#### Please solve RD Sharma Class 12 Chapter 24 Vector or Cross Product Exercise Very Short Answer Question, question 9 Maths textbook solution.

$\pm 12$

HINT:

Using trigonometric identity  $\sin ^2 \theta + \cos^2 \theta =1$

GIVEN:

\begin{aligned} &\text { ie, } \begin{aligned} |\vec{a}| &=10 \\ |\vec{b}| &=2 \\ \vec{a} \times \vec{b} &=16 \end{aligned} \end{aligned}

SOLUTION:

\begin{aligned} &\text { ie, } \begin{aligned} |\vec{a}| &=10 \\ |\vec{b}| &=2 \\ \vec{a} \times \vec{b} &=16 \end{aligned} \end{aligned}

we know,

$|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \operatorname{\sin} \theta$

Using trignometric identities

$\sin ^2 \theta + \cos^2 \theta =1$

$\left(\frac{4}{5}\right)^{2}+\operatorname{Cos}^{2} \theta=1$

\begin{aligned} &\therefore \operatorname{Cos}^{2} \theta=1-\frac{16}{25} \\ &\therefore \operatorname{Cos}^{2} \theta=\frac{9}{25} \\ &\therefore \operatorname{Cos} \theta=\pm \frac{3}{5} \end{aligned}

We know,

\begin{aligned} \vec{a} \cdot \vec{b} &=|\vec{a}| \cdot|\vec{b}| \operatorname{Cos} \theta \\ &=10(2)\left(\pm \frac{3}{5}\right) \end{aligned}

$=\pm 12$