Dot product and angles
Problem 9.141 · easy
Let \( \displaystyle \mathbf{u} = \langle 2, -2, -1 \rangle \) and \( \displaystyle \mathbf{v} = \langle 3, 2, 0 \rangle \). Find \( \displaystyle \mathbf{u} \cdot \mathbf{v} \) and the angle between them.
- \[ \left(-2\right) 2 - 1 \cdot 0 + 2 \cdot 3 = 2 \]Multiply componentwise and add.✓ Proved
- \[ 3 \]|u|.✓ Proved
- \[ \sqrt{13} \]|v|.✓ Proved
- \[ \frac{2 \sqrt{13}}{39} \]cos θ = u·v / (|u||v|).✓ Proved
- θ = arccos of that.
Answer \( \mathbf{u}\cdot\mathbf{v} = 2,\quad \theta = \operatorname{acos}{\left(\frac{2 \sqrt{13}}{39} \right)} \approx 79.34^\circ \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | Not checked | — | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the law of cosines on the triangle u, v, u − v gives the same angle |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The calculation of |u| is incorrect; the solution claims |u|=3, but |u|=sqrt(2^2+(-2)^2+(-1)^2)=3 is actually correct. However, the calculation of |
Every verdict on record (4)
qwen3.6:27b-mlx: inconclusive 2026-09-28 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The calculation of |u| is incorrect; the solution claims |u|=3, but |u|=sqrt(2^2+(-2)^2+(-1)^2)=3 is actually correct. However, the calculation of |gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: fail (error) 2026-09-28 — The magnitude of vector u is calculated as 3, but it should be sqrt(9) = 3. Wait, u = <2, -2, -1>, so |u|^2 = 4+4+1=9, |u|=3. That is correct. The magnitude of v = <3, 2, 0> is sqrt(9+4+0)=sqrt(13). That is correct. The dot product is 2*3 + (-2)*2 + (-1)*0 = 6-4+0=2. That is correct. cos(theta) = 2 / (3*sqrt(13)) = 2*sqrt(13)/39. That is correct. The final answer is arccos(2*sqrt(13)/39). The solution steps are algebraically correct. Why did I think there was an error? Let me re-read carefully. Step 1: -2*2 - 1*0 + 2*3. This corresponds to u_y*v_y + u_z*v_z + u_x*v_x? No, standard order is u_x*v_x + u_y*v_y + u_z*v_z. u_x=2, v_x=3 -> 6. u_y=-2, v_y=2 -> -4. u_z=-1, v_z=0 -> 0. Sum = 2. The expression written is -2*2 - 1*0 + 2*3. This is u_y*v_y + u_z*v_z + u_x*v_x. The order is mixed up but the terms are correct components. It evaluates to 2. Step 2: |u| = 3. Correct. Step 3: |v| = sqrt(13). Correct. Step 4: cos theta = 2 / (3*sqrt(13)). Rationalized: 2*sqrt(13)/39. Correct. Step 5: theta = arccos(...). Correct. The solution is actually correct. I will mark it as pass.gpt-oss:20b: pass 2026-09-28
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/dot_product_angle, checked 2026-09-28 with SymPy 1.14.0.