Question

If the dc's of two lines parallel lines are given by $2l+3m+kn=0$ and $l^2-m^2+5n^2=0$ then the values of k are:

• A. $5$
• B. $3$
• C. $-5$
• D. $-3$

Answer 1

Answer: (A), (C)

$-5$

If the D.C's of two lines are connected by the relations $al+bm+cn=0$ and $u{l}^2+v{m}^2+w{n}^2=0$ then the lines are perpendicular if $\sum a^2(v+w)=0$ and parallel if $\frac{a^2}{u}+\frac{b^2}{v}+\frac{c^2}{w}=0$
Given relations are $2l+3m+kn=0$ and $l^2-m^2+5n^2=0$: Here $a=2,b=3,c=k,u=1,v=-1,w=5$
Lines are parallel,
So, $\frac{a^2}{u}+\frac{b^2}{v}+\frac{c^2}{w}=0$
$\frac{4}{1}+\frac{9}{-1}+\frac{k^2}{5}=0$
$-5+\frac{k^2}{5}=0$
$k^2=25$
$k=\pm 5$