If two straight lines whose direction cosines are given by the relations l + m - n = 0, 3l…

Question

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 :

Accepted Answer

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 $m 
eq 0$, 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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