Equation of
a Plane
All forms of the equation of a plane — vector, Cartesian, intercept, normal, through three points, and more.
Concept of a Plane
A plane is a flat, two-dimensional surface that extends infinitely in all directions. In 3D space, a plane is uniquely determined by:
- A point on the plane and a normal vector (perpendicular to the plane), or
- Three non-collinear points that lie on the plane, or
- Two intersecting lines lying in the plane
Normal Form & Vector Equation
Let the plane have a normal vector $\vec{n}$ and pass through point $A$ with position vector $\vec{a}$. For any point $P(\vec{r})$ on the plane, $\overrightarrow{AP}$ is perpendicular to $\vec{n}$:
If $\hat{n}$ is the unit normal to the plane and $d$ is the perpendicular distance from the origin to the plane:
where $(l,m,n)$ are direction cosines of the normal and $d > 0$ is the distance from origin
General Form
$(a, b, c)$ is the normal vector to the plane. Not all of $a, b, c$ can be zero.
- Normal vector: $\vec{n} = a\hat{i}+b\hat{j}+c\hat{k}$
- Distance from origin: $\dfrac{|d|}{\sqrt{a^2+b^2+c^2}}$
- Every first-degree equation in $x, y, z$ represents a plane
- Plane passes through origin $\iff d = 0$
| Plane | Equation | Normal direction |
|---|---|---|
| Parallel to $xy$-plane | $z = k$ | $z$-axis $(0,0,1)$ |
| Parallel to $yz$-plane | $x = k$ | $x$-axis $(1,0,0)$ |
| Parallel to $xz$-plane | $y = k$ | $y$-axis $(0,1,0)$ |
| $xy$-plane itself | $z = 0$ | $(0,0,1)$ |
Intercept Form
If a plane cuts the $x$-axis at $a$, $y$-axis at $b$, and $z$-axis at $c$ (intercepts), its equation is:
The plane passes through $(a,0,0)$, $(0,b,0)$, $(0,0,c)$
Plane Through Three Non-Collinear Points
Given three points $A(x_1,y_1,z_1)$, $B(x_2,y_2,z_2)$, $C(x_3,y_3,z_3)$:
The normal to the plane through $A$, $B$, $C$ is $\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}$.
$\overrightarrow{AB} = \vec{b}-\vec{a}$, $\overrightarrow{AC} = \vec{c}-\vec{a}$
$$\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}, \quad \text{then} \quad \vec{r}\cdot\vec{n} = \vec{a}\cdot\vec{n}$$Plane Through a Point with Given Normal
Plane passing through $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$:
Expanding this gives the general form $ax+by+cz+d=0$ where $d = -(ax_1+by_1+cz_1)$.
Family of Planes & Related Results
If $P_1: a_1x+b_1y+c_1z+d_1=0$ and $P_2: a_2x+b_2y+c_2z+d_2=0$, any plane through their line of intersection is:
Choose $\lambda$ to satisfy any additional condition (e.g., passes through a specific point)
A plane parallel to $ax+by+cz+d=0$ has the form:
$$ax+by+cz+k = 0 \quad (\text{same normal direction, different constant})$$Between $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$:
$$\text{Distance} = \frac{|d_1 - d_2|}{\sqrt{a^2+b^2+c^2}}$$| Condition | Equation |
|---|---|
| Planes $P_1 \parallel P_2$ | $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ (normals parallel) |
| Planes $P_1 \perp P_2$ | $a_1a_2+b_1b_2+c_1c_2=0$ (normals perpendicular) |
All Forms at a Glance
- Given normal + point → Point-normal form
- Given three points → Determinant (3-point) form
- Given intercepts → Intercept form
- General manipulation → General form $ax+by+cz+d=0$
- Through intersection of planes → Family $P_1+\lambda P_2=0$
Solved Examples
Find the equation of the plane passing through $(2, 3, -1)$ with normal vector $\hat{i}+2\hat{j}-3\hat{k}$.
Using point-normal form with $(x_1,y_1,z_1)=(2,3,-1)$ and $(a,b,c)=(1,2,-3)$:
$$1(x-2)+2(y-3)+(-3)(z+1)=0$$ $$x-2+2y-6-3z-3=0$$ $$\boxed{x+2y-3z-11=0}$$Find the equation of the plane with intercepts $2$, $3$, $4$ on the $x$, $y$, $z$-axes.
Multiplying through by $12$: $\boxed{6x+4y+3z=12}$
Find the equation of the plane through $A(1,1,0)$, $B(1,2,1)$, $C(-2,2,-1)$.
$\overrightarrow{AB}=(0,1,1)$, $\overrightarrow{AC}=(-3,1,-1)$
$$\vec{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&1&1\\-3&1&-1\end{vmatrix}$$ $$=\hat{i}(-1-1)-\hat{j}(0+3)+\hat{k}(0+3)=-2\hat{i}-3\hat{j}+3\hat{k}$$Using point $A(1,1,0)$: $-2(x-1)-3(y-1)+3(z-0)=0$
$$\boxed{-2x-3y+3z+5=0} \quad\text{or}\quad 2x+3y-3z=5$$Find the equation of the plane through the line of intersection of $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to $x$-axis.
Family of planes: $(x+y+z-1)+\lambda(2x+3y-z+4)=0$
$$\Rightarrow (1+2\lambda)x+(1+3\lambda)y+(1-\lambda)z+(-1+4\lambda)=0$$Normal to this plane is $(1+2\lambda, 1+3\lambda, 1-\lambda)$.
For the plane to be parallel to $x$-axis, its normal must be perpendicular to $\hat{i}=(1,0,0)$:
$$\vec{n}\cdot\hat{i}=0 \implies 1+2\lambda=0 \implies \lambda=-\frac{1}{2}$$ $$\text{Plane: } 0\cdot x+\left(1-\frac{3}{2}\right)y+\left(1+\frac{1}{2}\right)z+\left(-1-2\right)=0$$ $$\Rightarrow -\frac{1}{2}y+\frac{3}{2}z-3=0 \implies \boxed{y-3z+6=0}$$Convert $\vec{r}\cdot(2\hat{i}-3\hat{j}+5\hat{k})=7$ to Cartesian form. Find its distance from origin.
Substituting $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$:
$$\boxed{2x-3y+5z=7}$$Distance from origin: $\dfrac{|7|}{\sqrt{4+9+25}} = \dfrac{7}{\sqrt{38}}$
- Normal form $lx+my+nz=d$ uses direction cosines, not direction ratios — divide by magnitude!
- Intercept form: make sure each intercept is non-zero (otherwise the plane passes through that axis)
- Three-point form via determinant — expand along the first row for clarity
- Most common 5-mark: "Find equation of plane through three points" or "plane through line of intersection"
- Family of planes $P_1+\lambda P_2=0$ is a powerful tool — practice finding $\lambda$ from conditions
- Parallel to an axis means its normal is perpendicular to that axis direction
- Foot of perpendicular from a point to a plane: use parametric form of the line through the point along the normal
- Image of a point in a plane: foot = midpoint of point and its image
- Using direction ratios instead of direction cosines in the normal form $lx+my+nz=d$
- In the three-point determinant, forgetting to subtract $x_1, y_1, z_1$ from the first row
- Confusing which constant is $d$ when converting vector to Cartesian form
- Planes parallel to $xy$-plane have equation $z=k$, NOT $z+k=0$ (sign matters)