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The obtuse angle between the lines whose direction ratios are determined by the equations $a+b+c=0,2 a b+2 a c-b c=0$ is
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Verified Answer
The correct answer is:
$\frac{2 \pi}{3}$
We have, $a+b+c=0$ and $2 a b+2 a c-b c=0$
$\begin{aligned}
& a=-(b+c) \\
& \therefore \quad-2(b+c) b+-2(b+c) c-b c=0 \\
& \Rightarrow \quad-2 b^2-2 b c-2 b c-2 c^2-b c=0 \\
&
\end{aligned}$
$\begin{gathered}\Rightarrow \quad 2 b^2+5 b c+2 c^2=0 \Rightarrow 2 b^2+4 b c+b c+2 c^2=0 \\ \Rightarrow \quad(2 b+c)(b+2 c)=0 \Rightarrow b=-2 c, \quad b=-c / 2 \\ a=-(-2 c+c)=c \Rightarrow a=-\left(\frac{-c}{2}+c\right)=\frac{-c}{2} \\ \therefore \quad a_1=1, b_1=-2, c_1=1, a_2=\frac{-1}{2}, b_2=\frac{-1}{2}, c_2=1 \\ a_1=1, b_1=-2, c_1=1, a_2=1, b_2=1, c_2=-2 \\ \cos \theta=\frac{(1)(1)+(-2)(1)+(1)(-2)}{\sqrt{1+4+1} \sqrt{1+1+4}} \\ \cos \theta=\frac{-1}{2} \Rightarrow \theta=\frac{2 \pi}{3}\end{gathered}$
$\begin{aligned}
& a=-(b+c) \\
& \therefore \quad-2(b+c) b+-2(b+c) c-b c=0 \\
& \Rightarrow \quad-2 b^2-2 b c-2 b c-2 c^2-b c=0 \\
&
\end{aligned}$
$\begin{gathered}\Rightarrow \quad 2 b^2+5 b c+2 c^2=0 \Rightarrow 2 b^2+4 b c+b c+2 c^2=0 \\ \Rightarrow \quad(2 b+c)(b+2 c)=0 \Rightarrow b=-2 c, \quad b=-c / 2 \\ a=-(-2 c+c)=c \Rightarrow a=-\left(\frac{-c}{2}+c\right)=\frac{-c}{2} \\ \therefore \quad a_1=1, b_1=-2, c_1=1, a_2=\frac{-1}{2}, b_2=\frac{-1}{2}, c_2=1 \\ a_1=1, b_1=-2, c_1=1, a_2=1, b_2=1, c_2=-2 \\ \cos \theta=\frac{(1)(1)+(-2)(1)+(1)(-2)}{\sqrt{1+4+1} \sqrt{1+1+4}} \\ \cos \theta=\frac{-1}{2} \Rightarrow \theta=\frac{2 \pi}{3}\end{gathered}$
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