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

$\frac{\pi }{2}$

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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: $3m(-2m - 3n) - 4n(-2m - 3n) + mn = 0$ $-6m^2 - 9mn + 8mn + 12n^2 + mn = 0$ $-6m^2 + 12n^2 = 0$ $m^2 = 2n^2$ $m = \pm \sqrt{2}n$. Case $1$: If $n = 1$,then $m = \sqrt{2}$. $l = -2(\sqrt{2}) - 3(1) = -2\sqrt{2} - 3$. Direction ratios of the first line are $(-2\sqrt{2} - 3, \sqrt{2}, 1)$. Case $2$: If $n = 1$,then $m = -\sqrt{2}$. $l = -2(-\sqrt{2}) - 3(1) = 2\sqrt{2} - 3$. Direction ratios of the second line are $(2\sqrt{2} - 3, -\sqrt{2}, 1)$. Let the direction vectors be $\vec{v_1} = (-2\sqrt{2} - 3, \sqrt{2}, 1)$ and $\vec{v_2} = (2\sqrt{2} - 3, -\sqrt{2}, 1)$. Calculating the dot product $\vec{v_1} \cdot \vec{v_2}$: $(-2\sqrt{2} - 3)(2\sqrt{2} - 3) + (\sqrt{2})(-\sqrt{2}) + (1)(1)$ $= (-(2\sqrt{2} + 3)(2\sqrt{2} - 3)) - 2 + 1$ $= -(8 - 9) - 1 = 1 - 1 = 0$. Since the dot product is $0$,the lines are perpendicular. Thus,the angle between them is $\frac{\pi}{2}$.

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

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