已知a,b是平面上两个不共线的非零向量,若a、1/2b、t(a+b)(t∈R)三向量的起点相同,则t为何值时,这三个向量的终点在同一直线上?我想知道为什么我这样做不对：∵这三个向量终点在同一直线上

已知a,b是平面上两个不共线的非零向量,若a、1/2b、t(a+b)(t∈R)三向量的起点相同,则t为何值时,这三个向量的终点在同一直线上?我想知道为什么我这样做不对：∵这三个向量终点在同一直线上
已知a,b是平面上两个不共线的非零向量,若a、1/2b、t(a+b)(t∈R)三向量的起点相同,则t为何值时,这三个向量的终点在同一直线上?
我想知道为什么我这样做不对：
∵这三个向量终点在同一直线上
∴b-a=λ[t(a+b)-a]
b-a=tλa-λa+tλb
b-a=(tλ-λ)a+tλb
∴ -1=tλ-λ →t=1/2
1=tλ →λ=2
∴当t为1/2时这三个向量终点在同一直线上
答案t是1/3

已知a,b是平面上两个不共线的非零向量,若a、1/2b、t(a+b)(t∈R)三向量的起点相同,则t为何值时,这三个向量的终点在同一直线上?我想知道为什么我这样做不对：∵这三个向量终点在同一直线上
错就错在b-a,题目明明要求a、1/2b、t(a+b)(t∈R)三个向量的终点在同一直线上,你为什么要改成b呢,1/2b-a=λ[t(a+b)-a]
∴ 1/2=tλ →λ=3/2
-1=tλ-λ →t=1/3

二、三式是一样的

其实很简单的。你画个最简单的图好了。画个坐标轴，假设a,b向量都是从原点出发的，a向量是伸向y正轴，长度为2，b向量伸向x正轴，长度也为2，那么，a+b向量就应该在对角线上。现在a=2，b=1，但a+b还在对角线上。所以画了个图，你看看吧，还看不懂再问我。

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When shopping for basketball shoes, you should really try to find those that are comfortable, that offer durability and that help you play at your best. These things are much more important than what ...

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