现在利用数量积来给出一个重要的概念. 定义6.若a·b=0就说a与b正交. 因为a·b=|a||b|cosθ,所以a与b正交的必要充分条件是,θ=π/2,3π/2(这就是通常的正交性)或者a与b中有一个是0.所以,0正交于任意向量.当然,从前面的规定0可以有任意方向来看,说0与任意的a正交自然无不可,但是说0与a的交角不是π/2而例如是π/3亦无不可;前面甚至还说0与任意向量平行.所 Now we use the scalar product to give an important concept. Definition 6. If a·b=0 we say that a and b are orthogonal. Because a·b=|a||b|cosθ, the need for a and b to be orthogonal The sufficient condition is that θ=π/2, 3π/2 (this is the usual orthogonality) or one of a and b is 0. So, 0 is orthogonal to any vector. Of course, from the previous provision 0 can have Looking from an arbitrary direction, it is natural to say that 0 is orthogonal to any a, but it is said that the angle between 0 and a is not π/2 and that it is not π/3; the front even says that 0 is parallel to any vector.