#### Please solve RD Sharma class 12 chapter Straight Line in Space exercise 24.1 question 27 sub question (i) maths textbook solution

$\frac{5}{3}(32 \hat{i}-\hat{j}-14 \hat{k})$

Hint: To solve this we use determinant method

Given:

\begin{aligned} &\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k} \\\\ &\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k} \\\\ &\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k} \\\\ &\vec{c} \cdot \vec{d}=15 \end{aligned}

Solution:

D is perpendicular to a and b both. Hence, parallel to a*b

\begin{aligned} &\vec{a} \times \vec{b}=\vec{c} \\\\ &\vec{a} \times \vec{b}=\vec{d} \\\\ &\vec{d}=\lambda\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{array}\right| \end{aligned}

\begin{aligned} &\vec{d}=\lambda(32 \hat{i}-\hat{j}-14 \hat{k}) \\\\ &\vec{c} \cdot \vec{d}=15 \\\\ &\lambda(2 \hat{i}-\hat{j}+4 \hat{k})(32 \hat{i}-\hat{j}-14 \hat{k})=15 \end{aligned}

\begin{aligned} &\lambda(64+1-56)=15 \\\\ &9 \lambda=15 \\\\ &\lambda=\frac{5}{3} \\\\ &\vec{d}=\frac{5}{3}(32 \hat{i}-\hat{j}-14 \hat{k}) \end{aligned}