Question

已知m=(3/5,-4/5),n=(cosα,sinα),|3m-2n|=3,求3m+n的绝对值的值

Answer 1

|m|=|n|=1.

|3m-2n|^2=9|m|^2+4|n|^2-12m*n=13-12m*n=9、m*n=1/3.

|3m+n|^2=9|m|^2+|n|^2+6m*n=10+2=12、|3m+n|=2√3.

a*b=(3m-2n)*(3m+n)=9m^2-3m*n-2n^2=6.

cos=a*b/(|a||b|)=6/(6√3)=√3/3.