若△ABC的三个顶点A.B.C及平面内一点P满足向量PA+PB+PC=0.且实数λ满足向量AB+AC=λAP求实数λ的值
问题描述:
若△ABC的三个顶点A.B.C及平面内一点P满足向量PA+PB+PC=0.且实数λ满足向量AB+AC=λAP求实数λ的值
答
AB+AC=AP+PB+AP+PC=2AP+(PB+PC)=2AP+(-PA)=2AP+AP=3AP ,
因此 λ=3 .