Mathematics - Vector Question with Solution | TestHub

Given $$\begin{array}{rl}& l+3m+5n=0\\ & \text{ and }5lm-2mn+6nl=0\end{array}$$ From eq. (1) we have $$l=-3m-5n$$ Put the value of $l$ in eq. (2), we get; $$\begin{array}{rl}& 5(-3m-5n)m-2mn+6n(-3m-5n)=0\\ & \Rightarrow 15m^2+45mn+30n^2=0\\ & \Rightarrow m^2+3mn+2n^2=0\\ & \Rightarrow m^2+2mn+mn+2n^2=0\\ & \Rightarrow (m+n)(m+2n)=0\\ & \therefore m=-n\text{ or }m=-2n\\ & \text{ For }m=-n,l=-2n\\ & \text{ And for }m=-2n,l=n\end{array}$$ $$\begin{array}{rl}& \therefore (l,m,n)=(-2n,-n,n)\text{ Or }(l,m,n)\\ & =(n,-2n,n)\\ & \Rightarrow (l,m,n)=(-2,-1,1)\text{ Or }(l,m,n)\\ & =(1,-2,1)\end{array}$$ Therefore, angle between the lines is given as: $$\begin{array}{rl}& \cos (\theta )=\frac{(-2)(1)+(-1)\cdot (-2)+(1)(1)}{\sqrt{6}\cdot \sqrt{6}}\\ & \Rightarrow \cos (\theta )=\frac{1}{6}\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{6}\right)\end{array}$$