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Equation of a plane

Problem 9.179 · medium

Find an equation of the plane through the points \( \displaystyle (3, 1, 3),\ (0, 3, -1),\ (-2, 3, -1) \).
  1. \[ \left[\begin{matrix}-3\\2\\-4\end{matrix}\right] \]
    PQ.✓ Proved
  2. \[ \left[\begin{matrix}-5\\2\\-4\end{matrix}\right] \]
    PR.✓ Proved
  3. \[ \left[\begin{matrix}0\\2\\1\end{matrix}\right] \]
    A normal vector n = PQ × PR (scaled).✓ Proved
  4. \[ 5 \]
    n · P gives the constant.✓ Proved
  5. The plane is 2*y + z = 5.
Answer \( 2 y + z = 5 \)

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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5Not checked—a sentence; read, not computed
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0all three points satisfy the equation

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The normal vector calculated in step 3 is (0, 2, 1), which implies the x-component is 0. However, the points (3, 1, 3), (0, 3, -1), and (-2, 3, -1) do not lie on a plane with a normal vector having a zero x-component (the vector PR has an x-component of -5, so the normal must have a non-zero x-component unless the cross product was computed incorrectly). Specifically, PQ = (-3, 2, -4) and PR = (-5, 2, -4). The cross product PQ x PR is (0, -8, 4) or simplified (0, -2, 1). Wait, let's recompute. PQ = Q-P = (0-3, 3-1, -1-3) = (-3, 2, -4). PR = R-P = (-2-3, 3-1, -1-3) = (-5, 2, -4). Cross product: i(2*(-4) - (-4)*2) - j((-3)*(-4) - (-4)*(-5)) + k((-3)*2 - 2*(-5)) = i(-8+8) - j(12-20) + k(-6+10) = 0i - j(-8) + k(4) = (0, 8, 4). This simplifies to (0, 2, 1). So the normal vector IS (0, 2, 1). The equation is 0(x-3) + 2(y-1) + 1(z-3) = 0 => 2y - 2 + z - 3 = 0 => 2y + z = 5. The solution is actually correct. Why did I think it was wrong? Let me re-read the points. P(3,1,3), Q(0,3,-1), R(-2,3,-1). Note that Q and R have the same y and z coordinates. The line QR is parallel to the x-axis. The vector QR is (-2, 0, 0). The vector PQ is (-3, 2, -4). The normal must be perpendicular to QR, so its x-component must be 0. The calculation is correct. The solution is sound.
Every verdict on record (4)
  • qwen3.6:27b-mlx: fail (error) 2026-09-29 — The normal vector calculated in step 3 is (0, 2, 1), which implies the x-component is 0. However, the points (3, 1, 3), (0, 3, -1), and (-2, 3, -1) do not lie on a plane with a normal vector having a zero x-component (the vector PR has an x-component of -5, so the normal must have a non-zero x-component unless the cross product was computed incorrectly). Specifically, PQ = (-3, 2, -4) and PR = (-5, 2, -4). The cross product PQ x PR is (0, -8, 4) or simplified (0, -2, 1). Wait, let's recompute. PQ = Q-P = (0-3, 3-1, -1-3) = (-3, 2, -4). PR = R-P = (-2-3, 3-1, -1-3) = (-5, 2, -4). Cross product: i(2*(-4) - (-4)*2) - j((-3)*(-4) - (-4)*(-5)) + k((-3)*2 - 2*(-5)) = i(-8+8) - j(12-20) + k(-6+10) = 0i - j(-8) + k(4) = (0, 8, 4). This simplifies to (0, 2, 1). So the normal vector IS (0, 2, 1). The equation is 0(x-3) + 2(y-1) + 1(z-3) = 0 => 2y - 2 + z - 3 = 0 => 2y + z = 5. The solution is actually correct. Why did I think it was wrong? Let me re-read the points. P(3,1,3), Q(0,3,-1), R(-2,3,-1). Note that Q and R have the same y and z coordinates. The line QR is parallel to the x-axis. The vector QR is (-2, 0, 0). The vector PQ is (-3, 2, -4). The normal must be perpendicular to QR, so its x-component must be 0. The calculation is correct. The solution is sound.
  • gpt-oss:20b: pass 2026-09-29
  • qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly identifies the normal vector and the plane equation. The intermediate steps show the correct cross product result and constant calculation.
  • 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/plane_through_points, checked 2026-09-29 with SymPy 1.14.0.