MATH 126, Section 12.5 | Calculus III | Stewart 9th ed.
Difficulty: Intermediate | Prerequisites: Sections 12.1–12.5 concepts (dot product, cross product, normal vectors)
This section is about doing things with lines and planes: writing their equations, finding where they intersect, computing angles between planes, and measuring distances from points to planes. Every problem boils down to extracting a normal vector or a direction vector, plugging into a standard form, and solving a short system of equations.
Parametric equations of a line
Through point (x₀, y₀, z₀) with direction vector ⟨a, b, c⟩:
x = x₀ + at, y = y₀ + bt, z = z₀ + ct
Equation of a plane
Through point (x₀, y₀, z₀) with normal vector ⟨a, b, c⟩:
a(x − x₀) + b(y − y₀) + c(z − z₀) = 0
which simplifies to ax + by + cz = d, where d = ax₀ + by₀ + cz₀.
Angle between two planes
If two planes have normal vectors n₁ and n₂, the angle θ between them satisfies:
cos θ = |n₁ · n₂| / (|n₁| |n₂|)
The absolute value ensures 0 ≤ θ ≤ 90° (the acute angle between the planes).
Distance from a point to a plane
From point (x₁, y₁, z₁) to plane ax + by + cz + d = 0:
D = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²)
Note: the plane equation must be written as ax + by + cz + d = 0 (move everything to one side) before applying.
The normal vector of the plane is the direction vector of the line.
Example: line through (4, 1, 8) perpendicular to x − y + 4z = 1.
The plane's normal is n = ⟨1, −1, 4⟩. Use this as the direction vector.
Parametric equations: x = 4 + t, y = 1 − t, z = 8 + 4t.
To find where this line hits a coordinate plane, set the appropriate coordinate to zero and solve for t.
xy-plane (z = 0): 8 + 4t = 0 gives t = −2, so the point is (2, 3, 0).
yz-plane (x = 0): 4 + t = 0 gives t = −4, so the point is (0, 5, −8).
xz-plane (y = 0): 1 − t = 0 gives t = 1, so the point is (5, 0, 12).
Two non-parallel planes intersect in a line. To find its parametric equations:
The direction vector of the line is n₁ × n₂ (the cross product of the two normals).
To find a point on the line, set one variable (often t = 0 for one coordinate) and solve the resulting 2×2 system.
Example: planes x + y + z = 7 and x + 6y + 6z = 7. Normals: ⟨1, 1, 1⟩ and ⟨1, 6, 6⟩. Cross product: ⟨(1)(6) − (1)(6), (1)(1) − (1)(6), (1)(6) − (1)(1)⟩ = ⟨0, −5, 5⟩. Simplify to ⟨0, −1, 1⟩. Set z = 0: x + y = 7 and x + 6y = 7. Subtracting: 5y = 0, so y = 0, x = 7. Point: (7, 0, 0). Parametric equations: x = 7, y = −t, z = t.
This requires combining two constraints.
The line's direction vector v must be perpendicular to the plane's normal (v · n = 0) so the line is parallel to the plane.
v must also be perpendicular to the given line's direction vector d (v · d = 0).
Compute v = n × d. This gives a vector perpendicular to both.
Example: through (0, 1, 2), parallel to x + y + z = 4, perpendicular to x = 1 + t, y = 1 − t, z = 2t. Normal: n = ⟨1, 1, 1⟩. Direction of given line: d = ⟨1, −1, 2⟩. Cross product: n × d = ⟨(1)(2) − (1)(−1), (1)(1) − (1)(2), (1)(−1) − (1)(1)⟩ = ⟨3, −1, −2⟩. Parametric equations: x = 3t, y = 1 − t, z = 2 − 2t.
You need a normal vector. Get two direction vectors that lie in the plane:
One is the line's direction vector.
The other connects a point on the line to the given point.
Cross these two vectors to get the normal.
Example: plane through (3, 6, −1) containing x = 4 − t, y = 2t − 1, z = −3t. The line's direction is d = ⟨−1, 2, −3⟩. A point on the line (t = 0) is (4, −1, 0). Vector from (4, −1, 0) to (3, 6, −1) is ⟨−1, 7, −1⟩. Normal = d × ⟨−1, 7, −1⟩ = ⟨(2)(−1) − (−3)(7), (−3)(−1) − (−1)(−1), (−1)(7) − (2)(−1)⟩ = ⟨19, 4, −5⟩ (verify your arithmetic). Plane: 19(x − 3) + 4(y − 6) − 5(z + 1) = 0.
Any plane through the line of intersection of planes P₁: a₁x + b₁y + c₁z = d₁ and P₂: a₂x + b₂y + c₂z = d₂ can be written as:
(a₁x + b₁y + c₁z − d₁) + λ(a₂x + b₂y + c₂z − d₂) = 0
This is a family of planes parameterised by λ. Use a third condition (perpendicularity to another plane, or passing through a specific point) to pin down λ.
Example: plane through the intersection of x − z = 3 and y + 3z = 1, perpendicular to x + y − 2z = 4. Write (x − z − 3) + λ(y + 3z − 1) = 0, giving normal ⟨1, λ, −1 + 3λ⟩. For perpendicularity to the third plane (normal ⟨1, 1, −2⟩), set dot product to zero: 1 + λ + (−1 + 3λ)(−2) = 0 → 1 + λ + 2 − 6λ = 0 → λ = 3/5. Substitute back to get the equation.
Find the point of intersection by solving the parametric equations simultaneously.
The direction vectors of the two lines both lie in the plane.
Cross the two direction vectors to get the normal.
Example: lines r = ⟨1, 3, 0⟩ + t⟨3, −3, 3⟩ and r = ⟨4, 0, 3⟩ + s⟨−3, 3, 0⟩. Setting equal: 1 + 3t = 4 − 3s, 3 − 3t = 3s, 3t = 3. From the third: t = 1, then s = 0 from the second. Intersection: (4, 0, 3). Normal = ⟨3, −3, 3⟩ × ⟨−3, 3, 0⟩ = ⟨(−3)(0) − (3)(3), (3)(−3) − (3)(0), (3)(3) − (−3)(−3)⟩ = ⟨−9, −9, 0⟩. Simplify: ⟨1, 1, 0⟩. Plane: (x − 4) + (y − 0) = 0 → x + y = 4.
Substitute the parametric equations of the line into the plane equation, solve for t, then compute the coordinates.
Example: line x = 2 − 2t, y = 3t, z = 1 + t; plane x + 2y − z = 4. Substitute: (2 − 2t) + 2(3t) − (1 + t) = 4 → 2 − 2t + 6t − 1 − t = 4 → 1 + 3t = 4 → t = 1. Point: (0, 3, 2).
Example: line through (−3, 1, 0) and (−1, 3, 4); plane 2x + y − z = −2. Direction: (−1) − (−3), 3 − 1, 4 − 0) = ⟨2, 2, 4⟩. Parametric: x = −3 + 2t, y = 1 + 2t, z = 4t. Substitute: 2(−3 + 2t) + (1 + 2t) − 4t = −2 → −6 + 4t + 1 + 2t − 4t = −2 → −5 + 2t = −2 → t = 3/2. Point: (0, 4, 6).
Use the formula cos θ = |n₁ · n₂| / (|n₁| |n₂|). Always report the acute angle.
Example: planes x + y + z = 7 and x + 6y + 6z = 7. Normals: n₁ = ⟨1, 1, 1⟩, n₂ = ⟨1, 6, 6⟩. Dot product: 1 + 6 + 6 = 13. Magnitudes: |n₁| = √3, |n₂| = √(1 + 36 + 36) = √73. cos θ = 13 / (√3 · √73) = 13 / √219. θ = arccos(13/√219) ≈ 28.5°.
This is a direct formula application, but the sign conventions trip people up.
Given point (x₁, y₁, z₁) and plane ax + by + cz = d, rewrite the plane as ax + by + cz − d = 0 (so it reads ax + by + cz + D = 0 where D = −d), then:
Distance = |ax₁ + by₁ + cz₁ − d| / √(a² + b² + c²)
Example: point (1, −6, 3), plane 3x + 2y + 6z = 5. Numerator: |3(1) + 2(−6) + 6(3) − 5| = |3 − 12 + 18 − 5| = |4| = 4. Denominator: √(9 + 4 + 36) = √49 = 7. Distance = 4/7.
The absolute value is essential. The signed version of the numerator tells you which side of the plane the point sits on, but the distance is always positive.
Confusing normal vectors with direction vectors. The coefficients of x, y, z in a plane equation give the normal vector, not a direction along the plane. A line perpendicular to a plane uses the normal as its direction, but a line parallel to a plane has a direction perpendicular to the normal.
Forgetting absolute value in the distance formula. The numerator must be wrapped in absolute value bars. Without them, you may get a negative distance, which is not physically meaningful.
Using the wrong form of the plane equation in the distance formula. The formula assumes ax + by + cz + D = 0 (everything on one side). If the plane is written as 3x + 2y + 6z = 5, you must subtract 5 from both sides first.
Getting cross product signs wrong. The cross product is not commutative: a × b = −(b × a). Be consistent with the order and double-check each component.
Assuming two planes through a point and a line of intersection are unique. The family (P₁) + λ(P₂) = 0 generates infinitely many planes through the intersection line. You need one more condition (λ value) to pin down a single plane.
The plane 2x − 3y + z = 7 has normal vector ____. (⟨2, −3, 1⟩)
A line perpendicular to the plane 2x − 3y + z = 7 through (1, 0, 4) has parametric equations: x = ____, y = ____, z = ____. (x = 1 + 2t, y = −3t, z = 4 + t)
The distance from the origin to the plane x + 2y + 2z = 9 is ____. (9/3 = 3)
True or false: to find the line of intersection of two planes, cross their normal vectors. (True, that gives the direction vector)
The angle between two planes with normals ⟨1, 0, 0⟩ and ⟨0, 1, 0⟩ is ____. (90°)
Q: Find parametric equations for the line through (2, 5, −1) perpendicular to the plane 3x − y + 2z = 10.
A: Normal = ⟨3, −1, 2⟩. Parametric: x = 2 + 3t, y = 5 − t, z = −1 + 2t.
Q: The line x = 1 + 2t, y = −3t, z = 5 + t meets the plane 2x − y + z = 12 at what point?
A: Substitute: 2(1 + 2t) − (−3t) + (5 + t) = 12 → 2 + 4t + 3t + 5 + t = 12 → 8t = 5 → t = 5/8. Point: (1 + 10/8, −15/8, 5 + 5/8) = (18/8, −15/8, 45/8) = (9/4, −15/8, 45/8).
Q: Find the distance from (2, 1, −3) to the plane x − 2y + 2z = 6.
A: |2 − 2 − 6 − 6| / √(1 + 4 + 4) = |−12| / 3 = 4.
Q: Find the angle between planes 2x + y − z = 3 and x + 3y + 2z = 1.
A: n₁ = ⟨2, 1, −1⟩, n₂ = ⟨1, 3, 2⟩. Dot: 2 + 3 − 2 = 3. |n₁| = √6, |n₂| = √14. cos θ = 3/√84 = 3/(2√21). θ = arccos(3/(2√21)) ≈ 70.9°.
Q: Write the equation of the plane through (1, 2, 3) containing the line of intersection of x + 2y + 3z = 1 and 2x − y + z = −3.
A: Family: (x + 2y + 3z − 1) + λ(2x − y + z + 3) = 0. Plug in (1, 2, 3): (1 + 4 + 9 − 1) + λ(2 − 2 + 3 + 3) = 0 → 13 + 6λ = 0 → λ = −13/6. Substitute λ back and simplify.
The cross product (Section 12.4) is the workhorse of this section: every time you need a normal to a plane from two directions, or a direction vector for the intersection of two planes, you are computing a cross product. Comfort with the component formula pays off heavily here.
These techniques return in Sections 14.4 and 14.6 (tangent planes to surfaces), where the gradient vector plays the role of the normal vector, and in Section 15.6 (surface area and surface integrals), where the cross product of partial derivatives gives the normal to a parametric surface.
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