Question

The direction cosine of the two lines are determined by the relations $l-5m+3n=0$ and $7l^2+5m^2-3n^2=0$.Find the solution.

Answer 1

It is given that,

$l-5m+3n=0$

$l=5m-3n$

And,

$7l^2+5m^2-3n^2=0$

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

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

$180m^2+60n^2-210mn=0$

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

$6m^2-4mn-3mn+2n^2=0$

$2m\left(3m-2n\right)-n\left(3m-2n\right)=0$

$\left(2m-n\right)\left(3m-2n\right)=0$

$2m-n=0$

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

And,

$3m-2n=0$

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

So, when $m=\frac{n}{2}$, then, $n=2m$,

$l=5m-3\left(2m\right)$

$l=5m-6m$

$l=-m$

$\frac{l}{-1}=\frac{m}{1}$

And since $\frac{m}{1}=\frac{n}{2}$, so,

$\frac{l}{-1}=\frac{m}{1}=\frac{n}{2}$

The direction cosines will be,

$\frac{l}{\sqrt{l^2+m^2+n^2}}=\frac{-1}{\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2}}$

$=-\frac{1}{\sqrt{6}}$

$\frac{m}{\sqrt{l^2+m^2+n^2}}=\frac{1}{\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2}}$

$=\frac{1}{\sqrt{6}}$

$\frac{n}{\sqrt{l^2+m^2+n^2}}=\frac{2}{\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2}}$

$=\frac{2}{\sqrt{6}}$

Thus, direction cosines is $-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}$.

When, $m=\frac{2}{3}n$, then, $n=\frac{3}{2}m$,

$l=5m-3\left(\frac{3}{2}m\right)$

$l=5m-\frac{9}{2}m$

$l=\frac{m}{2}$

$\frac{l}{1}=\frac{m}{2}$

And since $\frac{m}{1}=\frac{2n}{3}$, then $\frac{m}{2}=\frac{n}{3}$, so,

$\frac{l}{1}=\frac{m}{2}=\frac{n}{3}$

The direction cosines will be,

$\frac{l}{\sqrt{l^2+m^2+n^2}}=\frac{1}{\sqrt{{\left(1\right)}^2+{\left(2\right)}^2+{\left(3\right)}^2}}$

$=\frac{1}{\sqrt{14}}$

$\frac{m}{\sqrt{l^2+m^2+n^2}}=\frac{2}{\sqrt{{\left(1\right)}^2+{\left(2\right)}^2+{\left(3\right)}^2}}$

$=\frac{2}{\sqrt{14}}$

$\frac{n}{\sqrt{l^2+m^2+n^2}}=\frac{3}{\sqrt{{\left(1\right)}^2+{\left(2\right)}^2+{\left(3\right)}^2}}$

$=\frac{3}{\sqrt{14}}$

Thus, direction cosines is $\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$.