#### Please solve RD Sharma class 12 chapter Vector or Cross Product exercise 24.1 question 5 maths textbook solution

Answer         : $\sqrt{504}$

Hint               : To solve this we multiply $\vec{b}$by 2 as given then magnitude formula.

Given            : $\vec{a}=4 \hat{\imath}+3 \hat{\jmath}+\hat{k}$

$\vec{b}=\hat{\imath}-2 \hat{k}$

Find $|2 \vec{b} \times \vec{a}|$

Solution       :

We need to find unit vector of b  ($\hat{b}$)

\begin{aligned} &\hat{b}=\frac{\vec{b}}{|\vec{b}|} \\\\ &\hat{b}=\frac{(\hat{i}-2 \hat{k})}{\sqrt{1^{2}+(-2)^{2}}} \\\\ &\hat{b}=\frac{1}{\sqrt{5}}(\hat{i}-2 \hat{k}) \end{aligned}

$\begin{array}{r} 2 \hat{b}=\frac{2}{\sqrt{5}}(\hat{\mathbf{i}}-2 \hat{k}) \\\\ 2 \vec{b} \times \vec{a}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \\\frac{2}{\sqrt{5}} & 0 & \frac{-4}{\sqrt{5}} \\\\ 4 & 3 & 1 \end{array}\right| \end{array}$

\begin{aligned} &=\hat{\mathrm{i}}\left[0.1-3 \cdot\left(\frac{-4}{\sqrt{5}}\right)\right]-\hat{\mathrm{j}}\left[\left(\frac{2}{\sqrt{5}}\right) 1-(4)\left(\frac{-4}{\sqrt{5}}\right)\right]+\mathrm{k}\left[\left(\frac{2}{\sqrt{5}}\right) 3-(4)(0)\right] \\\\ &2 \vec{b} \times \vec{a}=\frac{12}{\sqrt{5}} \hat{\mathrm{i}}-\frac{18}{\sqrt{5}} \hat{\mathrm{j}}+\frac{6}{\sqrt{5}} \hat{\mathrm{k}} \end{aligned}

\begin{aligned} |2 \vec{b} \times \vec{a}| &=\sqrt{\left(\frac{12}{\sqrt{5}}\right)^{2}+\left(\frac{-18}{\sqrt{5}}\right)^{2}+\left(\frac{6}{\sqrt{5}}\right)^{2}} \\\\ &=\sqrt{\frac{144}{5}+\frac{324}{5}+\frac{36}{5}} \\\\ |2 \vec{b} \times \vec{a}| &=\sqrt{\frac{504}{5}} \end{aligned}