#### Need solution for RD Sharma maths class 12 chapter Vector or Cross Product exercise 24.1 question 19

Answer        : $\frac{1}{\sqrt{165}}(10 \hat{\imath}+7 \hat{\jmath}-6 \hat{k})$

Hint             : To solve this equation , we use determination method

Given          : $A=(3,-1,2)$

\begin{aligned} &B=(1,-1,-3) \\\\ &C=(4,-3,1) \end{aligned}

Solution     : $\overrightarrow{A B}=-2 \hat{\imath}-5 \hat{k}$

\begin{aligned} &\overrightarrow{A C}=\hat{\imath}-2 \hat{\jmath}-\hat{k} \\\\ &\vec{C}=x \overrightarrow{A B}+y \overrightarrow{A C} \end{aligned}

If $\vec{a}$ is perpendicular to $\vec{c}$

$\vec{d}$ is perpendicular to $\overrightarrow{AB}$ ,$\vec{d}$ perpendicular to $\overrightarrow{AC}$

\begin{aligned} &\vec{d}=\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{array}\right| \\\\ &=-10 \hat{\imath}-7 \hat{\jmath}+4 \hat{k} \\\\ &\hat{d}=\frac{\vec{d}}{|\vec{d}|}=\frac{-10 \hat{\imath}-7 \hat{\jmath}+4 \hat{k}}{\sqrt{165}} \end{aligned}