Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Given equations are $l+m+n=0$ and $2mn+3ln-5lm=0$.

From the first equation, we have $n=-(l+m)$.

Substituting this into the second equation:

$2m(-l-m)+3l(-l-m)-5lm=0$

$-2lm-2m^2-3l^2-3lm-5lm=0$

$3l^2+10lm+2m^2=0$

Dividing by $m^2$, we get:

$3{\left(\frac{l}{m}\right)}^2+10\left(\frac{l}{m}\right)+2=0$

Let the roots of this quadratic equation be $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$. The product of the roots is:

$\frac{l_1{l}_2}{m_1{m}_2}=\frac{2}{3}\Rightarrow \frac{l_1{l}_2}{2}=\frac{m_1{m}_2}{3}$

Similarly, substituting $m=-(l+n)$ into the second equation:

$2(-l-n)n+3ln-5l(-l-n)=0$

$-2ln-2n^2+3ln+5l^2+5ln=0$

$5l^2+6ln-2n^2=0$

Dividing by $n^2$, we get:

$5{\left(\frac{l}{n}\right)}^2+6\left(\frac{l}{n}\right)-2=0$

The product of the roots is:

$\frac{l_1{l}_2}{n_1{n}_2}=\frac{-2}{5}\Rightarrow \frac{l_1{l}_2}{2}=\frac{n_1{n}_2}{-5}$

Combining the two relations, we get:

$\frac{l_1{l}_2}{2}=\frac{m_1{m}_2}{3}=\frac{n_1{n}_2}{-5}=k$

This gives $l_1{l}_2=2k$, $m_1{m}_2=3k$, and $n_1{n}_2=-5k$.

The angle $\theta$ between the lines is given by:

$\cos \theta =l_1{l}_2+m_1{m}_2+n_1{n}_2$

$\cos \theta =2k+3k-5k=0$

Therefore, $\theta =\frac{\pi }{2}$.

Answer: $\frac{\pi }{2}$