(1)a+tb=(m+tp,n+tq)故|a+tb|=√[(m+tp)^2+(n+tq)^2]=√[(p^2+q^2)t^2+2(mp+nq)t+m^2+n^2]令f(t)=(p^2+q^2)t^2+2(mp+nq)t+m^2+n^2显然delta(就是那个三角形符号)=-4(mq-np)^2=0故f(t)在实数范围里恒大于等于0.故当t=...