If the direction ratios of two lines are given by 3lm − 4ln + mn = 0 and l + 2m + 3n = 0, then the angle between the lines is

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Aicharya · Accepted Answer

Given, 3lm − 4ln + mn = 0 … (i) l + 2m + 3n = 0 … (ii) $\Rightarrow \quad l=-2 m-3 n$ From Eq. (i), we get 3(− 2m − 3n)m − 4(− 2m − 3n) n + mn = 0 $-6 m^2-9 n m+8 m n+12 n^2+m n=0$ $ \begin{gathered} \Rightarrow \quad 6 m^2=12 n^2 \Rightarrow m^2=2 n^2 \ m= \pm \sqrt{2} n \end{gathered} $ $	herefore\left(l_1, m_1, n_1ight)$ and $\left(l_2, m_2, n_2ight)$ are direction ratios where, $m_1=\sqrt{2} n_1$ and $m_2=\sqrt{2} n_2$ $\begin{aligned} l_1 & =-(2 \sqrt{2}+3) n_1 	ext { and } l_2=-(-2 \sqrt{2}+3) n_2 \ 	herefore \cos 	heta & =\frac{l_1 l_2+m_1 m_2+n_1 n_2}{\sqrt{l_1^2+m_1^2+n_1^2} \sqrt{l_2^2+m_2^2+n_2^2}} \ & =\frac{n_1 n_2[-2+(9-8)+1]}{\sqrt{l_1^2+m_1^2+n_1^2} \sqrt{l_2^2+m_2^2+n_2^2}}=0 \ \cos 	heta & =\cos \pi / 2 \Rightarrow 	heta=\pi / 2\end{aligned}$

If the direction ratios of two lines are given by 3lm − 4ln + mn = 0 and l + 2m + 3n = 0, then the angle…

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