Question

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

• A. $\theta ={\cos }^{-1}\left(\frac{1}{6}\right)$
• B. $\theta ={\cos }^{-1}\left(\frac{1}{3}\right)$
• C. $\theta ={\cos }^{-1}\left(\frac{1}{2}\right)$
• D. $\theta ={\cos }^{-1}\left(\frac{1}{8}\right)$

Answer 1

Answer: (A)

$\theta ={\cos }^{-1}\left(\frac{1}{6}\right)$

Given,

$3l+m+5n=0$

$\Rightarrow m=-5n-3l$ ………..$\left(1\right)$

and $6mn-2nl+5lm=0$

$\Rightarrow 6(-5n-3l)n-2nl+5l(-5n-3l)=0$ [from $\left(1\right)$]

$\Rightarrow -30n^2-18nl-2nl-25nl-15l^2=0$

$\Rightarrow -30n^2-45nl-15l^2=0$

$\Rightarrow -15(2n^2+3nl+l^2)=0$

$\Rightarrow 2n^2+3nl+l^2=0$

$\Rightarrow 2n^2+2nl+nl+l^2=0$

$\Rightarrow 2n(n+l)+l(n+l)=0$

$\Rightarrow (n+1)(2n+l)=0$

$\Rightarrow l=-n$ or $2n+l=0$

$\Rightarrow l=-2n$

$\therefore$ When $l=-n$,

$m=-5n-3(-n)$

$=-5n+3n$

$=-2n$

$\therefore l:m:n=-n:-2n:n$

$=-1:-2:1$

$\therefore$ When $l=-2n$

$m=-5n-3(-2n)$

$=-5n+6n$

$=n$

$\therefore l:m:n=-2n:n:n$

$=-2:1:1$

The direction ratios are $(-1,-2,1)$ and $(-2,1,1)$

Angle between the lines whose direction cosines are given by the equations.

$3l+m+5n=0$ and $6mn-2nl+5lm=0$

$\therefore \theta ={\cos }^{-1}\left\{\frac{(-1)\times (-2)+(-2)\times 1+1\times 1}{\sqrt{(-1{)}^2+(-2{)}^2+1^2}\sqrt{(-2{)}^2+1^2+1^2}}\right\}$

$={\cos }^{-1}\left\{\frac{2-2+1}{\sqrt{1+4+1}\sqrt{4+1+1}}\right\}$

$={\cos }^{-1}\left\{\frac{1}{\sqrt{6}\cdot \sqrt{6}}\right\}$

$={\cos }^{-1}\left(\frac{1}{6}\right)$.