✔ 最佳答案
設a向量=(1,1)、b向量=(2,6),試求|ta向量+b向量|的最小值
Sol
A=|ta向量+b向量|>=0
=|t(1,1)+(2,6)|
=|(t,t)+(2,6)|
==|(t+2,t+6)|
A^2=(t+2,t+6) dot (t+2,t+6)
=(t+2)^2+(t+6)^2
=(t^2+4t+4^2)+(t^2+12t+36)
=2t^2+16t+40
=2(t^2+8t)+40
=2(t^2+8t+16)+8
=2(t+4)^2+8
0<=(t+4)^2
0<=2(t+4)^2
A^2<=8
A<=2√2