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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?

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?

Grade:6

5 Answers

sahil
142 Points
3 years ago
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
sahil
142 Points
3 years ago
sol no 2) iF you will expand the determinant to find a×b you get a×b=2i-10j-7k as the required answer.
sahil
142 Points
3 years ago
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]
precious
14 Points
8 months ago
A.B=(1×2) +(2×-3) +(-3×4) = 2i-6j-12 =-16
 
A×B= I[(2×4)-(-3×-3)] - J[(1×4)-(-3×2)] + K[(1×-3)-(2×2)]
=(8-9) - (4+6) + (-3-4) = -1I -10J - 7K
 
precious
14 Points
8 months ago
C.

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