Question

已知向量a,b满足|a|=|b|=1,且|ka+b|=根号3|a-kb|（k>0),

Answer 1

1|a|=|b|=1,=π/3即：a·b=1/2|ka+b|=√3|a-kb|即：|ka+b|^2=3|a-kb|^2即：k^2|a|^2+|b|^2+2ka·b=3(|a|^2+k^2|b|^2-2ka·b)即：k^2+1+k=3+3k^2-3k即：4k=2k^2+2即：k^2-2k+2=(k-1)^2=0即：k=12|ka+b|=√3|a-kb|即：...