Question

The angle between the lines whose direction consider satisfy equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

• A. $\frac{\mathrm{\pi }}{3}$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\frac{\mathrm{\pi }}{6}$
• D. $\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{3}$

Explanation for correct answer Option(s)

Find the angle between the line.

Consider the given Equation as,

$l+m+n=0......\left(1\right)$

$l^2+m^2-n^2=0......\left(2\right)$

Now substract $m$ and $n$ on both sides of the equation (1).

$$\begin{gathered} l+m+n-m-n=-m-n \\ I=-m-n......\left(3\right) \end{gathered}$$

Substitute the value of $l$ in the Equation (2) becomes

$$\begin{gathered} \Rightarrow {\left(-m-n\right)}^2+m^2-n^2=0 \\ \Rightarrow {\left(-m\right)}^2+{\left(-n\right)}^2+2\left(-m\right)\left(-n\right)+m^2-n^2=0 \\ \Rightarrow m^2+n^2+2mn+m^2-n^2=0 \\ \Rightarrow 2m^2+2mn=0 \\ \Rightarrow 2m\left(m+n\right)=0 \\ \Rightarrow 2m=0m+n=0 \\ \Rightarrow m=0m=-n \end{gathered}$$

Substitute $m=0$ in the equation (3)

$$\begin{gathered} l=-0-n \\ l=-n \end{gathered}$$

Since, the Equation satisfies

$$\begin{gathered} \Rightarrow \frac{l_1}{-1}=\frac{m_1}{0}=\frac{n_1}{1} \\ \Rightarrow \left(l_1,m_1,n_1\right)=\left(-1,0,1\right) \end{gathered}$$

Substitute $m=-n$ in the Equation (3) becomes

$$\begin{gathered} l=-\left(-n\right)-n \\ l=n-n \\ l=0 \end{gathered}$$

Since, the Equation satisfies

$$\begin{gathered} \Rightarrow \frac{l_2}{0}=\frac{m_2}{-1}=\frac{n_2}{1} \\ \Rightarrow \left(l_2,m_2,n_2\right)=\left(0,-1,1\right) \end{gathered}$$

The Equation satisfies

$$\begin{gathered} \Rightarrow \frac{l_1}{-1}=\frac{m_1}{0}=\frac{n_1}{1} \\ \Rightarrow \left(l_1,m_1,n_1\right)=\left(-1,0,1\right) \end{gathered}$$

To find the value of $\theta$, We have

$\cos \theta =\frac{l_1{l}_2+m_1{m}_2+n_1{n}_2}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_2^2+m_2^2-n_2^2}}$

Substitute, $l_1=-1,m_1=0,n_1=1,l_2=0,m_2=-1,n_2=1$in the above Equation

$$\begin{gathered} \cos \theta =\frac{\left(-1\right)\left(0\right)+\left(0\right)\left(-1\right)+\left(1\right)\left(1\right)}{\sqrt{1+0+1}\sqrt{0+1+1}} \\ \cos \theta =\frac{0+0+1}{\sqrt{2}\sqrt{2}} \\ \cos \theta =\frac{1}{{\left(\sqrt{2}\right)}^2} \\ \cos \theta =\frac{1}{2} \\ \theta ={\cos }^{-1}\frac{1}{2} \\ \theta =\frac{\mathrm{\pi }}{3} \end{gathered}$$

Hence, the angel between the lines is $\frac{\mathrm{\pi }}{3}$

Therefore, the correct answer is Option A.