MathsISC Class 12 · Three-dimensional Geometry

Equation of a Plane: Every Form, Distances and Angles

Updated 15 October 2026 · Fact-checked

A plane is fixed by a point and a normal vector. For the plane through $\vec{a}$ with normal $\vec{n}$, the vector equation is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$, and in Cartesian form $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. The coefficients of x, y and z are the normal's direction ratios.

Any point P is in the plane when the in-plane vector r − a is perpendicular to the normal n, so their dot product is zero.

Illustration drawn for QuestionBuddy.

Understand Equation of a Plane

The key idea: the normal. Every plane has a direction perpendicular to it, its normal $n$ . A point $P (r)$ lies in the plane through $A (a)$ exactly when $A P = r - a$ is perpendicular to $n$ :
$(r - a) \cdot n = 0.$
With $n = (a, b, c)$ and $A (x_{1}, y_{1}, z_{1})$ this is the one-point form $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$ . Expanding gives the general form $a x + b y + cz + d = 0$ , and the coefficients $a, b, c$ are the direction ratios of the normal. Reading the normal straight from the equation is the single most useful habit in this topic.

Normal form. If $\overset{n}{^}$ is the unit normal and $p$ the perpendicular distance from the origin, then $r \cdot \overset{n}{^} = p$ , or $l x + m y + n z = p$ where $l, m, n$ are the direction cosines of the normal. To change $a x + b y + cz = d$ (with $d > 0$ ) into normal form, divide by $a^{2} + b^{2} + c^{2}$ ; then $p = \frac{d}{a ^{2} + b ^{2} + c ^{2}}$ .

Plane through three points. For non-collinear $A, B, C$ , the vector $n = A B \times A C$ is normal to the plane, so $(r - a) \cdot (A B \times A C) = 0$ . In Cartesian form this is the determinant
$x - x_{1} x_{2} - x_{1} x_{3} - x_{1} y - y_{1} y_{2} - y_{1} y_{3} - y_{1} z - z_{1} z_{2} - z_{1} z_{3} - z_{1} = 0.$
Expanding the determinant (or finding the cross product) replaces the error-prone "cross-multiplication" for $a, b, c$ that the ISC 2024 and 2025 examiners saw candidates get wrong.

Intercept form. A plane cutting the axes at $(α, 0, 0)$ , $(0, β, 0)$ , $(0, 0, γ)$ is $\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 1$ . Proof: put the general form $a x + b y + cz = d$ ( $d \neq = 0$ ) through the three points to get $a = \frac{d}{α}$ , $b = \frac{d}{β}$ , $c = \frac{d}{γ}$ , then divide by $d$ .

Planes with a condition on the normal.

Parallel to the x-axis: the normal is perpendicular to $\hat{i}$ , so $a = 0$ (no x term), for example $b y + cz = d$ . In general, a plane through $A$ and $B$ parallel to a line with direction $u$ has $n = A B \times u$ . Missing this was the main error in ISC 2025 Q16(ii).
Perpendicular to two given planes: $n = n_{1} \times n_{2}$ .
Perpendicular to a given line: the line's direction ratios are the normal's.
Coordinate planes: $z = 0$ (xy-plane), $x = 0$ (yz-plane), $y = 0$ (zx-plane).

Distance of a point from a plane. For $P (x_{1}, y_{1}, z_{1})$ and the plane $a x + b y + cz + d = 0$ ,
$D = \frac{∣ a x _{1} + b y _{1} + c z _{1} + d ∣}{a ^{2} + b ^{2} + c ^{2}} .$
Reason: drop the perpendicular from $P$ ; it is parallel to $n$ , and the distance is the component of $A P$ along $\overset{n}{^}$ for any point $A$ in the plane. In vector form, $D = \frac{∣ p \cdot n - q ∣}{∣ n ∣}$ for the plane $r \cdot n = q$ . Two parallel planes $a x + b y + cz + d_{1} = 0$ and $a x + b y + cz + d_{2} = 0$ are $\frac{∣ d _{1} - d _{2} ∣}{a ^{2} + b ^{2} + c ^{2}}$ apart (make the coefficients identical first).

Foot of the perpendicular and image. The perpendicular from $P$ is $r = p + t n$ . Substitute into the plane to find $t$ ; that point is the foot $F$ . The image (reflection) of $P$ is $2 F - P$ .

Angle between two planes is the angle between their normals: $cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣}$ . Perpendicular planes: $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ . Parallel planes: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ . The angle between a line with direction $b$ and a plane is the complement of the angle with the normal: $sin ϕ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$ .

Equation of a Plane: formulas to remember

One-point form: $(r - a) \cdot n = 0; a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$; a, b, c are direction ratios of the normal.
Normal form: $r \cdot \overset{n}{^} = p; l x + m y + n z = p$; p ≥ 0 is the distance from the origin.
Three points: $(r - a) \cdot [(b - a) \times (c - a)] = 0$; Cartesian: 3 × 3 determinant equal to zero.
Intercept form: $\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 1$; Intercepts α, β, γ on the axes.
Distance of a point: $D = \frac{∣ a x _{1} + b y _{1} + c z _{1} + d ∣}{a ^{2} + b ^{2} + c ^{2}}$; Origin: $\frac{∣ d ∣}{a ^{2} + b ^{2} + c ^{2}}$ .
Parallel planes: $\frac{∣ d _{1} - d _{2} ∣}{a ^{2} + b ^{2} + c ^{2}}$; Same a, b, c in both equations first.
Angle between planes: $cos θ = \frac{∣ a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2} ∣}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$; Angle between the normals.
Line and plane: $sin ϕ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$; Sine, because the line makes 90° − θ with the normal.

How to solve Equation of a Plane questions

Every plane question reduces to two things: a point on the plane and its normal.

1Find one point on the plane (given, or read from a line that lies in it).
2Find the normal from the condition: given directly; a cross product of two vectors lying in or parallel to the plane; the direction of a perpendicular line; or $n_{1} \times n_{2}$ for a plane perpendicular to two planes.
3Write $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$ and simplify to $a x + b y + cz = d$ .
4Check by substituting every given point; each must satisfy the equation.
5For distances, put the plane in the form $a x + b y + cz + d = 0$ and use $\frac{∣ a x _{1} + b y _{1} + c z _{1} + d ∣}{a ^{2} + b ^{2} + c ^{2}}$ .
6For the foot of a perpendicular, write $(x_{1} + a t, y_{1} + b t, z_{1} + c t)$ , substitute in the plane and solve for t.

Equation of a Plane in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Angle between a plane and the ground

Suppose a wheelchair ramp is a flat surface through the points (0, 0, 0), (4, 0, 0) and (0, 10, 1), in metres, with the z-axis pointing up and the floor being the plane z = 0.

Find the equation of the ramp and the angle it makes with the floor.

Try it first, then show the solutionSolution

Two vectors in the ramp: $(4, 0, 0)$ and $(0, 10, 1)$ . Normal: $(4, 0, 0) \times (0, 10, 1) = (0, - 4, 40)$ , or $(0, - 1, 10)$ .
Through the origin: $- y + 10 z = 0$ , that is $y = 10 z$ .
Angle with the floor (normal $\hat{k}$ ): $cos θ = \frac{10}{101}$ , so $tan θ = \frac{1}{10}$ and $θ \approx 5.7°$ .

Answer: $y = 10 z$ ; it rises 1 m for every 10 m, an angle of about 5.7° with the floor.

Memory hook: Angle between surfaces = angle between their normals.

Distance of a point from a plane

Imagine a drone hovering at the point (3, 4, 10) above a hillside that is modelled by the plane x + 2y + 2z = 8 (distances in metres).

How far is the drone from the hillside, measured perpendicular to it?

Try it first, then show the solutionSolution

Write the plane as $x + 2 y + 2 z - 8 = 0$ .
$D = \frac{∣3 + 8 + 20 - 8∣}{1 + 4 + 4} = \frac{23}{3}$ .

Answer: $\frac{23}{3} \approx 7.67$ m.

Memory hook: Substitute the point, take the modulus, divide by the length of the normal.

Tricks to remember Equation of a Plane

Short memory aids. Say each one aloud once, then use it on a question the same day.

Coefficients = normal
In $a x + b y + cz + d = 0$ , read $(a, b, c)$ as the normal immediately.
Use it when: Angles, parallel and perpendicular tests, distances.
Missing letter, parallel axis
No x term means the plane is parallel to the x-axis; no y term, the y-axis; and so on.
Use it when: Planes parallel to an axis.
Two in-plane vectors, cross them
Any two non-parallel vectors lying in (or parallel to) the plane give the normal by a cross product.
Use it when: Three points; two points and a parallel line.
Plane angle: cos; line-plane angle: sin
Plane with plane uses normals (cos). Line with plane uses the normal too, so the angle flips to sin.
Use it when: Angle questions.

Equation of a Plane: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Normal form · 1 mark · MCQBoard level

The perpendicular distance of the plane $2 x - 3 y + 6 z = 14$ from the origin and the direction cosines of its normal are:
(a) 2; $\frac{2}{7}, - \frac{3}{7}, \frac{6}{7}$
(b) 14; 2, −3, 6
(c) 2; 2, −3, 6
(d) $\frac{14}{5}$ ; $\frac{2}{5}, - \frac{3}{5}, \frac{6}{5}$

Try it first, then show the solutionSolution

$4 + 9 + 36 = 7$ ; divide by 7: $\frac{2}{7} x - \frac{3}{7} y + \frac{6}{7} z = 2$ .

Answer: (a).

Example 2: Plane perpendicular to a line · 2 marks · Short answerBoard level

Find the equation of the plane through $(2, - 1, 3)$ perpendicular to the line joining $(1, 2, 0)$ and $(4, 0, 1)$ .

Try it first, then show the solutionSolution

Normal = direction of the line: $(4 - 1, 0 - 2, 1 - 0) = (3, - 2, 1)$ . (1 mark)
$3 (x - 2) - 2 (y + 1) + (z - 3) = 0$ , so $3 x - 2 y + z = 11$ . (1 mark)

Answer: $3 x - 2 y + z = 11$ .

Example 3: Plane through three points · 3 marks · Short answerHarder board level

Find the equation of the plane through $P (1, 2, - 1)$ , $Q (2, 0, 1)$ and $R (0, 3, 2)$ . (In the style of ISC 2025 Q16(i).)

Try it first, then show the solutionSolution

$P Q = (1, - 2, 2)$ , $P R = (- 1, 1, 3)$ . (1 mark)
$n = P Q \times P R = ((- 2) (3) - (2) (1), (2) (- 1) - (1) (3), (1) (1) - (- 2) (- 1)) = (- 8, - 5, - 1)$ , or $(8, 5, 1)$ . (1 mark)
$8 (x - 1) + 5 (y - 2) + (z + 1) = 0$ , so $8 x + 5 y + z = 17$ . Check: Q gives $16 + 0 + 1 = 17$ ; R gives $0 + 15 + 2 = 17$ . (1 mark)

Answer: $8 x + 5 y + z = 17$ .

Example 4: Plane parallel to an axis · 2 marks · Short answerHarder board level

Find the equation of the plane through $A (1, - 2, 4)$ and $B (3, 1, - 2)$ that is parallel to the z-axis.

Try it first, then show the solutionSolution

The normal is perpendicular to $A B = (2, 3, - 6)$ and to $\hat{k}$ : $n = (2, 3, - 6) \times (0, 0, 1) = (3, - 2, 0)$ . (1 mark)
$3 (x - 1) - 2 (y + 2) = 0$ , so $3 x - 2 y = 7$ ; B: $9 - 2 = 7$ . (1 mark)

Answer: $3 x - 2 y = 7$ (no z term, as expected).

Example 5: Angle between two planes · 2 marks · Short answerHarder board level

Find the angle between the planes $x - y + 2 z = 3$ and $2 x + y + z = 5$ , and say whether the plane $x + y - z = 1$ is perpendicular to the first.

Try it first, then show the solutionSolution

$cos θ = \frac{∣2 - 1 + 2∣}{6 6} = \frac{1}{2}$ , so $θ = \frac{π}{3}$ . (1 mark)
$(1) (1) + (- 1) (1) + (2) (- 1) = - 2 \neq = 0$ , so not perpendicular. (1 mark)

Answer: $\frac{π}{3}$ ; not perpendicular.

Example 6: A mast on a sloping roof · 5 marks · Case-basedHOTS / competency

Suppose a roof surface cuts the x, y and z axes at 2, 3 and 6 (metres). A mast stands on the roof with its top at $T (4, 3, 5)$ , the mast being perpendicular to the roof.
(i) Find the equation of the roof. (1 mark)
(ii) Find the length of the mast. (2 marks)
(iii) Find the point where the mast meets the roof. (2 marks)

Try it first, then show the solutionSolution

(i) $\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1$ , that is $3 x + 2 y + z = 6$ .
(ii) $D = \frac{∣12 + 6 + 5 - 6∣}{9 + 4 + 1} = \frac{17}{14} \approx 4.54$ m.
(iii) Foot: $(4 + 3 t, 3 + 2 t, 5 + t)$ with $3 (4 + 3 t) + 2 (3 + 2 t) + 5 + t = 6$ , so $23 + 14 t = 6$ and $t = - \frac{17}{14}$ .
Foot $= (\frac{5}{14}, \frac{8}{14}, \frac{53}{14}) = (\frac{5}{14}, \frac{4}{7}, \frac{53}{14})$ ; check: $\frac{15 + 16 + 53}{14} = 6$ .

Answer: (i) $3 x + 2 y + z = 6$ ; (ii) $\frac{17}{14} \approx 4.54$ m; (iii) $(\frac{5}{14}, \frac{4}{7}, \frac{53}{14})$ .

Where students lose marks in Equation of a Plane

The slips examiners see most often, and what to write instead.

Mistake: Cross-multiplication slips
The ISC 2024 (mobile tower) and 2025 (three points) examiners reported errors in solving for a, b, c by cross-multiplication.
Do this: Use $A B \times A C$ written as a determinant, then substitute every given point to check.
Usually costs: 2 to 3 marks.
Mistake: Plane parallel to an axis
In 2025 candidates did not see that a plane parallel to the x-axis has a normal perpendicular to the x-axis.
Wrong working
Taking $\hat{i}$ itself as the normal.
Right working
$n = A B \times \hat{i}$ .
Do this: Translate "parallel to a line" into "normal perpendicular to that line".
Usually costs: 2 marks.
Mistake: Angle between planes from the wrong vectors
The 2025 examiners stressed that the angle between planes is the angle between their normals.
Do this: Read the normals from the coefficients and use the cosine formula with a modulus.
Usually costs: 1 mark.
Mistake: Parallel-plane distance with unequal coefficients
Subtracting constants before making the coefficients identical gives a wrong distance.
Do this: Divide one equation so that a, b, c match, then use $\frac{∣ d _{1} - d _{2} ∣}{a ^{2} + b ^{2} + c ^{2}}$ .
Usually costs: 1 mark.

Exam tips

Always end by substituting the given points into your plane; a quick check that catches most slips.
Keep the normal in its simplest integer form, for example (8, 5, 1) instead of (−8, −5, −1).
Give both vector and Cartesian forms when the question says "find the equation" without specifying.
For a foot of a perpendicular, write the general point on the normal line first; it earns a method mark.

Practice questions on Equation of a Plane

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The distance of the plane $2 x + y - 2 z = 9$ from the origin is:
(a) 9
(b) 3
(c) $\frac{9}{5}$
(d) 1
Hint
Divide by $4 + 1 + 4$ .
Answer
(b) 3.
Q2 · 1 mark · Very short answerBoard level
Find the intercepts made on the axes by the plane $4 x - 3 y + 2 z = 12$ .
Hint
Divide by 12.
Answer
3, −4 and 6.
Q3 · 3 marks · Short answerHarder board level
Find the equation of the plane through $(1, 1, - 1)$ perpendicular to the planes $x + 2 y + 3 z = 7$ and $2 x - 3 y + 4 z = 0$ .
Hint
$n = n_{1} \times n_{2}$ .
Answer
$n = (17, 2, - 7)$ ; $17 x + 2 y - 7 z = 26$ .
Q4 · 2 marks · Short answerHarder board level
Find the distance between the parallel planes $2 x - y + 2 z = 4$ and $4 x - 2 y + 4 z = - 10$ .
Hint
Halve the second equation first.
Answer
$2 x - y + 2 z = - 5$ ; distance $\frac{∣4 + 5∣}{3} = 3$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The plane $3 y + 4 z = 5$ is parallel to the x-axis.
Reason (R): The normal $(0, 3, 4)$ of the plane is perpendicular to $\hat{i}$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
A plane is parallel to a line when its normal is perpendicular to the line.
Answer
(a).
Q6 · 3 marks · Short answerHOTS / competency
Find the image of the point $(1, 2, 3)$ in the plane $x + 2 y + 2 z = 3$ .
Hint
Find the foot F, then image = 2F − P.
Answer
$t = - \frac{8}{9}$ , foot $(\frac{1}{9}, \frac{2}{9}, \frac{11}{9})$ , image $(- \frac{7}{9}, - \frac{14}{9}, - \frac{5}{9})$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a lamp hangs at $L (1, 2, 3)$ below a sloping roof $2 x + 2 y + z = 12$ (metres), and the floor is $z = 0$ .
(i) Find the distance of the lamp from the roof. (1 mark)
(ii) Find the point of the roof nearest the lamp. (2 marks)
(iii) Find the angle between the roof and the floor. (1 mark)
Hint
The roof normal is (2, 2, 1).
Answer
(i) $\frac{∣2 + 4 + 3 - 12∣}{3} = 1$ m. (ii) $t = \frac{1}{3}$ : $(\frac{5}{3}, \frac{8}{3}, \frac{10}{3})$ . (iii) $cos θ = \frac{1}{3}$ , $θ = cos^{- 1} \frac{1}{3} \approx 70.5°$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the one-point form of a plane.Show answer
$a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$ , with $(a, b, c)$ the normal.
2.Distance of $6 x - 3 y + 2 z = 14$ from the origin?Show answer
$\frac{14}{7} = 2$ .
3.Condition for planes to be perpendicular.Show answer
$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ .
4.Equation of the zx-plane.Show answer
$y = 0$ .
5.Normal to the plane through A, B, C?Show answer
$A B \times A C$ .
6.Distance of (1, 1, 1) from x + y + z = 6?Show answer
$\frac{3}{3} = 3$ .

Equation of a Plane: frequently asked questions

How do I find the normal of a plane?

Read it from the equation: in ax + by + cz + d = 0 the normal is (a, b, c). If you are building the plane, cross two vectors that lie in it, or use the direction of a line perpendicular to it.

What is the difference between the normal form and the general form?

The general form ax + by + cz + d = 0 has any normal. The normal form lx + my + nz = p uses the unit normal, so the constant p is the distance from the origin.

Why is the angle between a line and a plane found with sine?

The formula uses the normal, which is at 90° to the plane. The angle with the normal is the complement of the angle with the plane, and cos of the complement is sin.

Is the plane through the intersection of two planes in the ISC 2027 syllabus?

The 2027 syllabus lists the vector and Cartesian equations, one-point, normal and intercept forms, distance of a point, intersection of a line and a plane, and angles. This page covers exactly those forms.

How many marks do planes carry in ISC?

Three-dimensional geometry carries 6 marks in 2027. Plane questions appeared for 4 marks in 2024 (a tower on a hill) and 4 marks in 2025 (two plane equations), plus a 1-mark angle item.