向量积数量积问题

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向量积数量积问题向量积数量积问题向量积数量积问题AB=(5-2)i+(-2-1)j+(6-0)k=3i-3j+6kF所做的功=F*(AB)=4*3+2*(-3)+2*6=12-6+12=18F的长度(

向量积数量积问题
向量积数量积问题

向量积数量积问题
AB = (5-2)i + (-2-1)j + (6-0)k = 3i - 3j + 6k
F所做的功 = F*(AB) = 4*3 + 2*(-3) + 2*6 = 12 - 6 + 12 = 18
F的长度(模)||F|| = [4^2 + 2^2 + 2^2]^(1/2) = [16+4+4]^(1/2) = (24)^(1/2)
(AB)的长度(模)||AB|| =[3^2+(-3)^2+6^2]^(1/2) = [9+9+36]^(1/2)=(54)^(1/2)
F与(AB)间的夹角的余弦cos(C) = F*(AB)/[ ||F||*||AB|| ]
= 18/[24*54]^(1/2) = 18/[4*6*6*9]^(1/2) = 18/[2*6*3] = 1/2,
F与(AB)间的夹角C = arccos(1/2) = 60度