Question

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

• A. $\left[\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right]$
• B. $\left[\frac{-1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right]$
• C. $\left[\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right]$
• D. $\left[\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{-3}{\sqrt{14}}\right]$

Answer 1

Answer: (A)

From the first relation, $l=5m-3n$. Putting this value of l in second relation
$7{\left(5m-3n\right)}^2+5m^2-3n^2=0$
$\Rightarrow 180m^2-210mn+60n^2=0$
or $6m^2-7mn+2n^2=0$
Note that it, being quadratic in m, n, gives two sets of values of m, n, and hence gives the d.r.s. of two lines. Now, factorising it, we get
$6m^2-3mn+4mn+2n^2=0$
or $\left(2m-n\right)\left(3m-2n\right)=0$
$\Rightarrow either2m-n=0$, or $3m-2n=0$
Taking $2m-n=0$ we get $2m=n$.
Also putting $m=n/2$ in $l=5m-3n$, we get
$l=\left(5n/2\right)-3n\Rightarrow l=-n/2\Rightarrow n=-2l$
Thus, we get, $-2l=2m=n$ or $\frac{l}{-1}=\frac{m}{1}=\frac{n}{2}$
$\Rightarrow$ d.r.s. of one line are $-1,1,2$.
Hence, the d,c,s. of one line are
$\left[\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}\right]$ or $\left[\frac{1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{-2}{\sqrt{6}}\right]$
Taking $3m-2n=0$, we get
$3m=2n$ or $m=\frac{2n}{3}.$
Putting this value in $l=5m-3n$, we obtain
$l=5\times \frac{2n}{3}-3n=\frac{n}{2}$ or $n=3l$
Thus $3l=\frac{3m}{2}=n\Rightarrow \frac{l}{1}=\frac{m}{2}=\frac{n}{3}$
$\Rightarrow$ the d.r’.ss of the second line are $1,2,3$; and hence d.c.s. of second line are $\left[\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right]$
or $\left[\frac{-1}{\sqrt{14}},\frac{-2}{\sqrt{14}},\frac{-3}{\sqrt{14}}\right]$