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 $
$\Rightarrow \lambda\left(b_1-b_2+2 b_3\right)=0$
$\Rightarrow \left(b_1-b_2+2 b_3\right)=0$
$\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 $
$\Rightarrow\lambda\left(3 b_1+b_2+b_3\right)=0$
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.