If the direction ratios of two lines are give | vedclass

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(D) Given equations are $a+2b+c=0$ $(i)$ and $11bc+6ca-14ab=0$ (ii). From $(i)$, $b = \frac{-(a+c)}{2}$. Substituting this into (ii): $11\left(\frac{-

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$\frac{\pi}{2}$

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(D) Given equations are $a+2b+c=0$ $(i)$ and $11bc+6ca-14ab=0$ (ii). From $(i)$, $b = \frac{-(a+c)}{2}$. Substituting this into (ii): $11\left(\frac{-(a+c)}{2}\right)c + 6ac - 14a\left(\frac{-(a+c)}{2}\right) = 0$ $\Rightarrow -\frac{11}{2}ac - \frac{11}{2}c^2 + 6ac + 7a^2 + 7ac = 0$ Multiplying by $2$: $-11ac - 11c^2 + 12ac + 14a^2 + 14ac = 0$ $\Rightarrow 14a^2 + 15ac - 11c^2 = 0$ Factoring the quadratic: $(7a+11c)(2a-c) = 0$. Case $1$: $2a = c \Rightarrow a = 1, c = 2$. Then $b = \frac{-(1+2)}{2} = -1.5$. To avoid fractions, let $a=2, c=4, b=-3$. Direction ratios: $(2, -3, 4)$. Case $2$: $7a = -11c \Rightarrow a = -11, c = 7$. Then $b = \frac{-(-11+7)}{2} = 2$. Direction ratios: $(-11, 2, 7)$. Let the direction ratios be $\vec{n_1} = (2, -3, 4)$ and $\vec{n_2} = (-11, 2, 7)$. The dot product $\vec{n_1} \cdot \vec{n_2} = (2)(-11) + (-3)(2) + (4)(7) = -22 - 6 + 28 = 0$. Since the dot product is $0$, the lines are perpendicular. Therefore, the angle between the lines is $\frac{\pi}{2}$.

If the direction ratios of two lines are given by $a+2b+c=0$ and $11bc+6ca-14ab=0$, then the angle between these lines is

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