Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The equation of line passing through $(1,2$, 3) and parallel to $b$ is given by $r=(\hat{i}+2\hat{j}+3\hat{k})+\lambda \left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)$ ...(i) The equations of the given planes are $\begin{array}{rl}& r\cdot (\hat{i}-\hat{j}+2\hat{k})=5...(ii)\\ & \text{ and }r\cdot (3\hat{i}+\hat{j}+\hat{k})=6...(iii)\end{array}$ The line in eq. (i) and plane in eq. (ii) are parallel. Therefore, the normal to the plane of eq. (ii) and the given line are perpendicular. $\begin{array}{rl}& \therefore (\hat{i}-\hat{j}+2\hat{k})\cdot \lambda \left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)=0\\ & \Rightarrow \lambda \left(b_1-b_2+2b_3\right)=0\\ & \Rightarrow \left(b_1-b_2+2b_3\right)=0...(iv)\end{array}$ Similarly, $\begin{array}{rl}& (3\hat{i}+\hat{j}+\hat{k})\cdot \lambda \left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)=0\\ & \Rightarrow \lambda \left(3b_1+b_2+b_3\right)=0...(v)\end{array}$ From eqs. (iv) and (v), we obtain $\begin{array}{rl}& \frac{b_1}{(-1)\times 1-1\times 2}=\frac{b_2}{2\times 3-1\times 1}=\frac{b_3}{1\times 1-3(-1)}\\ & \Rightarrow \frac{b_1}{-3}=\frac{b_2}{5}=\frac{b_3}{4}\end{array}$ Therefore, the direction ratios of $b$ are $-3,5$ and 4 . $\therefore \mathrm{b}={\mathrm{b}}_1\hat{\mathrm{i}}+{\mathrm{b}}_2\hat{\mathrm{j}}+{\mathrm{b}}_3\hat{\mathrm{k}}=-3\hat{\mathrm{i}}+5\hat{\mathrm{j}}+4\hat{\mathrm{k}}$ Substituting the value of $b$ in eq. (i), we obtain $r=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (-3\hat{i}+5\hat{j}+4\hat{k})$ This is the equation of the required line.