Question

If $\theta$ is the angle between lines whose direction ratios is given by $a-b+c=0$ and $3ab+bc=0,$ then $7{\cos }^2\theta -1=$

Answer 1

Given :
$3ab+bc=0\cdots \left(i\right)$ and $a-b+c=0\cdots \left(ii\right)$
$\Rightarrow b(3a+c)=0$
$\Rightarrow b=0$ or $3a+c=0\cdots \left(iii\right)$
putting in $\left(ii\right)$
for $b=0\Rightarrow a+c=0$
$\Rightarrow \frac{a}{-1}=\frac{c}{1}$
$\Rightarrow (a_1,b_1,c_1)=(-1,0,1)$
for $3a+c=0,i.e.\frac{a}{-1}=\frac{c}{3}$
$\Rightarrow 2a+b=0$ from $\left(ii\right)$
hence,
$\Rightarrow \frac{a}{-1}=\frac{b}{2}=\frac{c}{3}$

$\Rightarrow (a_2,b_2,c_2)=(-1,2,3)$
then $\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\Rightarrow \cos \theta =\frac{1+0+3}{\sqrt{2}\sqrt{14}}=\frac{2}{\sqrt{7}}\therefore 7{\cos }^2\theta -1=3$