The acute angle between two lines such that t | vedclass

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(C) Given the equations for direction cosines: $l+m+n=0$ and $l^2+m^2-n^2=0$. From the first equation,$l+m = -n$. Squaring both sides,we get $l^2+m^2+

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$60$

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(C) Given the equations for direction cosines: $l+m+n=0$ and $l^2+m^2-n^2=0$. From the first equation,$l+m = -n$. Squaring both sides,we get $l^2+m^2+2lm = n^2$. Substituting $l^2+m^2 = n^2$ from the second equation into this,we get $n^2+2lm = n^2$,which implies $2lm = 0$,so $lm = 0$. This means either $l=0$ or $m=0$. Case $1$: If $l=0$,then $m+n=0 \Rightarrow m=-n$. Since $l^2+m^2+n^2=1$,we have $0^2+(-n)^2+n^2=1 \Rightarrow 2n^2=1 \Rightarrow n = \pm \frac{1}{\sqrt{2}}$. Thus,the direction ratios are $(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ or $(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$. Case $2$: If $m=0$,then $l+n=0 \Rightarrow l=-n$. Similarly,$l^2+0^2+n^2=1 \Rightarrow 2n^2=1 \Rightarrow n = \pm \frac{1}{\sqrt{2}}$. Thus,the direction ratios are $(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$ or $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$. Let the two lines have direction cosines $\vec{u_1} = (0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ and $\vec{u_2} = (-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$. The cosine of the angle $	heta$ between them is $|\vec{u_1} \cdot \vec{u_2}| = |(0)(-\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}})(0) + (\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}})| = |0 + 0 + \frac{1}{2}| = \frac{1}{2}$. Since $\cos 	heta = \frac{1}{2}$,the acute angle $	heta = 60^o$.

The acute angle between two lines such that the direction cosines $l, m, n$ of each of them satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is ............ $^o$

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