The equation of the straight line through the origin parallel to the line (b + c)x + (c + a)y + (a + b)z = k = (b – c)x + (c – a)y + (a – b)z is

by

Q: The equation of the straight line through the origin parallel to the line
(b + c)x + (c + a)y + (a + b)z = k = (b – c)x + (c – a)y + (a – b)z is

(A) $\large \frac{x}{b^2-c^2} = \frac{y}{c^2-a^2} = \frac{z}{a^2-b^2} $

(B) $\large \frac{x}{a} = \frac{y}{b} = \frac{z}{c} $

(C) $\large \frac{x}{a^2-bc} = \frac{y}{b^2-ca} = \frac{z}{c^2-ab} $

(D) none of these

Sol. Equation of straight line through origin is

$\large \frac{x-0}{l} = \frac{y-0}{m} = \frac{z-0}{n} $

where l((b+c) + m(c+a) + n(a + b) = 0

and l(b – c) + m(c -a) + n(a – b) = 0

On solving,

$\large \frac{l}{2(a^2-bc)} = \frac{m}{2(b^2-ca)} = \frac{n}{2(c^2-ab)} $

Equation of the straight line is

$\large \frac{x}{a^2-bc} = \frac{y}{b^2-ca} = \frac{z}{c^2-ab} $

Hence (C) is the correct answer.

Also Read :