An angle between the lines whose direction co | vedclass

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(B) Given equations are:$l + 3m + 5n = 0$  ....$(1)$$5lm - 2mn + 6nl = 0$  ....$(2)$From equation $(1)$,we have $l = -3m - 5n$.Substituting the value

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

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(B) Given equations are:$l + 3m + 5n = 0$  ....$(1)$$5lm - 2mn + 6nl = 0$  ....$(2)$From equation $(1)$,we have $l = -3m - 5n$.Substituting the value of $l$ in equation $(2)$:$5(-3m - 5n)m - 2mn + 6n(-3m - 5n) = 0$$-15m^2 - 25mn - 2mn - 18mn - 30n^2 = 0$$-15m^2 - 45mn - 30n^2 = 0$Dividing by $-15$,we get:$m^2 + 3mn + 2n^2 = 0$$(m + n)(m + 2n) = 0$So,$m = -n$ or $m = -2n$.Case $1$: If $m = -n$,then $l = -3(-n) - 5n = 3n - 5n = -2n$. The direction ratios are $(-2n, -n, n)$,which simplifies to $(-2, -1, 1)$.Case $2$: If $m = -2n$,then $l = -3(-2n) - 5n = 6n - 5n = n$. The direction ratios are $(n, -2n, n)$,which simplifies to $(1, -2, 1)$.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}$Therefore,$	heta = \cos^{-1}\left(\frac{1}{6}ight)$.

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

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