Question

The direction cosines of two lines are determined by the relation $l-5m+3n=0$ and $7l^2+5m^2-3n^2=0$, find them.

Answer 1

Consider the given equation, $l-5m+3n=0$

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

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

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

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

$175m^2+63n^2-210mn+5m^2-3n^2$

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

$30(6m^2+2n^2-7mn)=0$

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

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

Hence,

$3m-2n=0$ or $2m-n=0$

Case (1) : $3m-2n=0$ $⟹m=\frac{2n}{3}$

Substitute for $m$ in (1)

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

Hence, the direction cosines $(l,m,n)$ is $(\frac{n}{3},\frac{2n}{3},n)$

The direction ratio is proportional to $(1,2,3)$ ----- {multiplying by $3$}

We have the formula for Direction cosines, $(\frac{a}{\sqrt{a^2+b^2+c^2}},\frac{b}{\sqrt{a^2+b^2+c^2}},\frac{c}{\sqrt{a^2+b^2+c^2}})$

$\therefore (\frac{1}{\sqrt{1^2+2^2+3^2}},\frac{2}{\sqrt{1^2+2^2+3^2}},\frac{3}{\sqrt{1^2+2^2+3^2}})=(\pm \frac{1}{\sqrt{14}},\pm \frac{2}{\sqrt{14}},\pm \frac{3}{\sqrt{14}})$ are the direction cosines

Case (2) : $2m-n=0$ $⟹m=\frac{n}{2}$

Substitute for $m$ in (1)

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

Hence, the direction cosines $(l,m,n)$ is $(-\frac{n}{2},\frac{n}{2},n)$

The direction ratio is proportional to $(-1,1,2)$ ----- {multiplying by $2$}

We have the formula for Direction cosines, $(\frac{a}{\sqrt{a^2+b^2+c^2}},\frac{b}{\sqrt{a^2+b^2+c^2}},\frac{c}{\sqrt{a^2+b^2+c^2}})$

$\therefore (\frac{-1}{\sqrt{(-1{)}^2+1^2+2^2}},\frac{1}{\sqrt{(-1{)}^2+1^2+2^2}},\frac{2}{\sqrt{(-1{)}^2+1^2+2^2}})=(\pm \frac{1}{\sqrt{6}},\pm \frac{1}{\sqrt{6}},\pm \frac{2}{\sqrt{6}})$ are the direction cosines