Let the direction cosines of two lines satisf | vedclass

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(C) Given equations are $3l+2m+n=0$ ...$(1)$ and $2mn-3nl+5lm=0$ ...$(2)$.From equation $(1)$,$n = -(3l+2m)$.Substituting this into equation $(2)$:$2m

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$\frac{25}{\sqrt{2991}}$

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(C) Given equations are $3l+2m+n=0$ ...$(1)$ and $2mn-3nl+5lm=0$ ...$(2)$.From equation $(1)$,$n = -(3l+2m)$.Substituting this into equation $(2)$:$2m(-(3l+2m)) - 3l(-(3l+2m)) + 5lm = 0$$-6ml - 4m^2 + 9l^2 + 6lm + 5lm = 0$$9l^2 + 5lm - 4m^2 = 0$.Dividing by $m^2$ (assuming $m 
eq 0$),we get $9(\frac{l}{m})^2 + 5(\frac{l}{m}) - 4 = 0$.Let $t = \frac{l}{m}$,then $9t^2 + 5t - 4 = 0$.$(9t-4)(t+1) = 0$,so $t = \frac{4}{9}$ or $t = -1$.Case $1$: $t = -1 \Rightarrow l = -m$. Then $n = -(3(-m)+2m) = m$. The direction ratios are $(-1, 1, 1)$.Case $2$: $t = \frac{4}{9} \Rightarrow l = \frac{4}{9}m$. Then $n = -(3(\frac{4}{9}m)+2m) = -(\frac{4}{3}m+2m) = -\frac{10}{3}m$. The direction ratios are $(\frac{4}{9}, 1, -\frac{10}{3})$,which is equivalent to $(4, 9, -30)$.Let the direction vectors be $\vec{a} = (-1, 1, 1)$ and $\vec{b} = (4, 9, -30)$.$\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(-1)(4) + (1)(9) + (1)(-30)|}{\sqrt{(-1)^2+1^2+1^2} \sqrt{4^2+9^2+(-30)^2}}$$= \frac{|-4 + 9 - 30|}{\sqrt{3} \sqrt{16+81+900}} = \frac{|-25|}{\sqrt{3} \sqrt{997}} = \frac{25}{\sqrt{2991}}$.

Let the direction cosines of two lines satisfy the equations $3l+2m+n=0$ and $2mn-3nl+5lm=0$. If $\theta$ is the angle between these two lines,then $\cos \theta=$

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