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

The Correct Option is 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 $\left(2m-n\right)\left(3m-2n\right) = 0$ $\Rightarrow either 2m - n = 0$, 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, $-2l = 2m = n\quad$ 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? or $\left[\frac{-1}{\sqrt{14}}, \frac{-2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}\right]$