Question

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

Answer 1

Consider the required line be parallel to vector b given by , $b=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$
The position vector of the point (1, 2, 3) is a = $\hat{i}+2\hat{j}+3\hat{k}$
The equation of line passing through (1, 2, 3) and parallel to b is given by r = a + $\lambda$ b
$\therefore r=(\hat{i}+2j+3\hat{k})+\lambda (b_1\hat{i}+b_{2}j+b_3\hat{k})$ . . . (i)
The equations of the given planes are
$r.(\hat{i}-\hat{j}+2\hat{k})=5\cdots \left(ii\right)andr.(3\hat{i}+\hat{j}+\hat{k})=6\cdots \left(iii\right)$
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.
$\therefore (\hat{i}-\hat{j}+2\hat{k}).\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})=0\Rightarrow \lambda (b_1-b_2+2b_3)=0\Rightarrow (b_1-b_2+2b_3)=0\cdots \left(iv\right)$
Similarly, $(3\hat{i}+\hat{j}+2\hat{k}).\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})=0$
$\Rightarrow \lambda (3b_1+b_2+b_3)=0\cdots \left(v\right)$
From 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.
$\therefore b=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}=-3\hat{i}+5\hat{j}+4\hat{k}$
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 .
Note If any line is parallel to the plane, then normal to the plane is perpendicular to the line.