Question

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

• A. 6
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (A)

$1+m-n=0$
$3l^2+m^2+cl(1+m)=0$
$n=1+m$
$31^2+m^2+c{l}^2+clm=0$
$(3+c)l^2+clm+m^2=0$
$(3+c){\left(\frac{l}{m}\right)}^2+c\left(\frac{l}{m}\right)+1=0\dots \dots (1)$
$\because$ lies are parallel.
Roots of (1) must be equal
$\Rightarrow D=0$
$c^2-4(3+c)=0$
$c^2-4c-12=0$
$(c-6)(c+2)=0$
$c=6\text{ or }c=-2$
+ve value of $c=6$