问题
解答题
设向量
(Ⅰ)求
(Ⅱ)求|3
|
答案
(Ⅰ)设
与a
夹角为θ,∵向量b
,a
满足|b
|=|a
|=1及|3b
-2a
|=b
,7
∴9
2+4a
2-12b
•a
=7,∴9×1+4×1-12×1×1×cosθ=7,∴cosθ=b
.1 2
又θ∈[0,π],∴
与a
夹角为b
.π 3
(Ⅱ)∵|3
+a
|=b
=9
2+a
2+6b
•a b
=9×1+1+6×1×1×cos π 3
.13