A) \[\pi /3\]
B) \[\pi /4\]
C) \[\pi /2\]
D) 0
Correct Answer: A
Solution :
[a] Given d?c?s of two lines are \[\ell ,m,n\]connected by the relation \[\ell +m+n=0\] and \[\ell m=0\] Now, \[\ell +m+n=0\Rightarrow \ell =-m-n\] \[\Rightarrow \ell =-(m+n)\] And \[\ell m=0\Rightarrow -(m+n)m=0\Rightarrow -mn-mm=0\] \[mn=-mn;\] Therefore m and \[m+n=0\] Then \[\frac{{{\ell }_{1}}}{-1}=\frac{{{m}_{1}}}{0}=\frac{{{n}_{1}}}{1}\] and if \[\ell +m+n=0\]then\[\frac{{{\ell }_{2}}}{0}=\frac{{{m}_{2}}}{-1}=\frac{{{n}_{2}}}{1}\] \[({{\ell }_{1}},{{m}_{1}},{{n}_{1}})=(-1,0,1)\] and \[({{\ell }_{2}},{{m}_{2}},{{n}_{2}})=(0,-1,1)\] We know that angle between them \[\cos \theta =\frac{0+0+1}{\sqrt{1+0+1}\sqrt{0+1+1}}=\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}\] \[\cos \theta =\frac{1}{2}=\cos 60{}^\circ \Rightarrow \theta =60{}^\circ \Rightarrow \theta =\frac{\pi }{3}\]
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