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Grade: 6
        Given two vectors (a)  i+2j-3k (b)  2i-3j+4k. find (1). a.b (2) a*b (3)the angle between the two vectors?
7 months ago

Answers : (3)

sahil
138 Points
							answering one by one first question no 1)sol: a.b =(i+2j-3k).(2i-3j+4k)=2-6-12=-16. So -16 is required answer
						
7 months ago
sahil
138 Points
							sol no 2) iF you will expand the determinant to find a×b you get a×b=2i-10j-7k as the required answer.
						
7 months ago
sahil
138 Points
							sol no 3) A.b=-16 (solved already) now |a|=(14)^1/2 also  |b|=(29)^1/2 so cosine of angle between those two is -16/(14×29)^1/2 or angle between them two is cos^-1[-16/(406)^1/2]
						
7 months ago
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