Question

已知点A,B的坐标分别为(x1,y1,z1),(x2,y2,z2),O为坐标原点，向量OA=a,向量OB=b，向量OA与向量OB的夹角为@,

求/a-b/的值
求cos@的值
求证：a*b=x1x2+y1y2+z1z2
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Answer 1

1)|a-b|==√(x2-x1)^+(y2-y1)^+(z2-z1)^
2)cos@=(ab)/(|a||b|)=(x1*x2+y1*y2+z1*z2)/[(√x1^+y1^+z1^)(√x2^+y2^+z2^)]
3)设x轴的空间单位向量为i，y轴的空间单位向量为j,z轴的空间单位向量为k
所以ij=jk=ki=0,ii=jj=kk=1.
a*b=(x1i+y1j+z1k)*(x2i+y2j+z2k)=x1i*(x2i+y2j+z2k)+y1j*(x2i+y2j+z2k)+z1k*(x2i+y2j+z2k)=x1x2ii+x1y2ij+x1z2ik+y1x2ji+y1y2jj+y1z2jk+z1x2ki+z1y2kj+z1z2kk=x1x2+y1y2+z1z2