Question

Find the direction cosines of two lines if they satisfy $l+3n=5m$ and $7l^2+5m^2=3n^2$.

Answer 1

Given equations are $l-5m+3n=0.....\left(1\right)$

$7l^2+5m^2-3n^2=0......\left(2\right)$

From (1), $l=5m-3n$

Substituting in (2) $\Rightarrow 7(5m-3n{)}^2+5m^2-3n^2=0$

$\Rightarrow 7(25m^2+9n^2-30mn)+5m^2-3n^2=0$

$\Rightarrow 6m^2-7mn+2n^2=0$

$\Rightarrow (3m-2n)(2m-n)=0$

$\Rightarrow m=\frac{2}{3}n$ or $m=\frac{n}{2}$

$l=5\left(\frac{2}{3}\right)n-3n=\frac{1}{3}n$ or $l=\frac{5n}{2}-3n$

$=-\frac{n}{2}$

So, the direction cosines are $\frac{1}{3}n,\frac{2}{3}n,n$

or

$\frac{-n}{2},\frac{n}{2},n$

$\Rightarrow$ direction ratios are $1,2,3$ or $-1,1,2$

$\therefore$ The direction cosines are $\pm \frac{1}{\sqrt{14}},\pm \frac{2}{\sqrt{14}},\pm \frac{3}{\sqrt{14}}$

or $\mp \frac{1}{\sqrt{6}},\pm \frac{1}{\sqrt{6}},\pm \frac{2}{\sqrt{6}}$