# If vectors are A= i+j+k and vector B = -i-j-k,then find the angle between vector (A - B) and A.

1
by chaspavanC

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by chaspavanC

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= 2i+2j+2k

V(A)=i+j+k

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

hence

cosФ=1 at 0°,180°

hence

Ф=0° Ans