Question

高中向量题

设a1 和a2 为平面上两个长度为1的不共线向量，且它们和的模长满足|a1+a2|=根号3 。则 (2a1-5a2)*(3a1+a2)=?
其中a1 a2上面都有矢量箭头

求详解 谢谢！

Answer 1

由a1+a2=根号3平方，得
a1方+a2方+2a1*a2=3
则2a1*a2=1
a1*a2=0.5
即a1,a2夹角为60度
由第二式得
6a1方-5a2方-13a1*a2
=6-5-13*0.5
=-5.5

Answer 2

两边平方a1^2+a2^2+2a1a2=3,
a1a2=0.5原式=6a1^2-13a1a2-5a2^2=-5.5

Answer 3

很简单啦 |a1+a2| 的平方=3 展开 a1^2+a2^2+2a1a2 =2+2a1a2=3 所以 a1a2=1.5
求 6a1^2+2a1a2-15a1a2-5a2^2=6-5-13x1.5

Answer 4

｜a1+a2｜=√3 平方
a1^2+2a1a2+a2^2=3
1+2a1a2+1=3
a1a2=1/2
(2a1-5a2)*(3a1+a2)
=6a1^2+2a1a2-15a1a2-5a2^2
=6a1^2-13a1a2-5a2^2
=6*1-13*1/2-5*1
=6-13/2-5
=-11/2