If the direction ratios of two lines are give | vedclass

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(A) Given equations are $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$. From the second equation,$l = -2m - 3n$. Substituting this into the first equation

Vedclass · Accepted Answer

$\pi /2$

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(A) Given equations are $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$. From the second equation,$l = -2m - 3n$. Substituting this into the first equation: $3(-2m - 3n)m - 4(-2m - 3n)n + mn = 0$. $-6m^2 - 9mn + 8mn + 12n^2 + mn = 0$. $-6m^2 + 12n^2 = 0$,which implies $m^2 = 2n^2$,so $m = \pm \sqrt{2}n$. Case $1$: $m = \sqrt{2}n$. Then $l = -2(\sqrt{2}n) - 3n = -(2\sqrt{2} + 3)n$. Direction ratios are $(-(2\sqrt{2} + 3), \sqrt{2}, 1)$. Case $2$: $m = -\sqrt{2}n$. Then $l = -2(-\sqrt{2}n) - 3n = (2\sqrt{2} - 3)n$. Direction ratios are $((2\sqrt{2} - 3), -\sqrt{2}, 1)$. Let the direction ratios be $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$. $a_1 a_2 + b_1 b_2 + c_1 c_2 = (-(2\sqrt{2} + 3))(2\sqrt{2} - 3) + (\sqrt{2})(-\sqrt{2}) + (1)(1)$. $= -((2\sqrt{2})^2 - 3^2) - 2 + 1 = -(8 - 9) - 1 = -(-1) - 1 = 1 - 1 = 0$. Since the dot product of the direction ratios is $0$,the lines are perpendicular. Therefore,the angle between them is $\pi /2$.

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,what is the angle between them?

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