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The angle between the vectors (i ^+j^+k^) and (i^-j^-k^) is sin-1√8/3 Sin-1(1/3) Cos-1√8/3 Cos-1√(8/3)
Dear Studentcos x = Dot product of vector A and B/ product of their magnitudesdot product of A and B= 1-1-1= -1magnitude of vector A =magnitude of vector B= √3cos x = (-1)/(√3.√3) = -1/3sin^2 x = 1 – cos^2 xsin x = √8/3x = sin inverse (√8/3)Thanks
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