}

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