Question

The equation of the line passing through $(1,2,3)$ and parallel to the planes $r\cdot (\hat{i}-\hat{j}+2\hat{k})=5$ and $r\cdot (3\hat{i}+\hat{j}+\hat{k})=6$ is

• A. $r=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (2\hat{i}+3\hat{j}+4\hat{k})$
• B. $r=(-3\hat{i}+5\hat{j}+4\hat{k})+\lambda (\hat{i}+2\hat{j}+3\hat{k})$
• C. $r=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (-3\hat{i}+5\hat{j}+4\hat{k})$
• D. $r=\lambda (-3\hat{i}+5\hat{j}+4\hat{k})$

Answer 1

Answer: (C)

The equation of line passing through $(1,2,3)$ and parallel to $b$ is given by $r=a+\lambda b$
$\therefore r=(\hat{i}+2\hat{j}+3\hat{k})+\lambda \left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)$....(i)
The equations of the given planes are
$r\cdot (\hat{i}-\hat{j}+2\hat{k})=5$....(ii)
and $r\cdot (3\hat{i}+\hat{j}+\hat{k})=6$....(iiii)
The line in Eq. (i) and plane in Eq. (ii) are parallel. Therefore, the rinermal lo the plane of Ey. (ii) and the yiven lire are perpendicular.
$\therefore (\hat{i}-\hat{j}+2\hat{k})\cdot \lambda \left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)=0$
$\Rightarrow \lambda \left(b_1-b_2+2b_3\right)=0$
$\Rightarrow \left(b_1-b_2+2b_3\right)=0$
$\text{ Similarly, }(3\hat{i}+\hat{j}+\hat{k})\cdot \lambda \left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)=0$
$\Rightarrow \lambda \left(3b_1+b_2+b_3\right)=0$
Form Eqs. (iv) and (v), we obtain
$\frac{b_1}{(-1)\times 1-1\times 2}=\frac{b_2}{2\times 3-1\times 1}=\frac{b_3}{1\times 1-3(-1)}$
$\Rightarrow \frac{b_1}{-3}=\frac{b_2}{5}=\frac{b_3}{4}$
Therefore, the direction ratios of $b$ are $-3.5$ and 4 .
$\therefore b-D_1\hat{i}+D_2\hat{j}+D_3\hat{k}--3\hat{i}+b\hat{j}+4\hat{k}$
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.