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V(A-B)= (1+1)i +(1+1)j+ (1+1)k
          = 2i+2j+2k

Angle between two vectors is given as
cosФ= V(A-B).V(A) / IV(A-B)I IV(A)I
where V(A-B).V(A) means scalar product or the dot product and 
 IV(A-B)I IV(A)I means product of modulus of vector V(A-B) and V(A) respectively.

Hence from this we get 

cosФ= 2*1+2*1+2*1 / (√2²+2²+2²)(√1²+1²+1²)
cosФ= 6/(2√3)(√3)
cosФ= 6/6
cosФ=1 at 0°,180°
Ф=0° Ans
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