Question

# The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and ${{l}^{2}}={{m}^{2}}+{{n}^{2}}$ is(a) $\dfrac{\pi }{3}$(b) $\dfrac{\pi }{4}$(c) $\dfrac{\pi }{6}$(d) $\dfrac{\pi }{2}$

Hint: First, before proceeding for this, we will use the given conditions to get the value of m and n by substituting the value of l from given equations of line. Then, by using the identity of the whole square as ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to proceed and then we get the values of m and n. Then, by using the values of m and n, we get the direction ratios and then we get the desired angle between them.

In this question, we are supposed to find the angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and ${{l}^{2}}={{m}^{2}}+{{n}^{2}}$.
So, before proceeding for this, we will use the given conditions to get the value of m and n by substituting the value of l from given equations of line.
The, by substituting the value of l from equation $l+m+n=0$to equation ${{l}^{2}}={{m}^{2}}+{{n}^{2}}$, we get:
${{\left( m+n \right)}^{2}}={{m}^{2}}+{{n}^{2}}$
Now, by using the identity of the whole square as ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to proceed as:
\begin{align} & {{m}^{2}}+{{n}^{2}}+2mn={{m}^{2}}+{{n}^{2}} \\ & \Rightarrow {{m}^{2}}+{{n}^{2}}-{{m}^{2}}-{{n}^{2}}+2mn=0 \\ & \Rightarrow 2mn=0 \\ & \Rightarrow m=0,n=0 \\ \end{align}
So, the above expression gives the conclusion that either m=0 or n=0.
Now, we will check for both the conditions one by one starting with m=0 in the first equation of line $l+m+n=0$, we get:
\begin{align} & l+0+n=0 \\ & \Rightarrow l+n=0 \\ & \Rightarrow l=-n \\ \end{align}
So, we get the direction ratios for the equation $l+m+n=0$as:
$\dfrac{l}{1}=\dfrac{m}{0}=\dfrac{n}{-1}....\left( i \right)$
Similarly, when we substitute n=0 in the equation $l+m+n=0$, we get:
\begin{align} & l+m+0=0 \\ & \Rightarrow l+m=0 \\ & \Rightarrow l=-m \\ \end{align}
So, we get the direction ratios for the equation $l+m+n=0$as:
$\dfrac{l}{1}=\dfrac{m}{-1}=\dfrac{n}{0}.....\left( ii \right)$
Now, by using the direction ratios from equation (i) and (ii), we get:
$\left( 1,0,-1 \right),\left( 1,-1,0 \right)$
Now, to get the angle between them, we use the formula as:
$\cos \theta =\dfrac{1\times 1+0\times \left( -1 \right)+\left( -1 \right)\times 0}{\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{0}^{2}}}\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{0}^{2}}}}$
Now, by solving the above expression, we get:
\begin{align} & \cos \theta =\dfrac{1+0+0}{\sqrt{2}\times \sqrt{2}} \\ & \Rightarrow \cos \theta =\dfrac{1}{2} \\ & \Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{1}{2} \right) \\ & \Rightarrow \theta =\dfrac{\pi }{3} \\ \end{align}
So, we get the angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and ${{l}^{2}}={{m}^{2}}+{{n}^{2}}$ as $\dfrac{\pi }{3}$.

So, the correct answer is “Option A”.

Note: Now, to solve this type of questions we must know the angle condition of the direction ratios which is used above in the question as both the lines are perpendicular so the dot product is zero is given by the condition that:
$\cos \theta =\dfrac{\left| {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}} \right|}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}}$
Where $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)$ and $\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ are the points of the line.