Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The given equation is $x^3+x^2-4x-4=0$.

Factoring the equation:

$x^2(x+1)-4(x+1)=0$

$(x^2-4)(x+1)=0$

$(x-2)(x+2)(x+1)=0$

The roots are $2,-1,-2$.

Since $l>m>n$, we have $l=2$, $m=-1$, $n=-2$.

The direction ratios of the first line are $(l,m,n)=(2,-1,-2)$.

The direction ratios of the second line are $(m,n,l)=(-1,-2,2)$.

The angle $\theta$ between the lines is given by:

$\cos \theta =\frac{lm+mn+nl}{\sqrt{l^2+m^2+n^2}\sqrt{m^2+n^2+l^2}}$

Substituting the values:

$lm+mn+nl=(2)(-1)+(-1)(-2)+(-2)(2)=-2+2-4=-4$

$l^2+m^2+n^2=2^2+(-1{)}^2+(-2{)}^2=4+1+4=9$

Therefore, $\cos \theta =\frac{-4}{\sqrt{9}\sqrt{9}}=\frac{-4}{9}$.

$\theta ={\cos }^{-1}\left(\frac{-4}{9}\right)$

Answer: ${\cos }^{-1}\left(\frac{-4}{9}\right)$