正方体ABCD-A1B1C1D1的棱长为a,求对角线AC与BC1的距离
问题描述:
正方体ABCD-A1B1C1D1的棱长为a,求对角线AC与BC1的距离
答
连接三角形A1BC1各点,AC//A1C1,则AC//面A1BC1,则AC到BC1的距离即为AC到面A1BC的距离,设AC与BD交于点M,A1C1与B1D1交于点N,连接BN,过M做ME垂直于BN于E,ME即为AC到面A1BC的距离.求得ME=根3/3