A) \[4\sqrt{2}x+7y+z=2\]
B) \[\sqrt{2}x+y+z=2\]
C) \[3\sqrt{2}x-4y-3z=7\]
D) \[\sqrt{2}x-y-z=2\]
Correct Answer: B
Solution :
\[\because \] \[{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1\] \[\Rightarrow \] \[{{\cos }^{2}}45+{{\cos }^{2}}60+{{\cos }^{2}}\gamma =1\] \[\Rightarrow \] \[{{\cos }^{2}}\gamma =1-\frac{1}{2}-\frac{1}{4}=1-\frac{3}{4}=\frac{1}{4}\Rightarrow \]\[\cos \gamma =\frac{1}{2}\] \[\therefore \] Direction Ratio?s of normal to the plane is \[\angle \cos {{45}^{o}};\,\cos {{60}^{o}},\,\frac{1}{2}>=<\frac{1}{\sqrt{2}},\frac{1}{2},\,\frac{1}{2}>\] \[\therefore \] Equation of plane passing through \[(\sqrt{2},\,-1,\,1)\] \[\frac{1}{\sqrt{2}}(x-\sqrt{2})+\frac{1}{2}(y+1)+\frac{1}{2}(z-1)=0\] \[\Rightarrow \] \[2(x-\sqrt{2})+\sqrt{2}(y+1)+\sqrt{2}(z-1)=0\] \[\Rightarrow \] \[\sqrt{2}(x-\sqrt{2})+(y+1)+d(z-1)=0\] \[\Rightarrow \] \[\sqrt{2}x-2+y+1+z-1=0\] \[\Rightarrow \] \[\sqrt{2}x+y+z=2\]
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