The dr’s of two lines are given by $a + b + c = 0$ , $2 a b + 2 a c - b c = 0$ . Then the angle between the lines is

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$π$

$\frac{π}{3}$

$\frac{π}{6}$

$\frac{π}{2}$

answer is B.

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Detailed Solution

$a + b + c = 0 \to (1)$

$2 a b + 2 a c - b c = 0 \to (2)$

$\Rightarrow 2 a (b + c) - b c = 0$ , $[a = - (b + c)]$

$\Rightarrow (b + 2 c) (2 b + c) = 0$ $\Rightarrow c = - 2 b$ or $- \frac{b}{2}$

If $c = - 2 b$ $\Rightarrow a + b - 2 b = 0 \Rightarrow a = b$

$a : b : c = 1 : 1 : - 2$

If $c = - \frac{b}{2}$ then $a + b - \frac{b}{2} = 0$ $\Rightarrow a = - \frac{b}{2}$

$\Rightarrow a : b : c = - \frac{b}{2} : b : - \frac{b}{2} = - 1 : 2 : - 1 = 1 : - 2 : 1$

If $θ$ is an angle between the lines then $\cos θ = \frac{(1) (1) + (1) (- 2) + (- 2) (1)}{\sqrt{1 + 1 + 4} \sqrt{1 + 4 + 1}}$

$= - \frac{1}{2} \Rightarrow θ = \frac{2 π}{3}$ or $\frac{π}{3}$

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