OQ=(√3,-1),|OQ| = 2,幅角 arg OQ = arc tan (-1/√3) = -π/6
(1) OP垂直O(jiān)Q,所以,幅角相差 π/2,則 a= -π/6 + π/2 = π/3,所以,tan a = √3
(2) 由余弦定理 |PQ|^2 = |OQ|^2 + |OP|^2 - 2 |OQ|*|OP| cos (a+π/6) = 5 - 4 cos(a+π/6)
顯然,最大值當(dāng)且僅當(dāng) cos(a+π/6),即 a = 5π/6時,最大值為 |PQ|^2=9,|PQ|=3.