#### Need solution for RD sharma maths class 12 chapter 24 vector or cross product Exercise Multiple choice questions question 18 maths textbook solution

Answer: $12\sqrt{3}$

Given: $\left | \overrightarrow{a} \right |=8$,$\left | \overrightarrow{b} \right |=3$ & $\left | \overrightarrow{a}\times \overrightarrow{b} \right |=12$

Hint: Use result $\quad|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a} \cdot \vec{b}|^{2} \\$

Explanation: Here $\left | \overrightarrow{a} \right |=8$,$\left | \overrightarrow{b} \right |=3$ & $\left | \overrightarrow{a}\times \overrightarrow{b} \right |=12$

We know,

\begin{aligned} \quad|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a} \cdot \vec{b}|^{2} \end{aligned}

\begin{aligned} \\ \Rightarrow(12)^{2}=(8)^{2}(3)^{2}-(\vec{a} \cdot \vec{b})^{2}\end{aligned}

\begin{aligned} \\ \Rightarrow 144=64 \times 9-(\vec{a} \cdot \vec{b})^{2} \\ \end{aligned}

\begin{aligned} \Rightarrow(\vec{a} \cdot \vec{b})^{2}=576-144=432 \\ \\ \end{aligned}

\begin{aligned}\Rightarrow(\vec{a} \cdot \vec{b})^{2}=432 \\ \\ \end{aligned}

$\Rightarrow(\vec{a} \cdot \vec{b})=\sqrt{432}=12 \sqrt{3}$