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

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

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}

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