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If two straight lines whose direction cosines are given by the relations $l+\mathrm{m}-\mathrm{n}=0,3 l^{2}+\mathrm{m}^{2}+\mathrm{cn} l=0$ are parallel, then the positive value of $\mathrm{c}$ is :
Option: 1
$6$

Option: 2
$4$

Option: 3
$3$

Option: 4
$2$

Answers (1)

best_answer

$\mathrm{l+m=n }$ ..........(1) $\mathrm{\quad 3 l^{2}+m^{2}+c n l=0 }$ ..............(2)

$\mathrm{Substituting \; n=1+m \; in\; eqn (2)}$

$\mathrm{\Rightarrow \quad 3 l^{2}+m^{2}+c(l+m) l=0} \\$

$\mathrm{\Rightarrow \quad(3+c) l^{2}+c m l+m^{2}=0} \\$

$\mathrm{\Rightarrow \quad(3+c)\left(\frac{l}{m}\right)^{2}+c\left(\frac{l}{m}\right)+1=0}$

Since lines are parallel $\Rightarrow$ this quadratic will have repeated roots $\Rightarrow \mathrm{D= 0}$

$\mathrm{\Rightarrow c^{2}-4(3+c)=0} \\$

$\mathrm{\Rightarrow c^{2}-4 c-12=0 \Rightarrow c=6 \text { or } c=-2}$

So positive value of $\mathrm{ c=6}$

Hence the correct answer is option 1.

Posted by

Ritika Kankaria

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