The direction cosines of two lines are connec | vedclass

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(B) Given: $l-m+n=0$ $\Rightarrow m=l+n$ $(i)$ Substituting $m=l+n$ into $2lm-3mn+nl=0$: $2l(l+n)-3(l+n)n+nl=0$ $2l^2+2ln-3ln-3n^2+nl=0$ $2l^2-3n^2=0$

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$\frac{1}{\sqrt{19}}$

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(B) Given: $l-m+n=0$ $\Rightarrow m=l+n$ $(i)$ Substituting $m=l+n$ into $2lm-3mn+nl=0$: $2l(l+n)-3(l+n)n+nl=0$ $2l^2+2ln-3ln-3n^2+nl=0$ $2l^2-3n^2=0$ $l^2 = \frac{3}{2}n^2$ Let $n=1$,then $l^2 = \frac{3}{2} \Rightarrow l = \pm \sqrt{\frac{3}{2}}$. For $l_1 = \sqrt{\frac{3}{2}}$,$m_1 = \sqrt{\frac{3}{2}}+1$. For $l_2 = -\sqrt{\frac{3}{2}}$,$m_2 = -\sqrt{\frac{3}{2}}+1$. The direction ratios of the two lines are $\vec{a} = (\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}}+1, 1)$ and $\vec{b} = (-\sqrt{\frac{3}{2}}, 1-\sqrt{\frac{3}{2}}, 1)$. $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|} = \frac{|-\frac{3}{2} + (1-\frac{3}{2}) + 1|}{\sqrt{\frac{3}{2} + (\frac{3}{2}+1+2\sqrt{\frac{3}{2}}) + 1} \cdot \sqrt{\frac{3}{2} + (1+\frac{3}{2}-2\sqrt{\frac{3}{2}}) + 1}}$ $\cos 	heta = \frac{|-\frac{3}{2} - \frac{1}{2} + 1|}{\sqrt{4+2\sqrt{\frac{3}{2}}} \cdot \sqrt{4-2\sqrt{\frac{3}{2}}}} = \frac{|-1|}{\sqrt{16 - 4(\frac{3}{2})}} = \frac{1}{\sqrt{16-6}} = \frac{1}{\sqrt{10}}$. Wait,re-evaluating the dot product: $\vec{a} \cdot \vec{b} = -\frac{3}{2} + (1 - \frac{3}{2}) + 1 = -\frac{3}{2} - \frac{1}{2} + 1 = -1$. $|\vec{a}|^2 = \frac{3}{2} + 1 + \frac{3}{2} + 2\sqrt{\frac{3}{2}} + 1 = 4 + 2\sqrt{\frac{3}{2}}$. $|\vec{b}|^2 = \frac{3}{2} + 1 + \frac{3}{2} - 2\sqrt{\frac{3}{2}} + 1 = 4 - 2\sqrt{\frac{3}{2}}$. $|\vec{a}||\vec{b}| = \sqrt{16 - 4(\frac{3}{2})} = \sqrt{16-6} = \sqrt{10}$. Thus $\cos 	heta = \frac{1}{\sqrt{10}}$. Given the options,the calculation yields $\frac{1}{\sqrt{10}}$. If we re-check the original equation $2lm-3mn+nl=0$,substituting $m=l+n$ gives $2l^2+2ln-3ln-3n^2+nl=0 \Rightarrow 2l^2-3n^2=0$. The result is $\frac{1}{\sqrt{10}}$.

The direction cosines of two lines are connected by the relations $l-m+n=0$ and $2lm-3mn+nl=0$. If $\theta$ is the angle between these two lines,then $\cos \theta=$

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