• # question_answer The acute angle between two lines such that the direction cosines I, m, n of each of them satisfy the equations I + m + n = 0 and ${{\operatorname{I}}^{2}}+{{\operatorname{m}}^{2}}-{{\operatorname{n}}^{2}}=0$ is:     JEE Main  Online Paper (Held On 22 April 2013) A)  ${{15}^{0}}$                                        B)  ${{30}^{0}}$ C)  ${{60}^{0}}$                                        D)  ${{45}^{0}}$

Let ${{l}_{1}},{{m}_{1}},{{n}_{1}}$and ${{l}_{2}},{{m}_{2}},{{n}_{2}}$be the d. c of line 1 and 2 respectively, then as given ${{l}_{1}}+{{m}_{1}}+{{n}_{1}}=0$ and ${{l}_{2}}+{{m}_{2}}+{{n}_{2}}=0$ and $l_{1}^{2}+m_{1}^{2}-n_{1}^{2}=0$and $l_{2}^{2}+m_{2}^{2}-n_{2}^{2}=0$ ($\because$$l+m+n=0$and ${{l}^{2}}+{{m}^{2}}-{{n}^{2}}=0$) Angle between lines, $\theta$is $\cos \theta ={{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}$                                ?(1) As given ${{l}^{2}}+{{m}^{2}}={{n}^{2}}$and $l+m=-n$ $\Rightarrow$${{(-n)}^{2}}-2lm={{n}^{2}}\Rightarrow 2lm=0$or $lm=0$ So ${{l}_{1}}{{m}_{1}}=0,{{l}_{2}}{{m}_{2}}=0$ If ${{l}_{1}}=0,{{m}_{1}}\ne 0$the ${{l}_{1}}{{m}_{2}}=0$ If ${{m}_{1}}=0,{{l}_{1}}\ne 0$then ${{l}_{2}}{{m}_{1}}=0$ If ${{l}_{2}}=0,{{m}_{2}}\ne 0$then ${{l}_{2}}{{m}_{1}}=0$ If ${{m}_{2}}=0,{{l}_{2}}\ne 0$then ${{l}_{1}}{{m}_{2}}=0$ Also ${{l}_{1}}{{l}_{2}}=0$and ${{m}_{1}}{{m}_{2}}=0$ ${{l}^{2}}+{{m}^{2}}-{{n}^{2}}={{l}^{2}}+{{m}^{2}}+{{n}^{2}}-2{{n}^{2}}=0$ $\Rightarrow$$1-2{{n}^{2}}=0\Rightarrow n=\pm \frac{1}{\sqrt{2}}$ $\therefore$  ${{n}_{1}}=\pm \frac{1}{\sqrt{2}},{{n}_{2}}=\pm \frac{1}{\sqrt{2}}$ $\therefore$$\cos \theta =\frac{1}{2}\theta ={{60}^{o}}$(acute angle)