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The angle between the vectors (i^+j^+k^) and (i^-j^-k^) issin-1√8/3\tSin-1(1/3)\tCos-1√8/3\tCos-1√(8/3)

Neha singh , 4 Years ago
Grade 12th pass
anser 1 Answers
Harshit Singh

Last Activity: 4 Years ago



Dear Student

cos x = Dot product of vector A and B/ product of their magnitudes
dot product of A and B= 1-1-1= -1
magnitude of vector A =magnitude of vector B= √3

cos x = (-1)/(√3.√3)
= -1/3

sin^2 x = 1 – cos^2 x
sin x = √8/3
x = sin inverse (√8/3)

Thanks

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