✨ Best Answer ✨
求D到平面BCHE的距離:
| 1×3+(-2)×(-1)+2×3-5 | / √(1²+(-2)²+2²) = 2
DG長為D到平面BCHE的兩倍=4
正方形的邊長與對角線的關係為1:√2 (45° 45° 90°三角形的比例關係)
設AD長=x
所以AD:DG=AD:BE=1:√2=x:4 -> x=2√2
Mathematics
Senior High
Resolved
求解這題 答案是2 √2
d = -b
6. 如圖,正立方體 ABCD-EFGH 中,若平面 BCHE 所在的平面方程式為x-2y+2z = 5,且
D(3, -1, 3),試求正立方體的稜長AD為
3+2+6-3
K+3
J6K+12
=
2k+6 = 6k²+2
2DK
2.
K+3
J6JK
2.2K.6
12
-36 (2k+6)=
24
4k+ 24k
82k2
H
G
(2K)²
4
E
1x12
6x2
K²+4-4
3x4
K
K
*X-12
K 2
116k²+121
11x3 -2x(-1)+2x3-3| 4
n
3
K³
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