If a line lies in the octant OXYZ and it make | vedclass

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(B) Let the direction cosines of the line be $l, m, n$. Since the line makes equal angles with the axes,we have $\alpha = \beta = \gamma$. Thus,$l = \

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$l = m = n = \pm \frac{1}{\sqrt{3}}$

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(B) Let the direction cosines of the line be $l, m, n$. Since the line makes equal angles with the axes,we have $\alpha = \beta = \gamma$. Thus,$l = \cos \alpha$,$m = \cos \beta = \cos \alpha = l$,and $n = \cos \gamma = \cos \alpha = l$. We know that $l^2 + m^2 + n^2 = 1$. Substituting $l = m = n$,we get $l^2 + l^2 + l^2 = 1$,which implies $3l^2 = 1$. Therefore,$l^2 = \frac{1}{3}$,which gives $l = \pm \frac{1}{\sqrt{3}}$. Since $l = m = n$,we have $l = m = n = \pm \frac{1}{\sqrt{3}}$.

If a line lies in the octant $OXYZ$ and it makes equal angles with the axes,then

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