Question

Direction cosines $l,m,n$ of two lines are connected by the equation $l-5m+3n=0$ and $7l^2+5m^2-3n^2=0$. The direction cosines of one of the lines are

• A. $-1/\sqrt{6},1/\sqrt{6},2/\sqrt{6}$
• B. $2/\sqrt{6},1/\sqrt{6},-1/\sqrt{6}$
• C. $1/\sqrt{6},2/\sqrt{6},-1/\sqrt{6}$
• D. $1/\sqrt{6},1/\sqrt{6},2/\sqrt{6}$

Answer 1

The correct option is A $-1/\sqrt{6},1/\sqrt{6},2/\sqrt{6}$

Consider the problem

Given

$l-5m+3n=0$

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

And

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

From $\left(i\right)$ put the value of $l$ in equation $\left(ii\right)$

Then,

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

Hence

$30(2m-n)(3m-2n)=0$

thus,

$2m=n$ and $3m=2n$

therefore,

$\frac{m}{1}=\frac{n}{2}=\frac{(5m-3n)}{5-2.3}=\frac{1}{\sqrt{6}}$

And

$\frac{m}{2}=\frac{n}{3}=\frac{(5m-3n)}{5.2-3.3}=\frac{1}{\sqrt{14}}$

Hence the required Directions cosines of line are

$-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}$