问题
解答题
平面内给定三个向量
(1)求|3
(2)若(
|
答案
(1)由题意
=(3,2),a
=(-1,2),b
=(4,1)c
∴3
+a
-2b
=(0,6)⇒|3c
+a
-2b
|=6c
(2)由题意得,
+ka
=(4k+3,k+2),2c
-b
=(-5,2)a
由(
+ka
)⊥(2c
-b
)⇒-5(4k+3)+2(k+2)=0⇒k=-a
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