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

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Aicharya · Accepted Answer

$\because l, m$ and $n$ are the direction cosines of any line $	herefore \quad l^2+m^2+n^2=1$...(i) Given, $l-m+n=0$ $\begin{aligned} & \Rightarrow \quad l=m-n 	ext { and } l m+m n-4 n l=0 \ & \Rightarrow \quad(m-n) m+m n-4 n(m-n)=0 \ & \Rightarrow \quad m^2-m n+m n-4 m n+4 n^2=0 \ & \Rightarrow \quad m^2+4 n^2-4 m n=0 \Rightarrow(m-2 n)^2=0 \ & \Rightarrow \quad m=2 n \Rightarrow l=2 n-n=n\end{aligned}$ From Eq. (i), $n^2+4 n^2+n^2=1$ $\Rightarrow \quad n= \pm \frac{1}{\sqrt{6}}, m= \pm \frac{2}{\sqrt{6}}, l= \pm \frac{1}{\sqrt{6}}$ $	herefore$ Direction cosines are $\left(\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}ight)$ and $\left(-\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}ight)$.

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

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