Question

If the direction ratios of two lines are given by \[ l+m+n=0, \qquad mn-2ln+lm=0, \] then the angle between the lines is:

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{2}$
• D. $0$

Hint:

For pair-of-lines problems in direction ratios, the condition \[ \frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0 \] directly indicates perpendicular lines.

Answer 1

The Correct Option is C

Concept: The given equations represent two lines through the origin in terms of direction ratios $(l,m,n)$. To determine the angle between them, we reduce the equations and check whether the dot product condition for perpendicularity is satisfied.

Step 1: Using the linear relation.
Given, \[ l+m+n=0 \] Therefore, \[ n=-(l+m) \] Substitute this into the second equation: \[ mn-2ln+lm=0 \] \[ m(-(l+m))-2l(-(l+m))+lm=0 \] \[ -lm-m^2+2l^2+2lm+lm=0 \] Simplifying: \[ 2l^2+2lm-m^2=0 \]

Step 2: Forming the quadratic equation.
Divide throughout by $m^2$: \[ 2\left(\frac{l}{m}\right)^2 + 2\left(\frac{l}{m}\right) - 1 = 0 \] Let \[ t=\frac{l}{m} \] Then, \[ 2t^2+2t-1=0 \] The two roots correspond to the two different lines.

Step 3: Using the perpendicularity condition.
For equations of the form \[ al+bm+cn=0 \] and \[ fmn+gnl+hlm=0, \] the lines are perpendicular if \[ \frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0 \] Here, \[ a=b=c=1 \] and \[ f=1,\quad g=-2,\quad h=1 \] Therefore, \[ \frac11+\frac{-2}{1}+\frac11 = 1-2+1 = 0 \] Hence, the two lines are perpendicular. Therefore, the angle between them is \[ \boxed{ \frac{\pi}{2} } \]