MCQ
If two straight lines whose direction cosines are given by the relations $l+m-n=0,3l^{2}+m^{2}+c n l =0$ are parallel, then the positive value of $c$ is
- ✓$6$
- B$4$
- C$3$
- D$2$
$3 l^{2}+ m ^{2}+ cl (1+ m )=0$
$n =1+ m$
$3 l ^{2}+ m ^{2}+ cl ^{2}+ clm =0$
$(3+ c ) l ^{2}+ clm + m ^{2}=0$
$(3+c)\left(\frac{l}{m}\right)^{2}+c\left(\frac{l}{m}\right)+1=0 \ldots \ldots(1)$
$\because$ lies are parallel.
Roots of $(1)$ must be equal
$\Rightarrow D=0$
$c ^{2}-4(3+ c )=0$
$c ^{2}-4 c -12=0$
$( c -6)( c +2)=0$
$c =6$ or $c =-2$
$+ve$ value of $c =6$
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