4. 空間四點A,B,C,D中,AB= 2,BC = 2,CD = 2,∠ABC = 135°,∠BCD = 90°,AB
2x √[x 2 x (0345
X
解
:
與CD 的夾角為45°,求 AD之長=
2
|Ab1²= |AB + 15 C + CD1²
52
XT1=²X (2x2
2×2×2×0
108 (35
MARC = |AB| ²+ | BC l ² + 1 cb / ² + ² AB · Be + ₂ Bi⋅ CD + 2 AB - cB
= 2+4 + 4 + 2x52x2x-194/45 + 2×2×2×10590+2x52x2
0 x 10y45°
10+ 4 + 0 + 4
2
Z
Ans: 3√2