A) \[\sqrt{2},1,-1\]
B) \[2,-1,1\]
C) \[2,\sqrt{2},-\sqrt{2}\]
D) \[2\sqrt{3},1,-1\]
Correct Answer: C
Solution :
Let the equation of plane is \[a(x-0)+b(y+1)+c(z-0)=0\] \[\Rightarrow \]\[ax+by+cz+b=0\] It passes through (0, 0, 1) \[\therefore \]\[a(0)+b(0)+c(1)+b=0\] \[\Rightarrow \]\[b+c=0\]... (i) Also, \[\cos \left( \frac{\pi }{4} \right)=\frac{a(0)+b(1)+c(-1)}{\sqrt{{{(1)}^{2}}+{{(-1)}^{2}}}\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\] \[\therefore \]\[\frac{1}{\sqrt{2}}=\frac{b-c}{\sqrt{2}\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\] ?(ii) From (i) and (ii), we get\[{{a}^{2}}=-2bc\]and\[b=-c\Rightarrow {{a}^{2}}=2{{c}^{2}}\]\[\Rightarrow \]\[a=\pm \sqrt{2}c\] \[\therefore \]Dr?s of \[(a,b,c)=(\sqrt{2},-1,1)\]or\[(\sqrt{2},1,-1)\] or\[(2,\sqrt{2},-\sqrt{2})\]
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