问题
填空题
已知单位向量
|
答案
因为单位向量
,a
的夹角为120°b
所以|2
+xa
|2=4b
2+4xa
•a
+x2b
2b
=x2-2x+4=(x-1)2+3
∴当x=1时|2
+xa
|2取最小值,此时|2b
+xa
|(x∈R)取得最小值,b
故答案为:1
已知单位向量
|
因为单位向量
,a
的夹角为120°b
所以|2
+xa
|2=4b
2+4xa
•a
+x2b
2b
=x2-2x+4=(x-1)2+3
∴当x=1时|2
+xa
|2取最小值,此时|2b
+xa
|(x∈R)取得最小值,b
故答案为:1
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