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I had exam day after tomorrow. Plz solve the whole exercise 2 for me . All the questions from 1 to 9 . Questions are in the attachment.
Solving one by one first no 1.Sol:unit vector of vector 6i+2j+3k is 6i+2j+3k/7 and to 3i-2j is 3i-2j/(13)^/12
2:cos of angle between vector a and b is a.b/|a||b| here let a be 2i-2j+4k and b be 3i+j+2k .Solve to get a.b (6-2+8)=12 |a||b|=4(21)^1/2 now cos = (3/7)^1/2 now sin is (1-cos^2)^1/2 solve to get sin =(4/7)^1/2
3:a×(b×c)=(a.c)b-(a.b)c this is a simple formula and we have also express it as combination of a, b and c.
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