A) \[({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}}),\,\,({{n}_{1}}{{l}_{2}}-{{n}_{2}}{{l}_{1}}),\,({{l}_{1}}{{m}_{2}}-{{l}_{2}}{{m}_{1}})\]
B) \[({{l}_{1}}{{l}_{2}}-{{m}_{1}}{{m}_{2}}),\,({{m}_{1}}{{m}_{2}}-{{n}_{1}}{{n}_{2}}),\,({{n}_{1}}{{n}_{2}}-{{l}_{1}}{{l}_{2}})\]
C) \[\frac{1}{\sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}}},\frac{1}{\sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}},\frac{1}{\sqrt{3}}\]
D) \[\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\]
Correct Answer: A
Solution :
Let lines are \[{{l}_{1}}x+{{m}_{1}}y+{{n}_{1}}z+d=0\] ?..(i) and \[{{l}_{2}}x+{{m}_{2}}y+{{n}_{2}}z+d=0\] .....(ii) If \[lx+my+nz+d=0\] is perpendicular to (i) and (ii), then, \[l{{l}_{1}}+m{{m}_{1}}+n{{n}_{1}}=0,\,\,l{{l}_{2}}+m{{m}_{2}}+n{{n}_{2}}=0\,\] \[\Rightarrow \text{ }\,\frac{l}{{{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}}}=\frac{m}{{{n}_{1}}{{l}_{2}}-{{l}_{1}}{{n}_{2}}}=\frac{n}{{{l}_{1}}{{m}_{2}}-{{l}_{2}}{{m}_{1}}}=d\] Therefore, direction cosines are \[({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}}),\,\,({{n}_{1}}{{l}_{2}}-{{l}_{1}}{{n}_{2}}),\,({{l}_{1}}{{m}_{2}}-{{l}_{2}}{{m}_{1}})\].You need to login to perform this action.
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