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Даны векторы а { 1; -2; 0}; b {3; -6; 0}, с {0; - ...

생성일: 2025년 5월 16일

답변에 사용된 모델 O3 by Chat01

질문

Даны векторы а { 1; -2; 0}; b {3; -6; 0}, с {0; - 3; 4}
Найти координаты вектора р = 2а - 1/3b - c

사고

Computing the coordinates for vector p

The user has given the vectors a = (1, -2, 0), b = (3, -6, 0), and c = (0, -3, 4), and asked

Thought for 4 seconds

답변

Координаты вычислим по компонентам:

  1. 2a=2 (1, −2, 0)=(2, −4, 0)2\mathbf a = 2\,(1,\,-2,\,0)= (2,\,-4,\,0)
  2. 13b=13 (3, −6, 0)=(1, −2, 0)\frac13\mathbf b =\frac13\,(3,\,-6,\,0)= (1,\,-2,\,0)
  3. 2a−13b=(2−1, −4−(−2), 0−0)=(1, −2, 0)2\mathbf a-\frac13\mathbf b = (2-1,\,-4-(-2),\,0-0)= (1,\,-2,\,0)
  4. (\mathbf p = \bigl(2\mathbf a-\tfrac13\mathbf b\bigr)-\mathbf c = (1,,-2,,0);-;(0,,-3,,4) = (1,,-2+3,,-4))
p=(1;  1;  −4)\boxed{\mathbf p = (1;\;1;\;-4)}

질문

Вычислить угол между векторами АВ и CD, если
A (6; -4; 8); B (8; -2; 4); C (12; -6; 4); D (14; -6; 2).

Python

python
AB = sp.Matrix([2,2,-4])
CD = sp.Matrix([2,0,-2])
dot = AB.dot(CD)
mag_AB = sp.sqrt(AB.dot(AB))
mag_CD = sp.sqrt(CD.dot(CD))
dot, mag_AB, mag_CD, dot/(mag_AB*mag_CD)


(12, 2*sqrt(6), 2*sqrt(2), sqrt(3)/2)

사고

AB = sp.Matrix([2,2,-4]) CD = sp.Matrix([2,0,-2]) dot = AB.dot(CD) mag_AB = sp.sqrt(AB.dot(AB)) mag_CD = sp.sqrt(CD.dot(CD)) dot, mag_AB, mag_CD, dot/(mag_AB*mag_CD)

(12, 2sqrt(6), 2sqrt(2), sqrt(3)/2)

Thought for 23 seconds

답변

Сначала запишем координаты искомых векторов:

AB→=B−A=(8−6; −2−(−4);  4−8)=(2; 2; −4),CD→=D−C=(14−12; −6−(−6);  2−4)=(2; 0; −2).\overrightarrow{AB}=B-A=(8-6;\,-2-(-4);\;4-8)=(2;\,2;\,-4),\qquad \overrightarrow{CD}=D-C=(14-12;\,-6-(-6);\;2-4)=(2;\,0;\,-2).

1. Скалярное произведение

AB→⋅CD→=2⋅2+2⋅0+(−4)⋅(−2)=4+0+8=12.\overrightarrow{AB}\cdot\overrightarrow{CD}=2\cdot2+2\cdot0+(-4)\cdot(-2)=4+0+8=12.

2. Длины векторов

∣AB→∣=22+22+(−4)2=4+4+16=26,∣CD→∣=22+02+(−2)2=4+0+4=22.|\overrightarrow{AB}|=\sqrt{2^{2}+2^{2}+(-4)^{2}}=\sqrt{4+4+16}=2\sqrt{6},\qquad |\overrightarrow{CD}|=\sqrt{2^{2}+0^{2}+(-2)^{2}}=\sqrt{4+0+4}=2\sqrt{2}.

3. Косинус угла

cos⁡θ=AB→⋅CD→∣AB→∣ ∣CD→∣=12(26)(22)=1283=323=32.\cos\theta=\frac{\overrightarrow{AB}\cdot\overrightarrow{CD}} {|\overrightarrow{AB}|\,|\overrightarrow{CD}|} =\frac{12}{(2\sqrt6)(2\sqrt2)}=\frac{12}{8\sqrt3}=\frac{3}{2\sqrt3}=\frac{\sqrt3}{2}.

4. Угол

θ=arccos⁡ ⁣(32)=30∘  (или π6 рад).\theta=\arccos\!\bigl(\tfrac{\sqrt3}{2}\bigr)=30^{\circ}\; \left(\text{или } \frac{\pi}{6}\ \text{рад}\right). ∠(AB→,CD→)=30∘\boxed{\angle(\overrightarrow{AB},\overrightarrow{CD}) = 30^{\circ}}

질문

Дан куб ABCDA1В1С1D1. Найти угол между векторами AD1 и АМ, где М-середина ребра DD1

Python

python
cos = 3/math.sqrt(10)
theta = math.degrees(math.acos(cos))
theta

18.434948822922017

사고

Correcting angle estimation

I’ve been revisiting my calculations for the angle θ, and it seems that my earlier estimation of 47.9° was off due to earlier missteps. After refining my approach, I find that cos 18.4° corresponds more closely to 0.9487, which makes the correct angle approximately 18.4°. I realize now that a cube’s typical coordinate system should’ve been used in the calculations, with specific vector definitions for clarity. Time to fix the response with those in mind.

Verifying angle and vectors

It looks like the vectors are almost in the same direction, which is why the angle between them is small—around 18°. The difference between the two vectors was minimal, and the cosine value is about 0.948, confirming the small angle. Instead of trying to express it as arcsin, it seems better to stick with the cosine value to describe the angle more accurately. The calculation confirms that the angle is indeed around 18°, which makes sense given their similar orientations.

Thought for 1m 0s

답변

За-фиксируем куб с длиной ребра a=1a=1 в прямоугольной системе координат

A(0,0,0),  B(1,0,0),  C(1,1,0),  D(0,1,0),  A1(0,0,1),  B1(1,0,1),  C1(1,1,1),  D1(0,1,1).A(0,0,0),\; B(1,0,0),\; C(1,1,0),\; D(0,1,0),\; A_1(0,0,1),\; B_1(1,0,1),\; C_1(1,1,1),\; D_1(0,1,1).

1. Координаты нужных векторов

  • AD1→=D1−A=(0,  1,  1)\displaystyle\overrightarrow{AD_1}=D_1-A=(0,\;1,\;1);
  • точка MM — середина ребра DD1DD_1:
    M ⁣(0,  1,  12)M\!\left(0,\;1,\;{1\over2}\right);
  • AM→=M−A=(0,  1,  12).\displaystyle\overrightarrow{AM}=M-A=\Bigl(0,\;1,\;{1\over2}\Bigr).

2. Скалярное произведение и модули

AD1→ ⁣⋅ ⁣AM→=0⋅0+1⋅1+1⋅12=32.\overrightarrow{AD_1}\!\cdot\!\overrightarrow{AM}=0\cdot0+1\cdot1+1\cdot\frac12=\frac32. ∣AD1→∣=02+12+12=2,∣AM→∣=02+12+(12)2=1+14=54=52.\bigl|\overrightarrow{AD_1}\bigr|=\sqrt{0^2+1^2+1^2}=\sqrt2,\qquad \bigl|\overrightarrow{AM}\bigr| =\sqrt{0^2+1^2+\bigl(\tfrac12\bigr)^2} =\sqrt{1+\tfrac14}=\sqrt{\tfrac54}=\frac{\sqrt5}{2}.

3. Косинус и угол

cos⁡φ=AD1→⋅AM→∣AD1→∣  ∣AM→∣=322⋅52=310.\cos\varphi= \frac{\overrightarrow{AD_1}\cdot\overrightarrow{AM}} {\bigl|\overrightarrow{AD_1}\bigr|\;\bigl|\overrightarrow{AM}\bigr|} =\frac{\tfrac32}{\sqrt2\cdot\tfrac{\sqrt5}{2}} =\frac{3}{\sqrt{10}}. φ=arccos⁡ ⁣(310)≈18.4∘.\varphi=\arccos\!\Bigl(\frac{3}{\sqrt{10}}\Bigr)\approx18.4^{\circ}.
  ∠ ⁣(AD1→, AM→)≈18.4∘  \boxed{\;\angle\!\bigl(\overrightarrow{AD_1},\,\overrightarrow{AM}\bigr)\approx18.4^{\circ}\;}

질문

Масса вещества 36 грамм, масса по водороду 11 грамм, воды 4 грамма
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