已知向量a=(2cos(-θ),2sin(-θ)),向量b=(cos(90°-θ),sin(90°-θ))若存在不等于0的实数k和t,使向量x=向量a+(t²-3)向量b,向量y=-k向量a+t向量b满足向量a⊥向量b.试求此时(k+t²)/t
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![已知向量a=(2cos(-θ),2sin(-θ)),向量b=(cos(90°-θ),sin(90°-θ))若存在不等于0的实数k和t,使向量x=向量a+(t²-3)向量b,向量y=-k向量a+t向量b满足向量a⊥向量b.试求此时(k+t²)/t](/uploads/image/z/1963200-48-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%EF%BC%882cos%EF%BC%88-%CE%B8%EF%BC%89%2C2sin%EF%BC%88-%CE%B8%EF%BC%89%EF%BC%89%2C%E5%90%91%E9%87%8Fb%3D%EF%BC%88cos%EF%BC%8890%C2%B0-%CE%B8%EF%BC%89%2Csin%EF%BC%8890%C2%B0-%CE%B8%EF%BC%89%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%26%23178%3B-3%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%8Fa%E2%8A%A5%E5%90%91%E9%87%8Fb.%E8%AF%95%E6%B1%82%E6%AD%A4%E6%97%B6%EF%BC%88k%2Bt%26%23178%3B%EF%BC%89%2Ft)
已知向量a=(2cos(-θ),2sin(-θ)),向量b=(cos(90°-θ),sin(90°-θ))若存在不等于0的实数k和t,使向量x=向量a+(t²-3)向量b,向量y=-k向量a+t向量b满足向量a⊥向量b.试求此时(k+t²)/t
已知向量a=(2cos(-θ),2sin(-θ)),向量b=(cos(90°-θ),sin(90°-θ))
若存在不等于0的实数k和t,使向量x=向量a+(t²-3)向量b,向量y=-k向量a+t向量b满足向量a⊥向量b.试求此时(k+t²)/t的最小值
已知向量a=(2cos(-θ),2sin(-θ)),向量b=(cos(90°-θ),sin(90°-θ))若存在不等于0的实数k和t,使向量x=向量a+(t²-3)向量b,向量y=-k向量a+t向量b满足向量a⊥向量b.试求此时(k+t²)/t
首先指出:你给的条件“向量a⊥向量b”应该是“向量x⊥向量y”,否则缺条件,而且“向量a⊥向量b”由它们的坐标直接可以推出.
由已知化简得,向量a=(2cosθ,-2sinθ),向量b=(sinθ,cosθ),
∴向量x=(2cosθ+ (t²-3)sinθ,(t²-3)cosθ-2sinθ),向量y=(tsinθ-2kcosθ,tcosθ+2ksinθ),
由向量x⊥向量y,得x•y=0,
即[2cosθ+ (t²-3)sinθ](tsinθ-2kcosθ)+[(t²-3)cosθ-2sinθ)](tcosθ+2ksinθ)=0,
化简得t(t²-3)-4k=0,即k=t(t²-3)/4,
故(k+t²)/t =k/t+t=(t²-3)/4+t=[(t+2)²-7]/4,
所以,当t=-2,k=-1/2时,(k+t²)/t取最小值-7/4.