If the direction cosines of two lines are giv | vedclass

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(A) Given equations are: $l+3m+5n=0$  --- $(i)$ $5lm-2mn+6ln=0$ --- $(ii)$ From $(i)$,$l = -3m - 5n$. Substituting this into $(ii)$: $5(-3m-5n)m - 2mn

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

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(A) Given equations are: $l+3m+5n=0$  --- $(i)$ $5lm-2mn+6ln=0$ --- $(ii)$ From $(i)$,$l = -3m - 5n$. Substituting this into $(ii)$: $5(-3m-5n)m - 2mn + 6(-3m-5n)n = 0$ $-15m^2 - 25mn - 2mn - 18mn - 30n^2 = 0$ $-15m^2 - 45mn - 30n^2 = 0$ Dividing by $-15$: $m^2 + 3mn + 2n^2 = 0$ $(m+n)(m+2n) = 0$ Case $1$: $m = -n$. Then $l = -3(-n) - 5n = -2n$. Direction ratios are $(-2n, -n, n)$ or $(2, 1, -1)$. Case $2$: $m = -2n$. Then $l = -3(-2n) - 5n = n$. Direction ratios are $(n, -2n, n)$ or $(1, -2, 1)$. Let the direction ratios be $\vec{a} = (2, 1, -1)$ and $\vec{b} = (1, -2, 1)$. The angle $	heta$ between the lines is given by: $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(2)(1) + (1)(-2) + (-1)(1)|}{\sqrt{2^2+1^2+(-1)^2} \sqrt{1^2+(-2)^2+1^2}}$ $\cos 	heta = \frac{|2 - 2 - 1|}{\sqrt{6} \sqrt{6}} = \frac{|-1|}{6} = \frac{1}{6}$ Therefore,$	heta = \cos^{-1}\left(\frac{1}{6}ight)$.

If the direction cosines of two lines are given by $l+3m+5n=0$ and $5lm-2mn+6ln=0$,then the angle between the lines is

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