Question

已知|a|=4,|b|=3,a,b的夹角为120°

1若c=a+b,d=2a+b,求c·d
2若c=a+2b,d=2a+kb,当c⊥a时，求k的值

（a,b,c,d均为向量）

Answer 1

|a|=4,|b|=3,a,b的夹角为120°,则
a^2 = |a|^2 =16，
b^2 = |b|^2 = 9，ab= |a||b|cos120°= 4*3 * (-1/2) = -6

1、c·d = (a+b)·(2a+b) = 2a^2 + 3ab+ b^2 =2*16 -3*6 + 9 = 32 - 18 + 9 = 23

2、当c⊥a时?! 应该是“当c⊥d时”吧
c⊥d，c·d = (a+2b)·(2a+kb) = 2a^2 + (4+k)ab + 2kb^2 = 0，
即 2*16 - 6(4+k) + 2*9*k =0
32 -24 - 6k +18k=0，则12k= -8，k= -2/3