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金红石结构的RuO2中a约等于4.5埃，c约等于3.0埃，那么[101]晶向和[001]晶向差多少°...

作成日: 2026年3月2日

使用モデル GPT-5.2 Thinking by Chat01

質問

金红石结构的RuO2中a约等于4.5埃，c约等于3.0埃，那么[101]晶向和[001]晶向差多少°，[-101]晶向和[001]晶向差多少°

思考

Calculating angle between crystallographic directions

I’m looking at the angle between crystallographic directions in tetragonal rutile. To do this, I calculate the angle between vectors, where the components are based on the a and c unit cell dimensions. After calculating, I find that the angle between [101] and [001] is approximately 56.3°. The same applies for [-101] and [001], though there could be a difference in orientation.

Refining angle calculation

Hmm, I realized that the vector has a negative x component, but the z-axis direction remains the same. The angle is still acute because the dot product is positive. I'll show the general calculation, and the final result should be about 56.3°, with one decimal place. To be precise, I’ll use Python.

Thought for 21s

回答

在四方晶系（金红石 RuO₂）里，把晶向 [hkl][hkl] 当作实空间矢量

r⃗=(h a,  k a,  l c)\vec r = (h\,a,\;k\,a,\;l\,c)

（因为 a⊥ca\perp c，且 a=ba=b）。

1) [101][101] 与 [001][001] 的夹角

[101]→(a,0,c),[001]→(0,0,c)[101]\to (a,0,c),\quad [001]\to (0,0,c) cos⁡θ=(a,0,c)⋅(0,0,c)a2+c2    c=c2ca2+c2=ca2+c2\cos\theta=\frac{(a,0,c)\cdot(0,0,c)}{\sqrt{a^2+c^2}\;\;c} =\frac{c^2}{c\sqrt{a^2+c^2}}=\frac{c}{\sqrt{a^2+c^2}}

代入 a=4.5 A˚, c=3.0 A˚a=4.5\ \text{Å},\ c=3.0\ \text{Å}：

cos⁡θ=3.04.52+3.02=3.029.25≈0.5547\cos\theta=\frac{3.0}{\sqrt{4.5^2+3.0^2}}=\frac{3.0}{\sqrt{29.25}}\approx 0.5547 θ≈arccos⁡(0.5547)=56.31∘\theta \approx \arccos(0.5547)=56.31^\circ

2) [−101][-101] 与 [001][001] 的夹角

[−101]→(−a,0,c)[-101]\to (-a,0,c)

与 [001][001] 点积仍为 c2c^2，长度也不变，所以

θ=56.31∘\theta = 56.31^\circ

答案：

  • [101][101] 与 [001][001]：约 56.3°
  • [−101][-101] 与 [001][001]：约 56.3°

質問

我有个问题，如果[-101]和[010]组成平面，那么这个平面的法向量和[001]方向夹角多大

思考

import math a=4.5 c=3.0 cos=a/math.sqrt(aa+cc) theta=math.degrees(math.acos(cos)) cos,theta

Thought for 21s

回答

把晶向当作实空间矢量（四方晶系，a=ba=b，且与 cc 正交）：

  • [−101] → v⃗1=(−a, 0, c)[-101]\ \rightarrow\ \vec v_1 = (-a,\,0,\,c)
  • [010] → v⃗2=(0, a, 0)[010]\ \rightarrow\ \vec v_2 = (0,\,a,\,0)

它们张成的平面的法向量就是叉乘：

n⃗=v⃗1×v⃗2=∣i^j^k^−a0c0a0∣=(−ac, 0, −a2)\vec n=\vec v_1\times \vec v_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ -a & 0 & c\\ 0 & a & 0 \end{vmatrix} =(-ac,\ 0,\ -a^2)

方向正负不影响夹角，所以取 n⃗=(ac,0,a2)\vec n=(ac,0,a^2)。

[001][001] 方向就是 zz 轴方向（可取 z⃗=(0,0,1)\vec z=(0,0,1) 或 (0,0,c)(0,0,c) 都一样），于是夹角 θ\theta 满足：

cos⁡θ=n⃗⋅z⃗∣n⃗∣ ∣z⃗∣=a2(ac)2+(a2)2=aa2+c2\cos\theta=\frac{\vec n\cdot \vec z}{|\vec n|\,|\vec z|} =\frac{a^2}{\sqrt{(ac)^2+(a^2)^2}} =\frac{a}{\sqrt{a^2+c^2}}

代入 a=4.5 A˚, c=3.0 A˚a=4.5\ \text{Å},\ c=3.0\ \text{Å}：

cos⁡θ=4.54.52+3.02=4.529.25≈0.83205\cos\theta=\frac{4.5}{\sqrt{4.5^2+3.0^2}} =\frac{4.5}{\sqrt{29.25}} \approx 0.83205 θ≈arccos⁡(0.83205)=33.69∘\theta \approx \arccos(0.83205)=33.69^\circ

答案： 这个平面的法向量与 [001][001] 的夹角约 33.7°。

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