#### Explain solution RD Sharma class 12 chapter Vector or Cross Product exercise 24.1 question 36 maths

$\frac{2 \hat{i}-\hat{j}-\hat{k}}{\sqrt{6}},-\frac{3}{5} \hat{j}-\frac{4}{5} \hat{k}, \sqrt{404} \text { sq.units }$

Hint:

To solve this we use determinant method

Given:

\begin{aligned} &\text { Magnitude }=10 \sqrt{3} \\ &\vec{a}=\hat{i}+2 \hat{j}+\hat{k} \\ &\vec{b}=-\hat{i}+3 \hat{j}+4 \hat{k} \end{aligned}

Solution:

\begin{aligned} &\vec{a}=\hat{i}+2 \hat{j}+\hat{k} \\\\ &\vec{b}=-\hat{i}+3 \hat{j}+4 \hat{k} \\\\ &\vec{a}+\vec{b}=\vec{c}=4 \hat{i}-2 \hat{j}-2 \hat{k} \end{aligned}

\begin{aligned} &|\vec{a} \times \vec{b}|=\frac{4 \hat{i}-2 \hat{j}-2 \hat{k}}{\sqrt{(4)^{2}+(-2)^{2}+(-2)^{2}}} \\\\ &=\frac{4 \hat{i}-2 \hat{j}-2 \hat{k}}{\sqrt{16+4+4}}=\frac{4 \hat{i}-2 \hat{j}-2 \hat{k}}{\sqrt{24}} \end{aligned}

\begin{aligned} &=\frac{4 \hat{i}-2 \hat{i}-2 \hat{k}}{2 \sqrt{6}} \\\\ &=\frac{2 \hat{i}-j-\hat{k}}{\sqrt{6}} \end{aligned}

Similarly

\begin{aligned} &\vec{d}=\vec{a}-\vec{b}=0 \hat{i}-6 \hat{j}-8 \hat{k} \\\\ &|\vec{a}-\vec{b}|=\frac{-6 \hat{j}-8 \hat{k}}{\sqrt{(-6)^{2}+(-8)^{2}}} \\\\ &=\frac{-6 \hat{j}-8 \hat{k}}{\sqrt{36+64}} \end{aligned}

\begin{aligned} &=\frac{-6 \hat{j}-8 \hat{k}}{10} \\\\ &=\frac{-3}{5} \hat{j}-\frac{4}{5} \hat{k} \end{aligned}

Area of the parallelogram $=|\vec{a} \times \vec{b}|$

$=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & -5 \\ 2 & 2 & 3 \end{array}\right|$

\begin{aligned} &=|-2 \hat{i}-16 \hat{j}+12 \hat{k}| \\\\ &=\sqrt{(-2)^{2}+(-16)^{2}+(12)^{2}} \\\\ &=\sqrt{4+256+144} \\\\ &=\sqrt{404} \text { sq.units } \end{aligned}