#### Please solve RD Sharma class 12 chapter Straight Line in Space exercise 24.1 question 7 sub question (i) maths textbook solution

Answer          :  $42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k}$

Hint                : To solve this equation we solve $\vec{a} \times \vec{b} \text { then }|\vec{a} \times \vec{b}|$

Given              : $2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k} \text { and } 3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}$       magnitude=49

Solution        : $\frac{\vec{a} \times \vec{b}}{(\vec{a} \times \vec{b})}=49$

\begin{aligned} &\vec{a}=2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k} \\\\ &\vec{b}=3 \hat{\imath}-6 \hat{\jmath}+2 \hat{b} \\\\ &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2 \end{array}\right| \end{aligned}

\begin{aligned} &=\hat{\imath}(42)-\hat{\jmath}(-14)+\hat{k}(-2) \\\\ &=42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k} \\\\ &=7(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \end{aligned}

\begin{aligned} &|\vec{a} \times \vec{b}|=7 \sqrt{6^{2}+2^{2}+(-3)^{2}} \\\\ &=7 \sqrt{36+4+9} \\\\ &=7 \sqrt{49} \quad=49 \\\\ &42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k} \end{aligned}