Difficulty: Intermediate | Prerequisites: Section 12.1–12.4 (vectors, dot product, cross product, equations of lines)
Planes are the two-dimensional analogue of lines in three-dimensional space. Where a line in 3D is defined by a point and a direction vector, a plane is defined by a point and a normal vector. This section builds directly on cross products (used to find normals from three points) and dot products (used to find angles between planes and distances to planes). If you are not comfortable with dot and cross products, review those first.
A plane in 3D is pinned down by one point on it and a normal vector (a vector perpendicular to the plane). The equation of a plane always takes the form a(x - x0) + b(y - y0) + c(z - z0) = 0, where (a, b, c) is the normal vector and (x0, y0, z0) is the known point. Most problems in this section boil down to finding the right normal vector for the situation, then plugging into that formula.
Normal vector
A vector n = (a, b, c) that is perpendicular to every vector lying in the plane. Think of it as the arrow sticking straight out of the plane's surface. The normal vector's components become the coefficients in the plane's equation.
Scalar equation of a plane
The equation a(x - x0) + b(y - y0) + c(z - z0) = 0, where (a, b, c) is the normal vector and (x0, y0, z0) is a point on the plane. In simple terms, this is the standard form you expand and simplify into ax + by + cz = d.
Direction numbers (direction vector)
The coefficients of the parameter t in a parametric line equation. For the line x = x0 + at, y = y0 + bt, z = z0 + ct, the direction numbers are (a, b, c). In simple terms, these tell you which way the line points, and they double as the normal vector when a plane is perpendicular to that line.
Parallel planes
Two planes whose normal vectors are scalar multiples of each other (they point in the same or opposite direction). Parallel planes never intersect. In simple terms, if you can multiply one normal by a constant to get the other, the planes are parallel.
Perpendicular planes
Two planes whose normal vectors have a dot product of zero: n1 . n2 = 0. In simple terms, the planes meet at a right angle.
Angle between planes
The acute angle between two planes, found using the dot product of their normal vectors. Defined by cos(theta) = |n1 . n2| / (|n1| |n2|). Think of it as the angle you would measure if you stood at the line where the planes intersect and looked at the "dihedral" opening between them.
Distance from a point to a plane
The shortest (perpendicular) distance from a point to a plane, given by D = |ax1 + by1 + cz1 - d| / sqrt(a^2 + b^2 + c^2). In simple terms, drop a perpendicular from the point to the plane and measure its length.
Intercept form of a plane
The equation x/a + y/b + z/c = 1, where a, b, c are the x-, y-, and z-intercepts. In simple terms, this is a compact way to write a plane's equation when you know where it crosses each axis.
Read off the direction vector of the line. For x = x0 + at, y = y0 + bt, z = z0 + ct, the direction vector is (a, b, c).
Because the plane is perpendicular to the line, the line's direction vector IS the plane's normal vector.
Plug the normal (a, b, c) and the given point (x0, y0, z0) into a(x - x0) + b(y - y0) + c(z - z0) = 0.
Example: plane through (6, 0, 3), perpendicular to x = 4t, y = 6 - t, z = 5 + 3t. The direction vector is (4, -1, 3). The equation is 4(x - 6) - 1(y - 0) + 3(z - 3) = 0, which simplifies to 4x - y + 3z = 33.
Parallel planes share the same normal vector. Read the coefficients (a, b, c) directly from the given plane's equation.
Use those same coefficients with the new point.
Example: plane through (4, -8, -4), parallel to 4x - y - z = 8. Normal is (4, -1, -1). Equation: 4(x - 4) - 1(y + 8) - 1(z + 4) = 0, which simplifies to 4x - y - z = 28.
Form two vectors from the three points: v1 = P2 - P1, v2 = P3 - P1.
Compute the cross product n = v1 x v2. This is the normal vector.
Use any of the three points with the normal in the scalar equation.
Example: points (4, 1, 4), (5, -8, 6), (-4, -5, 1). v1 = (1, -9, 2), v2 = (-8, -6, -3). Cross product n = (-9)(-3) - (2)(-6), (2)(-8) - (1)(-3), (1)(-6) - (-9)(-8) = (27 + 12, -16 + 3, -6 - 72) = (39, -13, -78). Simplify by dividing by 13: (3, -1, -6). Equation: 3(x - 4) - (y - 1) - 6(z - 4) = 0, giving 3x - y - 6z = -13.
x-intercept: set y = 0 and z = 0, solve for x.
y-intercept: set x = 0 and z = 0, solve for y.
z-intercept: set x = 0 and y = 0, solve for z.
Example: 2x + 4y + z = 8. x-intercept = 4, y-intercept = 2, z-intercept = 8.
Extract normal vectors n1 and n2 from each plane equation.
Apply cos(theta) = |n1 . n2| / (|n1| |n2|). Use the absolute value in the numerator to get the acute angle.
Example: x + y + z = 0 and x + 4y + 2z = 6. Normals: (1, 1, 1) and (1, 4, 2). Dot product = 1 + 4 + 2 = 7. Magnitudes: sqrt(3) and sqrt(21). cos(theta) = 7 / sqrt(63) = 7 / (3*sqrt(7)) = sqrt(7)/3.
Parallel check: are the normals scalar multiples? If n1 = k * n2 for some constant k, the planes are parallel.
Perpendicular check: is n1 . n2 = 0? If yes, the planes are perpendicular.
Neither: compute the angle using cos(theta) = |n1 . n2| / (|n1| |n2|), then theta = arccos of that value.
Example 1: 7x + 28y - 21z = 1 and -6x + 12y + 14z = 0. Normals: (7, 28, -21) and (-6, 12, 14). Dot product = -42 + 336 - 294 = 0. The planes are perpendicular.
Example 2: x - y - 3z = 1 and 3x + y - z = 2. Normals: (1, -1, -3) and (3, 1, -1). Dot product = 3 - 1 + 3 = 5. Not zero, and (1, -1, -3) is not a multiple of (3, 1, -1). Neither. Angle: cos(theta) = 5 / (sqrt(11) * sqrt(11)) = 5/11. theta = arccos(5/11) which is roughly 63.0 degrees.
The plane passes through the midpoint of the two points.
The normal vector is the vector connecting the two points (or any scalar multiple of it).
Midpoint of (-6, 4, 1) and (2, 6, 5) = (-2, 5, 3). Direction vector = (2 - (-6), 6 - 4, 5 - 1) = (8, 2, 4), which simplifies to (4, 1, 2).
Equation: 4(x + 2) + 1(y - 5) + 2(z - 3) = 0, giving 4x + y + 2z = -3 + 5 + 6 = 9. So 4x + y + 2z = 9.
If a plane has x-intercept a, y-intercept b, z-intercept c (all non-zero), the equation is x/a + y/b + z/c = 1.
This is derived from the standard form by dividing through by the constant d.
Use D = |ax1 + by1 + cz1 - d| / sqrt(a^2 + b^2 + c^2), where ax + by + cz = d is the plane and (x1, y1, z1) is the point.
The plane must be in the form ax + by + cz - d = 0 (move d to the left side).
Example: distance from (-5, 5, 2) to x - 2y - 4z = 8. D = |(-5) - 2(5) - 4(2) - 8| / sqrt(1 + 4 + 16) = |-5 - 10 - 8 - 8| / sqrt(21) = 31 / sqrt(21) = 31*sqrt(21) / 21.
\text{Scalar equation: } a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\text{Linear form: } ax + by + cz = d\text{Intercept form: } \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1|\,|\mathbf{n}_2|}D = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}\mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2 \quad \text{(normal from two in-plane vectors)}Students often confuse the direction vector of a line with a vector lying in the plane. When a plane is perpendicular to a line, the line's direction vector is the plane's normal, not a vector in the plane.
When checking whether two planes are parallel, students sometimes compare the constant terms (the d values). The constants do not matter for parallelism. Only the normal vector ratios matter.
In the distance formula, students forget to move d to the left side of the equation before plugging in. The formula requires ax + by + cz - d = 0 form, so the numerator is |ax1 + by1 + cz1 - d|, not |ax1 + by1 + cz1 + d|.
Students sometimes drop the absolute value in the angle formula and get an obtuse angle. The absolute value ensures you always get the acute angle between the planes.
Planes come up repeatedly in Calculus III, from computing flux integrals (the normal vector reappears in surface integrals) to setting up limits of integration in triple integrals. This is also where you first practise combining dot products and cross products in a single problem, which is a skill tested heavily on exams.
The "find the equation of a plane" problem type appears on nearly every Calc III exam. Know all three setups: point + normal, point + parallel plane, three points.
Distance from a point to a plane is a common exam question because it combines the plane equation with the distance formula in one step.
Angle-between-planes questions test whether you remember to use the absolute value (acute angle convention).
True or false: two planes with normal vectors (2, 4, -6) and (-1, -2, 3) are parallel. ______
The x-intercept of the plane 3x + 6y + 2z = 12 is x = ______.
True or false: if the dot product of two planes' normal vectors is zero, the planes are parallel. ______
The distance formula for point-to-plane requires the plane in the form ax + by + cz = d. The numerator of the formula is |______|.
To find a normal vector from three points, you compute the ______ of two in-plane vectors.
Answers: 1. True (one is -1/2 times the other). 2. x = 4. 3. False (they are perpendicular). 4. |ax1 + by1 + cz1 - d|. 5. Cross product.
Q: Find the equation of the plane through (2, 1, -1) and perpendicular to the line x = 3 - 2t, y = t, z = 4 + 5t.
A: The direction vector of the line is (-2, 1, 5), which becomes the normal. Equation: -2(x - 2) + 1(y - 1) + 5(z + 1) = 0, giving -2x + y + 5z = -8.
Q: Find the equation of the plane through (1, 0, -3) and parallel to 2x + 5y - z = 10.
A: Normal is (2, 5, -1). Equation: 2(x - 1) + 5(y - 0) - 1(z + 3) = 0, giving 2x + 5y - z = 5.
Q: Are the planes 3x - 6y + 9z = 4 and x - 2y + 3z = 7 parallel, perpendicular, or neither?
A: Normals are (3, -6, 9) and (1, -2, 3). Since (3, -6, 9) = 3(1, -2, 3), the planes are parallel.
Q: Find the cosine of the angle between x + 2y + z = 0 and 2x - y + 3z = 1.
A: Normals: (1, 2, 1) and (2, -1, 3). Dot product = 2 - 2 + 3 = 3. Magnitudes: sqrt(6) and sqrt(14). cos(theta) = 3 / sqrt(84) = 3 / (2*sqrt(21)) = sqrt(21)/14.
Q: Find the distance from (1, -2, 4) to the plane 3x + 2y - z = 5.
A: D = |3(1) + 2(-2) - 1(4) - 5| / sqrt(9 + 4 + 1) = |3 - 4 - 4 - 5| / sqrt(14) = 10 / sqrt(14) = 10sqrt(14) / 14 = 5sqrt(14) / 7.
Q: Find the equation of the plane through (1, 2, 3), (0, 1, 1), and (2, 0, -1).
A: v1 = (-1, -1, -2), v2 = (1, -2, -4). Cross product: (-1)(-4) - (-2)(-2), (-2)(1) - (-1)(-4), (-1)(-2) - (-1)(1) = (4 - 4, -2 - 4, 2 + 1) = (0, -6, 3). Simplify to (0, -2, 1). Equation: 0(x - 1) - 2(y - 2) + 1(z - 3) = 0, giving -2y + z = -1, or 2y - z = 1.
Q: Find the equation of the plane equidistant from (0, 0, 0) and (4, 2, 6).
A: Midpoint = (2, 1, 3). Normal = (4, 2, 6), simplify to (2, 1, 3). Equation: 2(x - 2) + 1(y - 1) + 3(z - 3) = 0, giving 2x + y + 3z = 14.
The normal vector reappears in surface integrals (Section 16.7), where you integrate a function over a surface using the same cross-product technique to find the normal. The distance formula from this section is the foundation for the distance between parallel planes and between skew lines (Section 12.5 exercises and beyond). The dot-product angle formula generalises to any inner product space you may encounter in linear algebra.
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