Equation of a Plane

The forms of a plane's equation

A plane is fixed by a point on it and a direction perpendicular to it (its normal $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ ). Every form below is a different way of supplying that information.

Equation of a plane in each standard form

Given	Equation
Normal $\vec{n}$ and point with position vector $\vec{a}$	$(\vec{r} - \vec{a}) \cdot \vec{n} = 0$
Normal $(a, b, c)$ and point $(x_1, y_1, z_1)$	$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$
General (Cartesian) form	$ax + by + cz + d = 0$
Unit normal $\hat{n}$ , distance $p$ from origin	$\vec{r} \cdot \hat{n} = p$
Intercepts $a$ , $b$ , $c$ on the axes	$\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$
Through the intersection of planes $P_1 = 0$ and $P_2 = 0$	$P_1 + \lambda P_2 = 0$

In the general form, the coefficients $(a, b, c)$ of $x$ , $y$ and $z$ are always the direction ratios of the normal. That is the fastest way to read a plane's orientation off its equation.

The coordinate planes

The $xy$ -plane is every point where the height above it is zero, so its equation is simply $z=0$ . Likewise the $yz$ -plane is $x=0$ and the $xz$ -plane is $y=0$ .

Worked example: a point and a normal

Find the equation of the plane through $(1, -1, 2)$ with normal $2\hat{i} + 3\hat{j} - \hat{k}$ .

Solution: Use the point-normal form with $(a, b, c) = (2, 3, -1)$ :

2(x - 1) + 3(y + 1) - (z - 2) = 0 \;\Rightarrow\; 2x + 3y - z + 3 = 0

Check: $2(1) + 3(-1) - 2 + 3 = 0$ , so the given point lies on it.

Worked example: three points

Find the equation of the plane through $A(1, 1, 0)$ , $B(1, 2, 1)$ and $C(-2, 2, -1)$ .

Solution: Two vectors lying in the plane are $\overrightarrow{AB} = (0, 1, 1)$ and $\overrightarrow{AC} = (-3, 1, -1)$ . Their cross product is perpendicular to both, so it is a normal:

\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} = -2\hat{i} - 3\hat{j} + 3\hat{k}

Using point $A$ :

-2(x - 1) - 3(y - 1) + 3z = 0 \;\Rightarrow\; 2x + 3y - 3z = 5

Check with the other two points: $B$ gives $2 + 6 - 3 = 5$ and $C$ gives $-4 + 6 + 3 = 5$ .

Intercept form

A plane's equation can be rewritten in intercept form, $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ , where $a$ , $b$ , and $c$ are exactly where the plane crosses the $x$ -, $y$ -, and $z$ -axes.

To get there from $ax + by + cz = d$ , divide through by $d$ so the right-hand side is 1, then write each coefficient as a denominator. This only works when $d \neq 0$ : a plane through the origin has all three intercepts at zero and no intercept form.

Worked example: the sum of the intercepts

Find the sum of the intercepts cut off by the plane $2x+y-z=5$ .

Solution: Divide through by 5 to get the right-hand side to 1:

\dfrac{2x}{5}+\dfrac{y}{5}-\dfrac{z}{5}=1 \;\Rightarrow\; \dfrac{x}{5/2}+\dfrac{y}{5}+\dfrac{z}{-5}=1

So the intercepts are $a=\dfrac{5}{2}$ , $b=5$ , and $c=-5$ . Their sum:

\dfrac{5}{2}+5-5 = \dfrac{5}{2}

Worked example: a plane from its intercepts

Find the equation of the plane that cuts the axes at 2, $-3$ and 4.

Solution: Substitute straight into intercept form, then clear the fractions by multiplying by 12:

\dfrac{x}{2} + \dfrac{y}{-3} + \dfrac{z}{4} = 1 \;\Rightarrow\; 6x - 4y + 3z = 12

Worked example: a plane through an intersection

Find the vector equation of the plane through the intersection of $3x-y+2z-4=0$ and $x+y+z-3=0$ , passing through the point $(2,2,1)$ .

Solution: Any plane through that intersection has the form $(3x-y+2z-4) + \lambda(x+y+z-3) = 0$ for some constant $\lambda$ . Plugging in $(2,2,1)$ pins down $\lambda$ :

(6-2+2-4) + \lambda(2+2+1-3) = 0 \;\Rightarrow\; 2 + 2\lambda = 0 \;\Rightarrow\; \lambda = -1

Substituting back in:

(3x-y+2z-4) - (x+y+z-3) = 0 \;\Rightarrow\; 2x-2y+z-1 = 0

In vector form, this is:

\vec{r}\cdot(2\hat{i}-2\hat{j}+\hat{k}) = 1

Distance from a point to a plane

Once a plane is in general form, the perpendicular distance $D$ from a point $(x_1, y_1, z_1)$ to it comes straight from the coefficients:

D = \dfrac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}

Example: the distance from $(2, 5, -3)$ to $6x - 3y + 2z - 4 = 0$ is

\dfrac{|12 - 15 - 6 - 4|}{\sqrt{36 + 9 + 4}} = \dfrac{13}{7}

Common mistakes

Reading intercepts before the right side is 1. In $2x + y - z = 5$ the $x$ -intercept is $\dfrac{5}{2}$ , not 2 or 5.
Dropping the sign of an intercept. A $-z$ term gives a negative intercept, as with $c = -5$ above. Leaving it positive changes the sum.
Confusing the normal with a direction in the plane. $(a, b, c)$ is perpendicular to the plane. A vector in the plane, like $\overrightarrow{AB}$ , is perpendicular to the normal instead.
Using $\vec{r} \cdot \vec{n} = d$ as the distance form. The right side is the distance from the origin only when $\vec{n}$ is a unit vector. Divide by $|\vec{n}|$ first.

Where this comes up in exams

The JEE Advanced 2026 syllabus lists the equation of a plane and the distance of a point from a plane. Neither appears in the JEE Main 2026 syllabus or the 2025-26 CBSE Class XII syllabus, whose three-dimensional geometry units cover lines only.

Checked against JEE (Advanced) 2026 syllabus, JEE (Main) 2026 syllabus (NTA) and CBSE Mathematics curriculum, Classes XI-XII (2025-26). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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