Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Eliminating $n$ between the relations: $u{l}^2+v{m}^2+w{\left(\frac{al+bm}{-c}\right)}^2=0$, so $\left(c^{2}u+w{a}^2\right)\frac{l^2}{m^2}+2abw\frac{l}{m}+\left(b^{2}w+c^{2}v\right)=0$. This is a quadratic in $\frac{l}{m}$ with roots $\frac{l_1}{m_1},\frac{l_2}{m_2}$, so $\frac{l_1{l}_2}{m_1{m}_2}=\frac{b^{2}w+c^{2}v}{c^{2}u+w{a}^2}$. Hence $\frac{l_1{l}_2}{b^{2}w+c^{2}v}=\frac{m_1{m}_2}{c^{2}u+w{a}^2}=\frac{n_1{n}_2}{a^{2}v+b^{2}u}$. For perpendicular lines $l_1{l}_2+m_1{m}_2+n_1{n}_2=0$, so $a^2(v+w)+b^2(w+u)+c^2(u+v)=0$.