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angle between diagonals of a parallelogram whose side represented by vector a =2i +j +k and vector b = i -j - k. Options are A) cos^-1 (1/3) B) cos^-1 (1/2) C) cos^-1 (4/9) D) cos^-1 (5/9)

angle between diagonals of a parallelogram whose side represented by vector a =2i +j +k and vector b = i -j - k. Options are A) cos^-1 (1/3) B) cos^-1 (1/2) C) cos^-1 (4/9) D) cos^-1 (5/9)

Grade:11

2 Answers

Shailendra Kumar Sharma
188 Points
5 years ago
Cos(Theta) =A.B./ABSo the angle will be Cos (Theta) = (​2i +j +k). (i -j - k) /( \sqrt{4+1+1})*\sqrt{(1+1+1))}\Theta =(2-1-1)/\sqrt{6} X\sqrt{3}) =0So the angle is 90 Degree as cos 90 is zero
Subham
15 Points
2 years ago
We can find digonal vector a+b and a-b by this we can use formula cos theta equals to (a-b)(a+b)/|a+b||a-b| we got cos theta equals to 1/3

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