已知向量a=(cos(-θ),sin(-θ)),向量b=(cos(π/2-θ),sin(π/2-θ)),(1)求证:向量a⊥向量b(2)若存在不等于0的实数k和t,使向量x=向量a+(t^2+3)向量b,向量y=-k向量a+t向量b满足向量x⊥向量y,试求此时(k+t
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 02:34:00
![已知向量a=(cos(-θ),sin(-θ)),向量b=(cos(π/2-θ),sin(π/2-θ)),(1)求证:向量a⊥向量b(2)若存在不等于0的实数k和t,使向量x=向量a+(t^2+3)向量b,向量y=-k向量a+t向量b满足向量x⊥向量y,试求此时(k+t](/uploads/image/z/10474702-70-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%28cos%28-%CE%B8%29%2Csin%28-%CE%B8%29%29%2C%E5%90%91%E9%87%8Fb%3D%28cos%28%CF%80%2F2-%CE%B8%29%2Csin%28%CF%80%2F2-%CE%B8%29%29%2C%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%E5%90%91%E9%87%8Fa%E2%8A%A5%E5%90%91%E9%87%8Fb%EF%BC%882%EF%BC%89%E8%8B%A5%E5%AD%98%E5%9C%A8%E4%B8%8D%E7%AD%89%E4%BA%8E0%E7%9A%84%E5%AE%9E%E6%95%B0k%E5%92%8Ct%2C%E4%BD%BF%E5%90%91%E9%87%8Fx%3D%E5%90%91%E9%87%8Fa%2B%EF%BC%88t%5E2%2B3%EF%BC%89%E5%90%91%E9%87%8Fb%2C%E5%90%91%E9%87%8Fy%3D-k%E5%90%91%E9%87%8Fa%2Bt%E5%90%91%E9%87%8Fb%E6%BB%A1%E8%B6%B3%E5%90%91%E9%87%8Fx%E2%8A%A5%E5%90%91%E9%87%8Fy%2C%E8%AF%95%E6%B1%82%E6%AD%A4%E6%97%B6%EF%BC%88k%2Bt)
已知向量a=(cos(-θ),sin(-θ)),向量b=(cos(π/2-θ),sin(π/2-θ)),(1)求证:向量a⊥向量b(2)若存在不等于0的实数k和t,使向量x=向量a+(t^2+3)向量b,向量y=-k向量a+t向量b满足向量x⊥向量y,试求此时(k+t
已知向量a=(cos(-θ),sin(-θ)),向量b=(cos(π/2-θ),sin(π/2-θ)),
(1)求证:向量a⊥向量b
(2)若存在不等于0的实数k和t,使向量x=向量a+(t^2+3)向量b,向量y=-k向量a+t向量b满足向量x⊥向量y,试求此时(k+t^2)/t的值
已知向量a=(cos(-θ),sin(-θ)),向量b=(cos(π/2-θ),sin(π/2-θ)),(1)求证:向量a⊥向量b(2)若存在不等于0的实数k和t,使向量x=向量a+(t^2+3)向量b,向量y=-k向量a+t向量b满足向量x⊥向量y,试求此时(k+t
1.a=(cos(θ),-sin(θ)),b=(sin(θ),cos(θ)),a*b=0,故a⊥b
2.x*y=k*a^2+t*(t^2+3)b^2=k+t*(t^2+3)=0
(k+t^2)/t=-t^2-3+t=-(t-1/2)^2-11/4
因此最大值是-11/4 没有最小值