求距离的问题

正方ABCD-A1B1C1D1的棱长为a,点M分向量AC1的比为1/2,N为BB1的中点,则 向量MN的长度为

热心网友

三角形ANC1中，显然：AC1 = (genhao3)a；AN=NC1=(genhao5)a/2；cos角NAC1 =(genhao15)/5三角形AMN中：AM=AC1/3 = (genhao3)a/3MN^2 = AM^2 + AN^2 - 2*AM*AN*cos角NAC1== MN = (genhao21)a/6

热心网友

三角形ANC1中，显然：AC1 = (genhao3)a；AN=NC1=(genhao5)a/2；cos角NAC1 =(genhao15)/5三角形AMN中：AM=AC1/3 = (genhao3)a/3MN^2 = AM^2 + AN^2 - 2*AM*AN*cos角NAC1== MN = (genhao21)a/6