The obtuse angle between the lines whose dire | vedclass

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(B) Given equations are $a+b+c=0$ and $2ab+2ac-bc=0$. Substituting $a=-(b+c)$ into the second equation: $-2(b+c)b - 2(b+c)c - bc = 0$ $-2b^2 - 2bc - 2

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$\frac{2 \pi}{3}$

Vedclass · Answer

(B) Given equations are $a+b+c=0$ and $2ab+2ac-bc=0$. Substituting $a=-(b+c)$ into the second equation: $-2(b+c)b - 2(b+c)c - bc = 0$ $-2b^2 - 2bc - 2bc - 2c^2 - bc = 0$ $-2b^2 - 5bc - 2c^2 = 0$ $2b^2 + 5bc + 2c^2 = 0$ $(2b+c)(b+2c) = 0$. Case $1$: $b = -2c$. Then $a = -(-2c+c) = c$. Direction ratios are $(1, -2, 1)$. Case $2$: $b = -c/2$. Then $a = -(-c/2+c) = -c/2$. Direction ratios are $(-1/2, -1/2, 1)$, which is equivalent to $(1, 1, -2)$. Let the direction ratios be $(a_1, b_1, c_1) = (1, -2, 1)$ and $(a_2, b_2, c_2) = (1, 1, -2)$. Using the formula $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$: $\cos 	heta = \frac{|(1)(1) + (-2)(1) + (1)(-2)|}{\sqrt{1+4+1} \sqrt{1+1+4}} = \frac{|1 - 2 - 2|}{\sqrt{6} \sqrt{6}} = \frac{|-3|}{6} = \frac{1}{2}$. Since we need the obtuse angle, $\cos 	heta = -1/2$, which gives $	heta = \frac{2 \pi}{3}$.

The obtuse angle between the lines whose direction ratios are determined by the equations $a+b+c=0$ and $2ab+2ac-bc=0$ is

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