Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Normal vector of a plane would be perpendicular to both the given lines and parallel to their cross product. Now ${\bar{\ell }}_1\times {\bar{\ell }}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 6\\ 1& 1& -1\end{array}\right|$ $=-9\hat{\mathrm{i}}+8\hat{\mathrm{j}}-\hat{\mathrm{k}}$ i.e. $\Rightarrow 9,-8,1$ are d.r. of normal to a plane Let $\bar{a}=\hat{i}+3\hat{j}+4\hat{k}$ and $\bar{n}=9\hat{i}-8\hat{j}+\hat{k}$ $\therefore \overline{\mathrm{r}}\cdot (9\hat{\mathrm{i}}-8\hat{\mathrm{j}}+\hat{\mathrm{k}})=9(1)+(-8)(3)+4\times 1=9-24+4$ $\overline{\mathrm{r}}\cdot (9\hat{\mathrm{i}}-8\hat{\mathrm{j}}+\hat{\mathrm{k}})=-11\Rightarrow 9\mathrm{x}-8\mathrm{y}+\mathrm{z}+11=0$ is the equation of plane.