Planes in 3D
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 2.5: “Equations of Lines and Planes in Space” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/2-5-equations-of-lines-and-planes-in-space |
| Textbook used in class | Stewart, Calculus, Section 12.5: “Equations of Lines and Planes” (Examples 4, 5, 6, 8) |
Opening Scenario
A plane in 3D is completely determined by one point in the plane and one vector perpendicular to it (a normal vector). Any vector lying in the plane is perpendicular to the normal, and that perpendicularity condition is what gives the equation.
Quick Reference
Equation of a plane through $P_0 = (x_0, y_0, z_0)$ with normal $\mathbf{n} = \langle a, b, c\rangle$: \[ \mathbf{n}\cdot(\mathbf{r} - \mathbf{r}_0) = 0, \quad \text{i.e.,} \quad a(x-x_0) + b(y-y_0) + c(z-z_0) = 0. \]
Scalar (standard) form: $ax + by + cz = d$, where $d = ax_0 + by_0 + cz_0$.
Distance from point $Q = (x_1, y_1, z_1)$ to the plane $ax + by + cz = d$: \[ \text{dist} = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}. \]
Two planes are parallel if their normal vectors are parallel; perpendicular if their normal vectors are perpendicular (dot product zero).
Key Concepts
1. Finding the Equation of a Plane
Example 1. Find the equation of the plane through $P_0 = (1, 3, 2)$ with normal $\mathbf{n} = \langle 3, 0, -2\rangle$. (Stewart 12.5, Example 4.)
\[ 3(x - 1) + 0(y - 3) + (-2)(z - 2) = 0 \implies 3x - 2z = -1. \]
Example 2. Find the equation of the plane through $(1, 1, 0)$, $(-1, 0, 2)$, and $(0, 2, 1)$. (Stewart 12.5, Example 5.)
Use two vectors in the plane to find the normal: \[ \mathbf{v}_1 = \langle -2, -1, 2\rangle, \quad \mathbf{v}_2 = \langle -1, 1, 1\rangle. \] \[ \mathbf{n} = \mathbf{v}_1\times\mathbf{v}_2 = \langle (-1)(1)-(2)(1), (2)(-1)-(-2)(1), (-2)(1)-(-1)(-1)\rangle = \langle -3, 0, -3\rangle. \] Simplified normal: $\langle 1, 0, 1\rangle$. Plane through $(1,1,0)$: $(x-1) + (z-0) = 0 \implies x + z = 1$.
2. Distance From a Point to a Plane
Example 3. Find the distance from $Q = (1, 2, -1)$ to the plane $2x - y + 3z = 4$. (Stewart 12.5, Example 8.)
\[ \text{dist} = \frac{|2(1) + (-1)(2) + 3(-1) - 4|}{\sqrt{4 + 1 + 9}} = \frac{|2 - 2 - 3 - 4|}{\sqrt{14}} = \frac{7}{\sqrt{14}} = \frac{\sqrt{14}}{2}. \]
3. Angle Between Two Planes
The dihedral angle between planes with normals $\mathbf{n}_1$ and $\mathbf{n}_2$ satisfies $\cos\theta = |\mathbf{n}_1\cdot\mathbf{n}_2|/(|\mathbf{n}_1||\mathbf{n}_2|)$.
thinking a plane is determined by its normal vector alone. The normal vector tells you the orientation of the plane (tilt) but not its location. You also need one specific point that lies in the plane. Two parallel planes have the same normal vector but different equations.
Common Errors Summary
| Error | Correction |
|---|---|
| Forgetting to expand and simplify to $ax + by + cz = d$ | The answer is not complete until you write the standard scalar form |
| Using the wrong formula for distance (missing absolute value) | The distance formula requires an absolute value in the numerator |
Common Misconceptions
the normal vector alone determines the plane.
This is the concept-image-conflicts-definition error. The normal vector specifies the orientation (tilt) of the plane but does not fix its location in space. An entire family of parallel planes shares the same normal vector. A point in the plane is also required; the condition $\mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0$ uses both the normal and the reference point $\mathbf{r}_0$. Two planes with the same normal vector and different points in them are parallel but distinct.
Leveled Practice
Problem 1. Find the equation of the plane through $(2, -1, 4)$ with normal $\langle 5, 1, -3\rangle$.
Show answer
$5(x-2) + (y+1) - 3(z-4) = 0 \implies 5x + y - 3z = -3$.
Problem 2. Find the distance from the origin to the plane $3x - y + 2z = 6$.
Show answer
$\text{dist} = |3(0) - 0 + 2(0) - 6|/\sqrt{9+1+4} = 6/\sqrt{14}$.
Mastery Checklist
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