Source: MATH 163 Class Notes, Section 12.5
Difficulty: Intermediate | Prerequisites: Vectors in 3D (dot product, cross product, magnitude), parametric equations of lines (Section 12.5 earlier material).
This material sits at the transition from lines in 3D space to planes. You have already learned how to write parametric and symmetric equations for a line, and now the focus shifts to planes: how to define one, how to measure distances to one, and how planes and lines interact. These tools come back constantly in multivariable calculus (surface integrals, tangent planes, linear approximation), so the geometry here is foundational. If you are comfortable with dot products, cross products, and parametric lines, you are ready for this.
You can measure the distance from a point to a line using a cross product, and the distance from a point to a plane using a short plug-in formula. To define a plane, you need a normal vector and a point on it. Two planes can intersect along a line, and you find that line by crossing their normals (for direction) and solving simultaneously (for a starting point).
Normal vector (n)
A vector perpendicular to a plane. The coefficients A, B, C in the standard plane equation Ax + By + Cz = D are the components of the normal vector. Think of it as an arrow sticking straight out of the plane's surface.
Direction vector (v)
The vector that indicates which way a line runs. In a parametric line r(t) = P + vt, the direction vector is v. In simple terms, it tells you the line's orientation in space.
Standard form of a plane equation
Ax + By + Cz = D, where (A, B, C) is the normal vector and D is computed by plugging in any known point on the plane. Think of it as the "address" of the plane written in terms of x, y, and z.
Vector form of a plane equation
(r - r_0) · n = 0, where r = (x, y, z), r_0 is a known point on the plane, and n is the normal vector. In simple terms, this says "any point on the plane, minus a known point, gives a vector that is perpendicular to the normal."
Parametric form of a line
x(t) = x_0 + at, y(t) = y_0 + bt, z(t) = z_0 + ct. Each coordinate is expressed as a separate function of a single parameter t.
Symmetric form of a line
(x - x_0)/a = (y - y_0)/b = (z - z_0)/c. All three parametric equations set equal by solving each for t. Think of it as the "compact" way of writing a 3D line without explicitly using a parameter.
Traces
The curves you get by slicing a surface with planes parallel to coordinate planes (setting x, y, or z to a constant). For z = x² + y², the trace at z = k is a circle of radius √k. In simple terms, traces are cross-sections of a 3D surface.
The core formula is d = |w × v| / |v|, where v is the direction vector of the line and w is the vector from any point on the line to the external point.
Identify the direction vector v and a point P_0 on the line from its parametric equation r(t) = P_0 + vt.
Construct w by subtracting P_0 from the external point P: w = P - P_0.
Compute the cross product w × v.
Take the magnitude of that cross product and divide by |v|.
Worked example: Find the distance from (4, 9, 2) to the line r(t) = ⟨1, -2, 5⟩ + ⟨3, 1, 7⟩t.
v = ⟨3, 1, 7⟩, point on line = (1, -2, 5).
w = ⟨4 - 1, 9 - (-2), 2 - 5⟩ = ⟨3, 11, -3⟩.
w × v = ⟨(11)(7) - (-3)(1), (-3)(3) - (3)(7), (3)(1) - (11)(3)⟩ = ⟨80, -30, -30⟩.
d = |⟨80, -30, -30⟩| / |⟨3, 1, 7⟩| = √(6400 + 900 + 900) / √(9 + 1 + 49) = √8200 / √59 ≈ 11.79.
Given three non-collinear points, you can define a unique plane. The method has three steps.
Step 1: Build two vectors in the plane.
Pick one point as a base and subtract it from the other two.
Example: points (1, 2, 3), (-3, 5, 7), (4, -2, 1).
v_1 = ⟨-3 - 1, 5 - 2, 7 - 3⟩ = ⟨-4, 3, 4⟩
v_2 = ⟨4 - 1, -2 - 2, 1 - 3⟩ = ⟨3, -4, -2⟩
Step 2: Cross product to get the normal vector.
n = v_1 × v_2. Use the determinant method with i, j, k.
For the example: n = ⟨(3)(-2) - (4)(-4), (4)(3) - (-4)(-2), (-4)(-4) - (3)(3)⟩ = ⟨10, 4, 7⟩.
Step 3: Write the plane equation.
Vector form: (⟨x, y, z⟩ - ⟨1, 2, 3⟩) · ⟨10, 4, 7⟩ = 0.
Expand the dot product: 10(x - 1) + 4(y - 2) + 7(z - 3) = 0.
Standard form: 10x + 4y + 7z = 39.
General rule: If the normal vector is ⟨a, b, c⟩ and one point on the plane is (x_0, y_0, z_0), the standard form is ax + by + cz = d, where d = ax_0 + by_0 + cz_0.
For a point P = (x_1, y_1, z_1) and a plane Ax + By + Cz = D, the distance is:
d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}}Plug the point's coordinates into the left side of the plane equation, subtract D, take the absolute value, and divide by the magnitude of the normal vector. The derivation uses projection of a vector onto the normal, covered in Section 12.5 of the textbook.
Formula | What it gives you |
|---|---|
d = |w × v| / |v| | Distance from a point to a line |
n = v_1 × v_2 | Normal vector to a plane through three points |
ax + by + cz = d, where d = ax_0 + by_0 + cz_0 | Standard form of a plane |
(r - r_0) · n = 0 | Vector form of a plane |
d = |Ax_1 + By_1 + Cz_1 - D| / √(A² + B² + C²) | Distance from a point to a plane |
θ = cos⁻¹( (n_1 · n_2) / (|n_1| |n_2|) ) | Angle between two planes |
Direction of intersection line = n_1 × n_2 | Direction vector for the line where two planes meet |
Students often confuse the normal vector with the direction vector. The normal vector is perpendicular to the plane; the direction vector runs along a line. They serve completely different roles in the formulas.
When computing distance from a point to a line, students sometimes use the dot product instead of the cross product. The cross product is what gives the parallelogram area needed for this distance formula.
Students forget the absolute value in the point-to-plane distance formula and end up with negative distances, which are not meaningful here.
When finding a plane through three points, students occasionally use two vectors that share a direction (picking collinear points), which makes the cross product zero and the problem unsolvable. Always check that your three points are not on the same line.
⚠️ The point-to-line distance formula (cross product version) is a classic exam problem. You will almost certainly see one.
⚠️ Finding a plane equation through three points is a multi-step problem that tests cross products, vector subtraction, and dot products all at once. Know the full pipeline.
⚠️ The point-to-plane distance formula is short to apply but easy to misremember. Make sure you have the absolute value on the numerator and the square root on the denominator.
⚠️ Converting between vector form, parametric form, and symmetric form of a line is a common "show all three" exam question. Practice writing all three from the same starting point and direction.
True or false: The normal vector of a plane lies within the plane. (False, it is perpendicular to it.)
Fill in the blank: To find the distance from a point to a line, you compute |w × v| / ____. (|v|)
True or false: Two planes that have parallel normal vectors are always the same plane. (False, they are parallel but may be different planes.)
Fill in the blank: The direction vector for the line of intersection of two planes is found by taking the _____ of their normal vectors. (cross product)
True or false: In the point-to-plane distance formula, you can get a negative distance if the point is on the opposite side of the plane. (False, the absolute value ensures the distance is always non-negative.)
Q: Find the distance from the point (2, 0, -1) to the line r(t) = ⟨0, 1, 3⟩ + ⟨1, -1, 2⟩t.
A: w = ⟨2 - 0, 0 - 1, -1 - 3⟩ = ⟨2, -1, -4⟩. Cross product w × v = ⟨(-1)(2) - (-4)(-1), (-4)(1) - (2)(2), (2)(-1) - (-1)(1)⟩ = ⟨-6, -8, -1⟩. d = √(36 + 64 + 1) / √(1 + 1 + 4) = √101 / √6 ≈ 4.10.
Q: Find the equation of the plane containing (0, 0, 0), (1, 0, 2), and (0, 3, 1) in standard form.
A: v_1 = ⟨1, 0, 2⟩, v_2 = ⟨0, 3, 1⟩. n = v_1 × v_2 = ⟨(0)(1) - (2)(3), (2)(0) - (1)(1), (1)(3) - (0)(0)⟩ = ⟨-6, -1, 3⟩. Standard form: -6x - y + 3z = 0 (since the origin is on the plane, d = 0).
Q: Find the angle between the planes 2x + y - z = 4 and x - y + 3z = 2.
A: n_1 = ⟨2, 1, -1⟩, n_2 = ⟨1, -1, 3⟩. n_1 · n_2 = 2 - 1 - 3 = -2. |n_1| = √6, |n_2| = √11. θ = cos⁻¹(-2 / √66) ≈ 104.3° (or its supplement ≈ 75.7° if you want the acute angle between the planes).
Q: Find the point where the line x(t) = 3t + 1, y(t) = -2t, z(t) = 5t - 4 intersects the plane 5x - 4y + 2z = 3.
A: Substitute: 5(3t + 1) - 4(-2t) + 2(5t - 4) = 3. This gives 33t - 3 = 3, so t = 2/11. Plugging back in: (17/11, -4/11, -34/11).
Q: Find the distance from the point (1, 2, 3) to the plane 2x - y + 2z = 4.
A: d = |2(1) - 1(2) + 2(3) - 4| / √(4 + 1 + 4) = |2 - 2 + 6 - 4| / 3 = 2/3.
This material connects directly to tangent planes and linear approximation in Chapter 14. When you take partial derivatives of a function f(x, y) and build a tangent plane, you are using the same normal-vector-and-point framework from this section.
The cross product geometry here also appears in surface integrals and flux (Chapter 16), where normal vectors to surfaces are essential.
Traces (Section 12.6, briefly introduced here with z = x² + y²) lead into quadric surfaces, which are the 3D analogues of conic sections and appear throughout the rest of the course.
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