Question

If the direction cosines of two lines are given by $l+m+n=0$ and $l^2-5m^2+n^2=0$, then the angle between them is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is D $\frac{\pi }{3}$

$l+m+n=0$
$\Rightarrow l+n=-m$...(i)
$l^2-5m^2+n^2=0$

Substituting the equation from eq (i)

$l^2+n^2-5(-l-n{)}^2=0$

$l^2+n^2-5(l^2+n^2+2ln)=0$

$-4l^2-4n^2-10ln=0$

$2l^2+2n^2+5ln=0$

$2l^2+5ln+2n^2=0$

$l=\frac{-5n\pm n\sqrt{25-16}}{4}$

$l=\frac{-5n\pm 3n}{4}$

$l=-2n$ and $l=\frac{-n}{2}$

Hence $l=\frac{-n}{2},m=\frac{-n}{2}$
If $l=-2n,m=n$

Hence the Dr's are

$(\frac{-n}{2},\frac{-n}{2},n)$ and $(-2n,n,n)$

Therefore Dc's are
$(\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}})$ and $(\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}})$

Hence $\cos \theta =(\frac{-2}{\sqrt{6}}.\frac{-1}{\sqrt{6}})+(\frac{-1}{\sqrt{6}}.\frac{1}{\sqrt{6}})+(\frac{2}{\sqrt{6}}.\frac{1}{\sqrt{6}})$

$=\frac{2-1+2}{6}$

$=\frac{3}{6}$

$=\frac{1}{2}$

Hence $\theta =\frac{\pi }{3}$