Question

If the direction ratios of two lines are given by \[l+2m+3n=0\] and \[3lm+mn-4nl=0\], then angle between the lines is: -
(a) \[\dfrac{\pi }{2}\]
(b) \[\dfrac{\pi }{3}\]
(c) \[\dfrac{\pi }{4}\]
(d) \[\dfrac{\pi }{6}\]

Answer 1

Hint: Assume the equations, \[l+2m+3n=0\] and \[3lm+mn-4nl=0\] as equation (i) and (ii) respectively. Substitute the value of l, found in terms of m and n, from equation (i) in equation (ii). Solve the quadratic equation containing m and n. Find the value of m in terms of n and substitute each value of m to get a corresponding value of l. Use the relation: - \[\cos \theta =\dfrac{{{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}}{\sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}}.\sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}\] to find the value of \[\theta \].

Complete step by step answer:
We have been given two lines: -
\[\Rightarrow l+2m+3n=0\] - (i)
\[\Rightarrow 3lm+mn-4nl=0\] - (ii)
Substituting the value of l from equation (i) in (ii), we get,
\[\begin{align}
  & \Rightarrow 3m\left( -2m-3n \right)+mn-4n\left( -2m-3n \right)=0 \\
 & \Rightarrow -6{{m}^{2}}-9mn+mn+8mn+12{{n}^{2}}=0 \\
 & \Rightarrow 6{{m}^{2}}=12{{n}^{2}} \\
 & \Rightarrow {{m}^{2}}=2{{n}^{2}} \\
\end{align}\]
Taking square root both sides, we get,
\[\Rightarrow m=\pm \sqrt{2}n\]
Case (i): - When \[m=\sqrt{2}n\],
\[\begin{align}
  & \Rightarrow l=-\left( 2m+3n \right) \\
 & \Rightarrow l=-\left( 2\sqrt{2}n+3n \right) \\
 & \Rightarrow l=-\left( 2\sqrt{2}+3 \right)n \\
\end{align}\]
Case (ii): - When \[m=-\sqrt{2}n\],
\[\begin{align}
  & \Rightarrow l=-\left( 2m+3n \right) \\
 & \Rightarrow l=-\left( -2\sqrt{2}n+3n \right) \\
 & \Rightarrow l=-\left( -2\sqrt{2}+3 \right)n \\
\end{align}\]
Therefore, the two set of values of l, m and n can be represented as: - \[\left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right)\] and \[\left( {{l}_{2}},{{m}_{2}},{{n}_{2}} \right)\].
Here, \[\left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right)=\left( -\left( 2\sqrt{2}+3 \right),\sqrt{2},1 \right)\]
And, \[\left( {{l}_{2}},{{m}_{2}},{{n}_{2}} \right)=\left( -\left( -2\sqrt{2}+3 \right),-\sqrt{2},1 \right)\]
Now, the angle between two lines are given as: -
\[\Rightarrow \cos \theta =\dfrac{{{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}}{\sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}}.\sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}\]
Substituting all the values, we get,
\[\begin{align}
  & \Rightarrow \cos \theta =\dfrac{\left[ -\left( 2\sqrt{2}+3 \right) \right]\times \left[ -\left( -2\sqrt{2}+3 \right) \right]+\sqrt{2}\times \left( -\sqrt{2} \right)+1\times 1}{\sqrt{{{\left[ -\left( 2\sqrt{2}+3 \right) \right]}^{2}}+{{\left( \sqrt{2} \right)}^{2}}+{{1}^{2}}}.\sqrt{{{\left[ -\left( -2\sqrt{2}+3 \right) \right]}^{2}}+{{\left( -\sqrt{2} \right)}^{2}}+{{1}^{2}}}} \\
 & \Rightarrow \cos \theta =\dfrac{\left( 3+2\sqrt{2} \right)\times \left( 3-2\sqrt{2} \right)-2+1}{\sqrt{{{\left( 3+2\sqrt{2} \right)}^{2}}+2+1}.\sqrt{{{\left( 3-2\sqrt{2} \right)}^{2}}+2+1}} \\
 & \\
 & \Rightarrow \cos \theta =\dfrac{9-8-2+1}{\sqrt{{{\left( 3+2\sqrt{2} \right)}^{2}}+3}.\sqrt{{{\left( 3-2\sqrt{2} \right)}^{2}}+3}} \\
\end{align}\]
\[\Rightarrow \cos \theta =0\], since the numerator will become 0.
We know that, \[\cos {{90}^{\circ }}=\dfrac{\pi }{2}=0\].
\[\Rightarrow \theta =\dfrac{\pi }{2}\]

So, the correct answer is “Option A”.

Note: One may note that after solving the quadratic equation we have considered both the values of m and no values are neglected. This is because we have to find two values of each m, n and l so that we can apply the formula to find the angle. We need not to find exact values of l, m and n because they are ratios and will get cancelled when we are finding angles.