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(a) $\dfrac{\pi }{3}$

(b) $\dfrac{\pi }{4}$

(c) $\dfrac{\pi }{6}$

(d) $\dfrac{\pi }{2}$

Answer
Verified

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 $\dfrac{\pi }{3}$.

$\cos \theta =\dfrac{\left| {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}} \right|}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}}$

Where $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)$ and $\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ are the points of the line.