Let the direction cosines of two lines satisf | vedclass

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(D) Given equations for direction cosines $(l, m, n)$ are:$4l + m - n = 0 \implies n = 4l + m$ ... $(1)$$2mn + 10nl + 3lm = 0$ ... $(2)$Substitute $n

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$\frac{10}{3\sqrt{38}}$

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(D) Given equations for direction cosines $(l, m, n)$ are:$4l + m - n = 0 \implies n = 4l + m$ ... $(1)$$2mn + 10nl + 3lm = 0$ ... $(2)$Substitute $n = 4l + m$ into equation $(2)$:$2m(4l + m) + 10l(4l + m) + 3lm = 0$$8lm + 2m^2 + 40l^2 + 10lm + 3lm = 0$$40l^2 + 21lm + 2m^2 = 0$$(8l + m)(5l + 2m) = 0$Case $1$: $m = -8l$. Then $n = 4l - 8l = -4l$. Direction ratios are $(l, -8l, -4l)$ or $(1, -8, -4)$.Case $2$: $m = -\frac{5}{2}l$. Then $n = 4l - \frac{5}{2}l = \frac{3}{2}l$. Direction ratios are $(l, -\frac{5}{2}l, \frac{3}{2}l)$ or $(2, -5, 3)$.Let the direction vectors be $\vec{a} = (1, -8, -4)$ and $\vec{b} = (2, -5, 3)$.$\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(2) + (-8)(-5) + (-4)(3)|}{\sqrt{1^2 + (-8)^2 + (-4)^2} \sqrt{2^2 + (-5)^2 + 3^2}}$$\cos 	heta = \frac{|2 + 40 - 12|}{\sqrt{1 + 64 + 16} \sqrt{4 + 25 + 9}} = \frac{30}{\sqrt{81} \sqrt{38}} = \frac{30}{9 \sqrt{38}} = \frac{10}{3 \sqrt{38}}$.

Let the direction cosines of two lines satisfy the equations: $4l+m-n=0$ and $2mn+10nl+3lm=0$. Then the cosine of the acute angle between these lines is:

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