The direction cosines of two lines are connec | vedclass

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(B) Given relations are $l+m-n=0$ and $lm-2mn+nl=0$. From the first relation,$n=l+m$. Substituting this into the second relation: $lm-2m(l+m)+(l+m)l=0

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$1/\sqrt{7}$

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(B) Given relations are $l+m-n=0$ and $lm-2mn+nl=0$. From the first relation,$n=l+m$. Substituting this into the second relation: $lm-2m(l+m)+(l+m)l=0$. $lm-2ml-2m^2+l^2+lm=0$. $l^2-2m^2=0$,which gives $l^2=2m^2$,so $l=\pm \sqrt{2}m$. Case $1$: If $l=\sqrt{2}m$,then $n=l+m=(\sqrt{2}+1)m$. The direction ratios are $(\sqrt{2}m, m, (\sqrt{2}+1)m)$,so the vector is $\vec{a_1} = \sqrt{2}\hat{i} + \hat{j} + (\sqrt{2}+1)\hat{k}$. Case $2$: If $l=-\sqrt{2}m$,then $n=l+m=(1-\sqrt{2})m$. The direction ratios are $(-\sqrt{2}m, m, (1-\sqrt{2})m)$,so the vector is $\vec{a_2} = -\sqrt{2}\hat{i} + \hat{j} + (1-\sqrt{2})\hat{k}$. Now,$\cos 	heta = \frac{|\vec{a_1} \cdot \vec{a_2}|}{|\vec{a_1}| |\vec{a_2}|}$. $\vec{a_1} \cdot \vec{a_2} = (\sqrt{2})(-\sqrt{2}) + (1)(1) + (\sqrt{2}+1)(1-\sqrt{2}) = -2 + 1 + (1-2) = -2$. $|\vec{a_1}|^2 = 2 + 1 + (\sqrt{2}+1)^2 = 3 + 2 + 1 + 2\sqrt{2} = 6+2\sqrt{2}$. $|\vec{a_2}|^2 = 2 + 1 + (1-\sqrt{2})^2 = 3 + 1 + 2 - 2\sqrt{2} = 6-2\sqrt{2}$. $|\vec{a_1}| |\vec{a_2}| = \sqrt{(6+2\sqrt{2})(6-2\sqrt{2})} = \sqrt{36-8} = \sqrt{28} = 2\sqrt{7}$. $\cos 	heta = \frac{|-2|}{2\sqrt{7}} = \frac{1}{\sqrt{7}}$.

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

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