Question

The d.rs of two lines are given by the equations $a+b+c=0$ and $2ab+2ac-bc=0$. Then the angle between the lines is

• A. $\pi$
• B. $\frac{2\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is B $\frac{2\pi }{3}$

Given, $(a+b+c)=0$
$(a+b+c{)}^2=0$
$a^2+b^2+c^2+2ab+2bc+2ac=0$
Taking the modulus/distance form origin as $1$, we get
$a^2+b^2+c^2=1$
$1+2ab+2bc+2ac=0$
$1+bc+2bc=0$
$3bc=-1$
$bc=\frac{-1}{3}$ ...(i)
Also given, $2ab+2ac-bc=0$
$2a(b+c)-bc=0$
$2a(b+c)+\frac{1}{3}=0$ ...(from i)
$a(-a)=\frac{-1}{6}$ ...(from original equation)
$-a^2=\frac{-1}{6}$
$a=\pm \frac{1}{\sqrt{6}}$
Now $a+b+c=0$
$b+c=\mp \frac{1}{\sqrt{6}}$
Let us consider $b+c=\frac{1}{\sqrt{6}}$
And $bc=\frac{-1}{3}$
From this we get a quadratic equation in (c)
$2\sqrt{6}{c}^2+3c-\sqrt{6}=0$
$c=\frac{-2}{\sqrt{6}}$ and $c=\frac{1}{\sqrt{6}}$ therefore
$b=\frac{-1}{\sqrt{6}}$ and $b=\frac{2}{\sqrt{6}}$
Hence summing up, we get the vector equations of the lines as
$\frac{1}{\sqrt{6}}{i}^{\prime }+\frac{-1}{\sqrt{6}}{j}^{\prime }+\frac{-2}{\sqrt{6}}{k}^{\prime }$ and
$\frac{1}{\sqrt{6}}{i}^{\prime }+\frac{2}{\sqrt{6}}{j}^{\prime }+\frac{1}{\sqrt{6}}{k}^{\prime }$
Therefore using dot, product, we get
$\left(1\right)\left(1\right)\cos \theta =\frac{1-2-2}{6}$
$\cos \theta =\frac{-1}{2}$
$\theta ={120}^0$