An angle between the  lines whose direction cosines are given by the equation l+3m+5n=0 and· 5lm−2mn+6nl=0, is

# An angle between the  lines whose direction cosines are given by the equation $l+3m+5n=0$ and· $5lm-2mn+6nl=0$, is

1. A

${\mathrm{cos}}^{-1}\left(1/8\right)$

2. B

${\mathrm{cos}}^{-1}\left(1/3\right)$

3. C

${\mathrm{cos}}^{-1}\left(1/4\right)$

4. D

${\mathrm{cos}}^{-1}\left(1/6\right)$

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### Solution:

The given equations are

$l+3m+5n=0$                                 …(i)

$5lm-2mn+6nl=0$                         …(ii)

From (i) $l=-3m-5n$

Putting this value of $I$ in (ii), we have

$5\left(-3m-5m\right)m-2mn+6n\left(-3m-5n\right)=0$

$\begin{array}{l}⇒-15{m}^{2}-30{n}^{2}-45mn=0⇒{m}^{2}+2{n}^{2}+3mn=0\\ ⇒{m}^{2}+3mn+2{n}^{2}=0⇒m\left(m+2n\right)+n\left(m+2n\right)=0\end{array}$

either $m=-n$ or $m=-2n$

For, $m=-n,l=-2n$; for $;m=-2n,l=n$

$\therefore$ Direction ratios of two lines are

$⟨-2n,-n,n⟩$ and $⟨n,-2n,n⟩$

i.e. $⟨-2,-1,1⟩$ and $⟨1,-2,1⟩$

$\therefore$ The required angle is

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