If the direction ratios of two lines are give | vedclass

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(D) We have the equations for the direction ratios $(l, m, n)$ as:$3lm - 4ln + mn = 0$ $\ldots$ $(i)$$l + 2m + 3n = 0$ $\ldots$ (ii)From (ii),we get $

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$\frac{\pi}{2}$

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(D) We have the equations for the direction ratios $(l, m, n)$ as:$3lm - 4ln + mn = 0$ $\ldots$ $(i)$$l + 2m + 3n = 0$ $\ldots$ (ii)From (ii),we get $l = -(2m + 3n)$.Substituting this into $(i)$:$-3(2m + 3n)m + 4(2m + 3n)n + mn = 0$$-6m^2 - 9mn + 8mn + 12n^2 + mn = 0$$-6m^2 + 12n^2 = 0 \Rightarrow m^2 = 2n^2 \Rightarrow m = \pm \sqrt{2}n$.Case $1$: $m = \sqrt{2}n$. Then $l = -(2\sqrt{2} + 3)n$. The direction ratios are $(l_1, m_1, n_1) = (-(3 + 2\sqrt{2}), \sqrt{2}, 1)$.Case $2$: $m = -\sqrt{2}n$. Then $l = -(-2\sqrt{2} + 3)n = (2\sqrt{2} - 3)n$. The direction ratios are $(l_2, m_2, n_2) = (-(3 - 2\sqrt{2}), -\sqrt{2}, 1)$.For the angle $	heta$ between the lines:$\cos 	heta = \frac{l_1l_2 + m_1m_2 + n_1n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}$The numerator is $(3 + 2\sqrt{2})(3 - 2\sqrt{2}) + (\sqrt{2})(-\sqrt{2}) + (1)(1) = (9 - 8) - 2 + 1 = 1 - 2 + 1 = 0$.Since the numerator is $0$,$\cos 	heta = 0$,which implies $	heta = \frac{\pi}{2}$.

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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