NDA Maths · 3D Geometry

The Plane

A plane is fixed by a point and a normal direction; its equation, distances, and angles all read off the normal ⟨a, b, c⟩.

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Why this matters

Fourteen PYQs spanning EASY to HARD. The plane's normal vector is the master key: it gives the equation, the angle between two planes, the distance from a point, and the foot of a perpendicular. Six concepts cover the lot — equation forms, intercepts, the three-point plane, distances, angles, and the plane-through-intersection pencil.

Concept 1 of 6

Equation of a plane and its normal

Intuition

The coefficients in

ax + by + cz = d

ARE the components of the plane's normal vector

\langle a, b, c\rangle

. So building a plane is mostly about finding its normal: a plane perpendicular to a given line just borrows that line's direction ratios as its normal.

Definition

The general plane is $ax + by + cz = d$ with normal $\langle a, b, c\rangle$ . Point-normal form: the plane through $(x_0,y_0,z_0)$ with normal $\langle a,b,c\rangle$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ . Parallel planes share the same normal (only $d$ differs).

Point-normal form

a(x - x_0) + b(y - y_0) + c(z - z_0) = 0

$\langle a,b,c\rangle$ normal direction ratios
$(x_0,y_0,z_0)$ a point on the plane

Diagram · plane, normal & distance from origin (drag to rotate)

Shortest path from O to the plane runs along the normal to the foot N; its length is |d| / √(a²+b²+c²).

Worked example

Find the equation of the plane through

(2, -1, 3)

perpendicular to the line with direction ratios

\langle 1, 4, 2\rangle

Perpendicular to that line → the plane's normal IS $\langle 1, 4, 2\rangle$ .
Point-normal form: $1(x-2) + 4(y+1) + 2(z-3) = 0$ .
Expand: $x + 4y + 2z - 4 = 0$ .

Answer:

x + 4y + 2z = 4

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 13D GeometryEASY

What is the equation of the plane passing through the point

(1,1,1)

and perpendicular to the line whose direction ratios are

\langle 3,2,1 \rangle

[Q65 · Apr · 2025]

Drill 2 more on equation of a plane and its normal

Concept 2 of 6

Intercept form and special planes

Intuition

If a plane cuts the axes at

a, b, c

, its equation is the clean intercept form

\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1

. And planes parallel to a coordinate plane are the simplest of all —

z = k

is everything at height

k

, a plane parallel to the XY-plane.

Definition

Intercept form: a plane with x-, y-, z-intercepts $a, b, c$ is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ . Special planes: $z = k$ is parallel to the XY-plane; $x = k$ parallel to the YZ-plane; $y = k$ parallel to the ZX-plane. The locus $z = 7$ is therefore a plane, not a line.

Intercept form

\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1

Worked example

Find the equation of the plane that cuts intercepts of 2, 3 and 4 on the x-, y- and z-axes respectively.

Intercept form: $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ .
Multiply through by 12: $6x + 4y + 3z = 12$ .

Answer:

6x + 4y + 3z = 12

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 23D GeometryEASY

What is the equation of the plane which cuts an intercept 5 units on the z-axis and is parallel to xy-plane?

[Q65 · Apr · 2020]

Drill 1 more on intercept form and special planes

Concept 3 of 6

Plane through three points

Intuition

Three non-collinear points fix a plane. Build two edge vectors from one point, take their cross product to get the normal, then use point-normal form. (The determinant formula packages the same computation.)

Definition

For points $A, B, C$ : form $\overrightarrow{AB}$ and $\overrightarrow{AC}$ ; the normal is $\vec n = \overrightarrow{AB} \times \overrightarrow{AC}$ ; then write $\vec n \cdot (\vec r - \vec A) = 0$ . Equivalently, the plane is the determinant equation below.

Determinant form through three points

\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0

Worked example

Find the equation of the plane through

A(1,1,0)

B(2,0,1)

and

C(0,1,2)

Form two edge vectors: $\overrightarrow{AB} = B - A = \langle 1, -1, 1\rangle$ and $\overrightarrow{AC} = C - A = \langle -1, 0, 2\rangle$ .
The normal is their cross product: $\vec n = \overrightarrow{AB} \times \overrightarrow{AC} = \langle (-1)(2)-(1)(0),\ (1)(-1)-(1)(2),\ (1)(0)-(-1)(-1)\rangle = \langle -2, -3, -1\rangle$ .
Point-normal form through $A$ : $-2(x-1) - 3(y-1) - 1(z-0) = 0$ .
Simplify: $2x + 3y + z = 5$ . (Check: $A$ gives $2+3+0=5$ ✓.)

Answer:

2x + 3y + z = 5

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 33D GeometryHARD

What is the equation of the plane through (-2,6,-6), (-3,10,-9) and (-5,0,-6)?

[Q59 · Apr · 2018]

Concept 4 of 6

Distance from a point and the foot of the perpendicular

Intuition

The perpendicular distance from a point to a plane plugs the point into

ax+by+cz-d

and divides by the normal's length. For two PARALLEL planes, make the normals identical and the distance is just the gap in the constants over that same length. The foot of the perpendicular is reached by stepping from the point along the unit normal.

Definition

Distance from $(x_1,y_1,z_1)$ to $ax+by+cz+d=0$ is the formula below. For parallel planes $ax+by+cz+d_1 = 0$ and $ax+by+cz+d_2 = 0$ (SAME coefficients), the distance is $\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$ — scale one plane first so the normals match.

Distance from a point to a plane

\text{distance} = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}

Diagram · plane, normal & distance from origin (drag to rotate)

Shortest path from O to the plane runs along the normal to the foot N; its length is |d| / √(a²+b²+c²).

Worked example

Find the distance of the point

(2, 3, 4)

from the plane

3x - 6y + 2z + 11 = 0

Plug into $|ax_1+by_1+cz_1+d|$ : $|3(2) - 6(3) + 2(4) + 11| = |6 - 18 + 8 + 11| = |7| = 7$ .
Normal length: $\sqrt{9 + 36 + 4} = \sqrt{49} = 7$ .
Distance $= \tfrac{7}{7} = 1$ .

Answer:

1

unit.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 43D GeometryMODERATE

The distance between the parallel planes

4x-2y+4z+9=0

and

8x-4y+8z+21=0

[Q64 · Apr · 2019]

Scale parallel planes to a common normal BEFORE subtracting constants

4x-2y+4z+9=0

and

8x-4y+8z+21=0

look like they differ by 12 in the constant — but the normals differ by a factor of 2. Halve the second plane first; only then is

|d_1-d_2|/|n|

valid.

Always divide by $\sqrt{a^2+b^2+c^2}$ — the numerator alone is NOT the distance

The distance is

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

. Plugging the point into

|ax_1+by_1+cz_1+d|

and stopping there forgets the normalisation by the normal's length — it only equals the distance when

\sqrt{a^2+b^2+c^2}=1

. For

(1,2,2)

and

2x+y+2z+5=0

: numerator

=13

, but the distance is

13/3

, not

13

Drill 2 more on distance from a point and the foot of the perpendicular

Concept 5 of 6

Angle between two planes

Intuition

The angle between two planes is the angle between their normals — same dot-product formula as two lines, applied to

\langle a_1,b_1,c_1\rangle

and

\langle a_2,b_2,c_2\rangle

. Perpendicular planes have perpendicular normals; parallel planes have proportional normals.

Definition

For planes with normals $\vec n_1, \vec n_2$ , the angle $\theta$ between them satisfies $\cos\theta = \frac{|\vec n_1 \cdot \vec n_2|}{|\vec n_1||\vec n_2|}$ . The planes are perpendicular iff $\vec n_1 \cdot \vec n_2 = 0$ and parallel iff the normals are proportional.

Angle between planes (via normals)

\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}

Worked example

Find the angle between the planes

2x - y + z = 1

and

x + y + 2z = 3

Normals: $\langle 2,-1,1\rangle$ and $\langle 1,1,2\rangle$ .
Dot: $2 - 1 + 2 = 3$ . Magnitudes: $\sqrt6$ and $\sqrt6$ .
$\cos\theta = \dfrac{3}{\sqrt6 \cdot \sqrt6} = \dfrac{3}{6} = \tfrac12$ .
So $\theta = 60° = \tfrac{\pi}{3}$ .

Answer:

\tfrac{\pi}{3}

(60°).

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 53D GeometryMODERATE

Consider the following statements: 1. The angle between the planes

2x-y+z=1

and

x+y+2z=3

\frac{\pi}{3}

. 2. The distance between the planes

6x-3y+6z+2=0

and

2x-y+2z+4=0

\frac{10}{9}

. Which of the above statements is/are correct?

[Q59 · Sep · 2018]

Concept 6 of 6

Plane through the line of intersection of two planes

Intuition

Every plane containing the line where

P_1 = 0

and

P_2 = 0

meet can be written as

P_1 + \lambda P_2 = 0

for some

\lambda

. Pick

\lambda

to satisfy one extra condition — passing through a point, or being perpendicular to a third plane — and you're done.

Definition

The family (pencil) of planes through the intersection of $P_1: a_1x+b_1y+c_1z+d_1 = 0$ and $P_2: a_2x+b_2y+c_2z+d_2 = 0$ is $P_1 + \lambda P_2 = 0$ . Determine $\lambda$ from the extra constraint (point on the plane, or perpendicularity to a given plane via the normal dot product). The direction of the line of intersection itself is $\vec{n_1} \times \vec{n_2}$ — perpendicular to both normals.

Pencil of planes

P_1 + \lambda P_2 = 0

Worked example

Find the plane through the intersection of

x + y + z = 3

and

2x - y + z = 4

that passes through

(3, 1, 1)

Family: $(x+y+z-3) + \lambda(2x-y+z-4) = 0$ .
Substitute $(3,1,1)$ : $(3+1+1-3) + \lambda(6-1+1-4) = 0 \Rightarrow 2 + 2\lambda = 0$ .
$\lambda = -1$ .
Plug back: $(x+y+z-3) - (2x-y+z-4) = -x + 2y + 1 = 0$ , i.e. $x - 2y = 1$ .

Answer:

x - 2y = 1

Practice this conceptself-check · 5 quick reps

From the bank · past-year question

Example 63D GeometryHARD

The equation of the plane passing through the intersection of the planes

2x+y+2z=9

4x-5y-4z=1

and the point

(3,2,1)

[Q63 · Apr · 2019]

Drill 3 more on plane through the line of intersection of two planes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (6)

Equation of a plane and its normal
Point-normal form
$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$
Intercept form and special planes
Intercept form
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
Plane through three points
Determinant form through three points
$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$
Distance from a point and the foot of the perpendicular
Distance from a point to a plane
$\text{distance} = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$
Angle between two planes
Angle between planes (via normals)
$\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}$
Plane through the line of intersection of two planes
Pencil of planes
$P_1 + \lambda P_2 = 0$

Watch out for (2)

Scale parallel planes to a common normal BEFORE subtracting constants
→ Distance from a point and the foot of the perpendicular
Always divide by $\sqrt{a^2+b^2+c^2}$ — the numerator alone is NOT the distance
→ Distance from a point and the foot of the perpendicular

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