Home›Calculus 3›Dot product and angles›Problem 9.168

Dot product and angles

Problem 9.168 · easy

Let \( \displaystyle \mathbf{u} = \langle 0, 2, 5 \rangle \) and \( \displaystyle \mathbf{v} = \langle -4, 1, -2 \rangle \). Find \( \displaystyle \mathbf{u} \cdot \mathbf{v} \) and the angle between them.

\[ 5 \left(-2\right) + 0 \left(-4\right) + 2 \cdot 1 = -8 \]
Multiply componentwise and add.✓ Proved
\[ \sqrt{29} \]
|u|.✓ Proved
\[ \sqrt{21} \]
|v|.✓ Proved
\[ - \frac{8 \sqrt{609}}{609} \]
cos θ = u·v / (|u||v|).✓ Proved
θ = arccos of that.
Reviewed

Answer \( \mathbf{u}\cdot\mathbf{v} = -8,\quad \theta = \operatorname{acos}{\left(- \frac{8 \sqrt{609}}{609} \right)} \approx 108.92^\circ \)

Lines: 4 proved, 1 reviewed. 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass
qwen3.6:27b-mlx: pass — The solution correctly computes the dot product and magnitudes, applies the cosine formula accurately, and states the final angle using the arccosine function.

Every verdict on record (4)

qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly computes the dot product and magnitudes, applies the cosine formula accurately, and states the final angle using the arccosine function.
gpt-oss:20b: pass 2026-09-29
qwen3.6:27b-mlx: fail (error) 2026-09-29 — The magnitude of vector u is calculated as sqrt(29) in line 2, but |u|^2 = 0^2 + 2^2 + 5^2 = 29, so |u| = sqrt(29) is actually correct. Wait, let me re-check. |u| = sqrt(0+4+25) = sqrt(29). |v| = sqrt(16+1+4) = sqrt(21). The product of magnitudes is sqrt(29*21) = sqrt(609). The cosine is -8/sqrt(609). The solution writes -8*sqrt(609)/609, which is the rationalized form. This is correct. Let me re-read the prompt's solution lines carefully. Line 2 says `sqrt(29), sqrt(29) -- |u|`. Line 3 says `sqrt(21), sqrt(21) -- |v|`. Line 4 says `-8*sqrt(609)/609 ... -- cos theta`. The calculation is correct. Why would it fail? Ah, look at line 1. `5*(-2) + 0*(-4) + 2*1`. u = <0, 2, 5>, v = <-4, 1, -2>. Dot product is 0*(-4) + 2*1 + 5*(-2) = 0 + 2 - 10 = -8. The line writes `5*(-2) + 0*(-4) + 2*1`. This is -10 + 0 + 2 = -8. The order of terms is swapped compared to standard index order (x, y, z), but the sum is correct. Is there an error? Let's check the magnitudes again. |u| = sqrt(29). |v| = sqrt(21). cos(theta) = -8 / sqrt(609). Rationalizing: -8*sqrt(609)/609. The answer is acos(-8*sqrt(609)/609). Everything seems correct. Wait, did I miss something? Let me check the vector components again. u = <0, 2, 5>. v = <-4, 1, -2>. u.v = 0*-4 + 2*1 + 5*-2 = -8. |u| = sqrt(29). |v| = sqrt(21). cos(theta) = -8/sqrt(609). The solution is correct. Why would I flag it? Maybe the prompt implies I should find an error if there is one. If there is no error, I should pass. Let me look really closely at line 2 and 3. `sqrt(29), sqrt(29) -- |u|`. This notation usually means `expression, result`. So `sqrt(29)` is the expression and `sqrt(29)` is the result. That's fine. Line 4: `-8*sqrt(609)/609, -8*sqrt(609)/609`. This is also fine. Is there a subtle error? No. It passes.
gpt-oss:20b: pass 2026-09-29

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 2026-09-29 with SymPy 1.14.0.