Question

平面向量a=(根号3,-1),b=(1/2,(根号3)/2)

若存在不同时为0的实数k和t使向量x=向量a+（t-3)*向量b，向量y=-k*向量a+t*向量b，且向量x垂直于向量y，试确定函数k=f(t) 并求出k=f(t)的单调性

大哥们！快来解围撒！

Answer 1

由向量a和向量b的坐标知，a*b=0
且向量x垂直于向量y，可知x*y=0
x*y=（a+（t-3)*b）*（-k*a+t*b）=-ka^2+ta*b-(t-3)a*b+t(t-3)b^2
由坐标得|a|^2=4,|b|^2=1,
则x*y=-4k+t(t-3)=0
k=1/4t(t-3)，即k=f(t)=11/4t(t-3)
这是一个二次函数，图像开口向上，对称轴是x=3/2,
所以（-&，3/2）是单调递减，（3/2,+&）是单调递增
。。。。我是姐姐，不是大哥

Answer 2

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追问

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Answer 3

向量x=（(t-3+2*根号3)/2，-1+(t-3)*(根号3)/2)
向量y=(t/2-根号3*k,k+根号3*t/2)
k=(t*t-3*t)/4,在 t<3/2递减，t>3/2递增

Answer 4

x和y相乘再利用数量积可得出k的函数再求单调性