A) \[\frac{1}{3}\times \frac{273}{298}\]
B) 1/2
C) 1
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
Since, the line lies on both the given planes, then the normal to the planes are perpendicular to the line L. If direction cosines of L are\[l,\text{ }m,\text{ }n,\]then \[2l+3m+n=0\] ...(i) And \[l+3m+2n=0\] ...(ii) On solving Eqs. (i) and (ii), we get \[\frac{l}{3}=\frac{m}{-3}=\frac{n}{3}\] \[\therefore \] \[l:m:n=\frac{1}{\sqrt{3}}:\left( -\frac{1}{\sqrt{3}} \right):\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[l=\frac{1}{\sqrt{3}}\]\[\Rightarrow \] \[\cos \alpha =\frac{1}{\sqrt{3}}\]
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