Question

The acute angle between the lines $\frac{x-1}{l}=\frac{y+1}{m}=\frac{z}{n}$ and $\frac{x+1}{m}=\frac{y-3}{n}=\frac{z-1}{l},$ where $l>m>n$ and $l,m,n$ are the roots of the cubic equation $x^3+x^2-4x-4=0,$ is

• A. ${\cos }^{-1}\frac{1}{9}$
• B. ${\cos }^{-1}\frac{2}{9}$
• C. ${\cos }^{-1}\frac{1}{3}$
• D. ${\cos }^{-1}\frac{4}{9}$

Answer 1

The correct option is D ${\cos }^{-1}\frac{4}{9}$

Given lines $\frac{x-1}{l}=\frac{y+1}{m}=\frac{z}{n}$ and $\frac{x+1}{m}=\frac{y-3}{n}=\frac{z-1}{l},$
Angle between lines is given by,
$\cos \theta =\frac{lm+mn+nl}{l^2+m^2+n^2}\cdots \left(i\right)$
Now, $l,m,n$ are the roots of $x^3+x^2-4x-4=0$
Thus, $l+m+n=-1$ and $lm+mn+ln=-4$
$\Rightarrow (l+m+n{)}^2=l^2+m^2+n^2+2(lm+mn+ln)\Rightarrow (-1{)}^2=l^2+m^2+n^2+2(-4)\Rightarrow l^2+m^2+n^2=9$
So, $\cos \theta =-\frac{4}{9}$
$\therefore$ Acute angle between the lines is ${\cos }^{-1}\frac{4}{9}$