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2003年天津市高考理科卷有这样一道向量形式的动点的轨迹题目:已知O是平面上一定点,A、B、C是平面上不共线的三点,动点P满足→→OP=OA+λ(→AB→|AB|+→AC→|AC|),λ∈[0,+∞),则动点P的轨迹一定通过ΔABC的().A.外心B.内心C.重心D.垂心向上的单位向量.易知,四边形AEGF是菱形,从而射线AG是∠BAC的内角平分线,它一定通过ΔABC的内心.故选B.变式1:已知O是平面上一定点,A、B、C是平面上不共线
2003 Tianjin Science Entrance Examination Science Volume has such a vector form of the trajectory of the problem: O is known to be a certain point on the plane, A, B, C is not plane three points, the moving point P to meet →→ OP = OA + λ (→ AB → | AB | + → AC → | AC |), λ∈ [0, + ∞), then the trajectory of moving point P must pass through . Center of gravity D. Vertical unit vector. It is easy to know that the quadrilateral AEGF is a rhombus so that the radiation AG is the bisector of the internal angle of the BAC, which must pass through the center of the ΔABC. Therefore, B. is selected. Variant 1: O is known to be flat A certain point, A, B, C are not collinear on the plane