已知向量xy,设a=x+3y,b=-6x+3y,q=2x-1/3y,用ab表示向量q

问题描述:

已知向量xy,设a=x+3y,b=-6x+3y,q=2x-1/3y,用ab表示向量q

设q =ka+tb = (kx+3ky) + (-6tx+3ty)= (k-6t)x+(3k+3t)y =2x-1/3y
所以 k-6t=2
3k+3t=-1/3
解得
t=-19/63
k=4/21
所以 q =4/21a-19/63b