A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{6}\]
C) \[\frac{\pi }{2}\]
D) 0
Correct Answer: A
Solution :
For the plane \[2x-y+z=6,\] \[{{a}_{1}}=2,{{b}_{1}}=-1,{{c}_{1}}=1\] and for the plane\[x+y+2z=3,\] \[{{a}_{2}}=1,{{b}_{2}}=1,{{c}_{2}}=2\] \[\therefore \]\[\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\] \[=\frac{2\times 1+1\times -1+2\times 1}{\sqrt{{{2}^{2}}+{{(-1)}^{2}}+{{1}^{2}}}\sqrt{{{1}^{2}}+{{1}^{2}}+{{2}^{2}}}}\] \[=\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}}\] \[=\frac{3}{6}=\frac{1}{2}\] \[\Rightarrow \] \[\cos \theta =\frac{1}{2}=\cos \frac{\pi }{3}\] \[\Rightarrow \] \[\theta =\frac{\pi }{3}\]
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