Question

An angle between the lines whose direction cosines are given by the equations, $l+3m+5n=0$ and $5lm-2mn+6nl=0$, is :

• A. ${\cos }^{-1}\left(\frac{1}{4}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• C. ${\cos }^{-1}\left(\frac{1}{6}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{8}\right)$

Answer 1

The correct option is C ${\cos }^{-1}\left(\frac{1}{6}\right)$

The equations are
$l+3m+5n=\mathrm{0...}\left(1\right)$
$5lm-2mn+6nl=\mathrm{0...}\left(2\right)$
Put $l=-3m-5n$ from equation $\left(1\right)$ into equation $\left(2\right)$
$\therefore 5(-3m-5n)m-2mn+6n(-3m-5n)=0$
$\Rightarrow m^2+3nm+2n^2=0$
$\Rightarrow m^2+(2n+n)m+2n^2=0$
$m=-2n,-n$
If $m=-2n$ then from equation $\left(1\right)$
$l=n$
If $m=-n$ then from equation $\left(1\right)$
$l=-2n$
Therefore, direction ratios are
$(l,m,n)\equiv (n,-2n,n)\text{and}(-2n,-n,n)$
$\Rightarrow (l,m,n)\equiv (1,-2,1)\text{and}(-2,-1,1)$
Angle between the lines
$\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$=\frac{\left(1\right)(-2)+(-2)(-1)+\left(1\right)\left(1\right)}{\sqrt{1^2+(-2{)}^2+1^2}\sqrt{(-2{)}^2+(-1{)}^2+1^2}}$
​​​​​​$\Rightarrow \cos \theta =\frac{1}{6}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{6}\right)$