问题
解答题
已知向量
(1)若(3
(2)若
|
答案
(1)∵3
-a
=3(1,-2)-(3,4)=(0,-10),b
+ka
=(1,-2)+k(3,4)=(1+3k,-2+4k),b
又(3
-a
)∥(b
+ka
),b
∴-10(1+3k)-0=0,解得k=-
.1 3
(2)m
-a
=m(1,-2)-(3,4)=(m-3,-2m-4),b
∵
⊥(ma
-a
),∴m-3-2(-2m-4)=0,b
解得m=-1.