设A(2,3,1)B(4,1,2)C(6,3,7)D(-5,-4,8)求D到平面ABC的距离谢谢了,
设A(2,3,1)B(4,1,2)C(6,3,7)D(-5,-4,8)求D到平面ABC的距离?
人气:289 ℃ 时间:2019-10-09 08:33:43
解答
设平面ABC的法向量n=(x,y,z),∵n AB =0,n AC =0,∴ (x,y,z)(2,-2,1)=0 (x,y,z)(4,0,6)=0 即 2x-2y+z=0 4x+6z=0 x=- 3 2 z y=-z.令z=-2,则n=(3,2,-2). ∴cos<n,AD >= 3×(-7)+2×(-7)-2×7 32+22+(-2)2 (-7)2+(-7)2+72 . ∴点D到平面ABC的距离为d,d=| AD ||cos<n,AD >|= 49 17 = 49 17 17 .
推荐
- 设A(2,3,1),B(4,1,2),C(6,3,7),D(-5,-4,8),则D到平面ABC的距离为?
- 已知A(2,3,1)B(4,1,2)C(6,3,7)D(-5,-4,8),求D到平面ABC的距离
- 设A (2,3,1),B(4,1,2),C(6,3,7),D(-5,-4,8),求点D到平面ABC的距离
- △ABC的三个顶点A、B、C到平面α的距离分别为2 cm、3 cm、4 cm,且它们在α的同侧,则△ABC的重心到平面α的距离为_.
- 设A(2,0,1),B(0,1,2),C(2,1,4),D(-1,2,6)则D到平面ABC的距离是多少.
- 石灰石的比重是多少
- Now,not only would they have to deal with whatever illness she had,but also with the...
- 0.625的十次方+8的11此方+2的12次方=?
猜你喜欢