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Q.

A point charge is placed at origin. Let EA→,EB→  and EC→ be the electric field at three points A (1,2,3), B (1,1,-1) and C (2,2,2) due to charge q1 then

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a

EA→⊥EB→

b

EA→∥EC→

c

|EB→|=4|EC→|

d

|EB→|=8|EC→|

answer is M.

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Detailed Solution

EA→  is along OA→  and           EB→  is along OB→ OA→=i+2j+3k  and           OB→=i+j−k Since OA.→OB→=0                  EA→⊥EB→ Further |E→| α1r2                    |OC→| =2|OB→|                                               |EB→|=4|EC→|
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