The angle between the lines frac{x-1}{l}=frac | vedclass

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(B) Given the equation $x^3+x^2-4x-4=0$. Factoring the equation: $x^2(x+1)-4(x+1)=0 \implies (x^2-4)(x+1)=0$. So,$(x-2)(x+2)(x+1)=0$. The roots are $2

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

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(B) Given the equation $x^3+x^2-4x-4=0$. Factoring the equation: $x^2(x+1)-4(x+1)=0 \implies (x^2-4)(x+1)=0$. So,$(x-2)(x+2)(x+1)=0$. The roots are $2, -1, -2$. Given $l > m > n$,we have $l=2, m=-1, n=-2$. The direction ratios of the first line are $\vec{a} = (l, m, n) = (2, -1, -2)$. The direction ratios of the second line are $\vec{b} = (m, n, l) = (-1, -2, 2)$. The angle $	heta$ between the lines is given by $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$. $\vec{a} \cdot \vec{b} = (2)(-1) + (-1)(-2) + (-2)(2) = -2 + 2 - 4 = -4$. $|\vec{a}| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4+1+4} = 3$. $|\vec{b}| = \sqrt{(-1)^2 + (-2)^2 + 2^2} = \sqrt{1+4+4} = 3$. $\cos 	heta = \frac{|-4|}{3 	imes 3} = \frac{4}{9}$. However,the angle between lines is usually defined as $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-4}{9}$. Thus,$	heta = \cos^{-1}\left(\frac{-4}{9}ight)$.

The 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 roots of the equation $x^3+x^2-4x-4=0$,is

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