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

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Solution

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$\to r . (^i -^j + 2^k) = 5 \to$ Plane I
$\to r . (3^i +^j +^k) = 6 \to$ Plane II
Let direction ratios of required line are $a, b, c$
Given Direction ratio of normal from plane 1 in $(1, - 1, 2)$
Direction ratio of normal from plane 2 in $(3, 1, 1)$
Since
Required line is 1 to normal of plane I
using condition of perpendicularity
$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$
$a - b + 2 c = 0.... (1)$
Same Required line is $⊥$ to normal of plane II
using condition
$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$
$3 a + b + c = 0 . . . . (2)$
Solving eq (1) & (2)
$\frac{a}{- 1 - 2} = \frac{b}{6 - 1} = \frac{c}{1 + 3}$
$\frac{a}{3} = \frac{b}{5} = \frac{c}{4} = λ$ (say)
$a = - 3 λ$ $b = 5 λ$ & $c = 4 λ$
Direction vector to line $\to b = λ (- 3^i + 5^j + 4^k)$
Required equation of line $\to r = \to a + λ \to b$
$\to r = (^i + 2^j + 3^k) + λ (- 3^i + 5^j + 4^k)$
Hence vector equation of line is $\to r = (^i + 2^j + 3^k) + λ (- 3^i + 5^j + 4^k)$

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