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(NONE) Given equations are $l+m-n=0$ and $l^{2}+m^{2}-n^{2}=0$.From the first equation,$n = l+m$.Substituting this into the second equation: $l^{2}+m^

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$\frac{5}{8}$

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(NONE) Given equations are $l+m-n=0$ and $l^{2}+m^{2}-n^{2}=0$.From the first equation,$n = l+m$.Substituting this into the second equation: $l^{2}+m^{2}=(l+m)^{2} = l^{2}+m^{2}+2lm$.This simplifies to $2lm = 0$,which means $l=0$ or $m=0$.Case $1$: If $l=0$,then $n=m$. Since $l^{2}+m^{2}+n^{2}=1$,we have $0^{2}+m^{2}+m^{2}=1$,so $2m^{2}=1$,$m = \pm \frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.Case $2$: If $m=0$,then $n=l$. Since $l^{2}+m^{2}+n^{2}=1$,we have $l^{2}+0^{2}+l^{2}=1$,so $2l^{2}=1$,$l = \pm \frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.Let the direction cosines be $\vec{u} = (0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ and $\vec{v} = (\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.The cosine of the angle $\alpha$ between them is $\cos \alpha = |\vec{u} \cdot \vec{v}| = |0 \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot 0 + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}| = \frac{1}{2}$.We need to find $\sin^{4} \alpha + \cos^{4} \alpha$.Since $\cos \alpha = \frac{1}{2}$,$\sin^{2} \alpha = 1 - \cos^{2} \alpha = 1 - \frac{1}{4} = \frac{3}{4}$.Then $\sin^{4} \alpha + \cos^{4} \alpha = (\frac{3}{4})^{2} + (\frac{1}{2})^{2} = \frac{9}{16} + \frac{1}{4} = \frac{9+4}{16} = \frac{13}{16}$.

Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $l+m-n=0$ and $l^{2}+m^{2}-n^{2}=0$. Then the value of $\sin^{4} \alpha + \cos^{4} \alpha$ is

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