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If two straight lines whose direction cosines are given by the relations $$l + m - n = 0$$, $$3{l^2} + {m^2} + cnl = 0$$ are parallel, then the positive value of c is :

JEE · Math · previous-year question

  1. A.6correct
  2. B.4
  3. C.3
  4. D.2

Answer

A. 6

Explanation

Given that the direction cosines satisfy $l + m - n = 0$, we find that $n = l + m$. The other equation is $3l^2 + m^2 + cnl = 0$, and substituting $n = l + m$ gives $3l^2 + m^2 + cl(l + m) = 0$. This simplifies to $(3 + c)l^2 + clm + m^2 = 0$. As the lines are parallel, they share the same direction ratios, so we can express $l$ in terms of $m$, say $l = km$. Substituting this into our equation gives $(3 + c)(km)^2 + ckm^2 + m^2 = 0$. This simplifies to $m^2[k^2(3 + c) + kc + 1] = 0$. Since $m \neq 0$, we must have $k^2(3 + c) + kc + 1 = 0$. Here, we consider the ratio $k = \frac{l}{m}$ to be constant, since the lines are parallel. The equation then becomes a quadratic equation in $k$. As the lines are parallel, the discriminant of the quadratic equation must be equal to zero for the equation to have equal roots. Hence, the discriminant $D = (c^2 - 4(3 + c)) = 0$. Solving this quadratic equation gives $c^2 - 4c - 12 = 0$. Factoring this equation gives $(c - 6)(c + 2) = 0$. Solving for $c$ gives $c = 6, -2$. However, we are looking for the positive value of $c$, so $c = 6$. Therefore, the correct answer is 6

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