The acute angle between the lines whose direction cosines are given by the equations $l+m+n=0$ and $2 l m+2 l n-m n=0$ is

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Aicharya · Accepted Answer

Given equations are,

On squaring both side of the Eq. (i). $ \begin{array}{cc} & l^2+m^2+n^2+2 l m+2 m n+2 n l=0 \\ \because & l^2+m^2+n^2=1, \end{array} $

From Eqs. (ii) and (iv), $ 3 m n=-1 \Rightarrow m n=-\frac{1}{3} $ From Eq. (iii) $6 l^2-1=0$, let root of this equation is $l_1$ and $l_2$, so $l_1 l_2=-\frac{1}{6}$. Now, from Eqs. (i) and (ii), $ 2 l m+(2 l-m)(-l-m)=0 $ $ \begin{array}{rlrl} \Rightarrow & 2 l^2-l m-m^2 =0 \\ \Rightarrow & 2\left(\frac{l}{m}\right)^2-\left(\frac{l}{m}\right)-1 =0 \\ \Rightarrow & \frac{l_1 l_2}{m_1 m_2} =\frac{-1}{2} \\ \Rightarrow & \frac{l_1 l_2}{-1 / 6} =\frac{m_1 m_2}{2 / 6} \\ & \text { Similarly, } \frac{l_1 l_2}{-1 / 6} =\frac{n_1 n_2}{2 / 6} \end{array} $ So, $\quad \cos \theta=\left|l_1 l_2+m_1 m_2+n_1 n_2\right|=\frac{3}{6}=\frac{1}{2}$ $ \Rightarrow \quad \theta=\frac{\pi}{3} $

The acute angle between the lines whose direction cosines are given by the equations $l+m+n=0$ and $2 l m+2…

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