If the direction cosines of a line satisfy th | vedclass

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(D) Given that $l, m, n$ are the direction cosines of a line,we have $l^2+m^2+n^2=1$ $(i)$.From the given relations $l-m+n=0$,we get $l=m-n$.Substitut

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$\left(\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$

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(D) Given that $l, m, n$ are the direction cosines of a line,we have $l^2+m^2+n^2=1$ $(i)$.From the given relations $l-m+n=0$,we get $l=m-n$.Substituting this into the second relation $lm+mn-4nl=0$:$(m-n)m + mn - 4n(m-n) = 0$$m^2 - mn + mn - 4mn + 4n^2 = 0$$m^2 - 4mn + 4n^2 = 0$$(m-2n)^2 = 0 \Rightarrow m=2n$.Substituting $m=2n$ into $l=m-n$,we get $l=2n-n=n$.Now,substitute $l=n$ and $m=2n$ into the identity $l^2+m^2+n^2=1$:$n^2 + (2n)^2 + n^2 = 1$$n^2 + 4n^2 + n^2 = 1$$6n^2 = 1 \Rightarrow n = \pm \frac{1}{\sqrt{6}}$.Thus,$l = \pm \frac{1}{\sqrt{6}}$ and $m = \pm \frac{2}{\sqrt{6}}$.The direction cosines are $\left(\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$ or $\left(-\frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}}\right)$.

If the direction cosines of a line satisfy the relations $l-m+n=0$ and $lm+mn-4nl=0$,then the direction cosines of the line are

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