Question

If $\theta$ is the angle between lines whose direction ratios is given by the relation $a+b+c=0$ and $2ac-3bc=0,$ then $5-76{\cos }^2\theta =$

Answer 1

Given relation of d.r's :
$a+b+c=0\cdots \left(i\right)$ and $2ac-3bc=0\cdots \left(ii\right)$
$\Rightarrow c(2a-3b)=0$
$\Rightarrow c=0$ or $2a=3b$
putting in $\left(i\right)$
For $c=0\Rightarrow a+b=0$
$\Rightarrow \frac{a}{-1}=\frac{b}{1}$
$\Rightarrow (a_1,b_1,c_1)=(-1,1,0)$
For $2a=3b;i.e.\frac{a}{3}=\frac{b}{2}:$
$\Rightarrow \frac{3b}{2}+b+c=0$ from $\left(i\right)$
$\Rightarrow \frac{b}{2}=\frac{c}{-5}\Rightarrow (a_2,b_2,c_2)=(3,2,-5)$
$\Rightarrow \cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\Rightarrow \cos \theta =\frac{-3+2+0}{\sqrt{2}\sqrt{38}}\Rightarrow 76{\cos }^2\theta =1$
$\therefore 5-76{\cos }^2\theta =4$