Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

(D) The vector perpendicular to both the given lines is given by $\left|\begin{array}{lll}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 4\\ 2& 2& 1\end{array}\right|=\hat{i}(-6)-\hat{j}(-7)+\hat{k}(-2)=-6\hat{i}+7\hat{j}-2\hat{k}$ Hence d.r.s. of required line are $6,-7,2$. Thus eq. of required line is $\frac{x-1}{6}=\frac{y-2}{-7}=\frac{2-3}{2}\text{ i.e. }\frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}$