The Plane

Q: What does the general equation of a plane tell you?

In ax + by + cz + d = 0, the coefficients a, b, c are direction ratios of the normal to the plane, and |d| divided by √(a² + b² + c²) is its distance from the origin. A missing variable means the plane is parallel to that axis, and d = 0 means it passes through the origin.

Q: How do you find the equation of a plane through three points?

Take two vectors in the plane, such as AB and AC, and find the normal AB × AC. Then use the point-normal form with point A. This gives the same result as setting the 3 × 3 determinant form of the three-point equation equal to zero.

Q: How do you find the distance of a point from a plane?

Substitute the point into the left side of ax + by + cz + d = 0, take the modulus, and divide by √(a² + b² + c²). The sign of the substituted value, before taking the modulus, also tells you which side of the plane the point lies on.

Q: How do you find the image of a point in a plane?

Move from the point along the normal (a, b, c) by a parameter λ, and require the midpoint of the point and its image to lie on the plane. This gives λ = −2(ax1 + by1 + cz1 + d)/(a² + b² + c²). Half of that value gives the foot of the perpendicular.

Q: What is the angle between a line and a plane?

It is the angle between the line and its projection on the plane. Because the dot product of the line direction with the normal gives the complementary angle, the formula uses sine: sin φ = |b · n| / (|b||n|). If this is zero, the line is parallel to the plane.

Q: How do you check whether two lines are coplanar?

Compute the scalar triple product (a2 − a1) · (b1 × b2), or the equivalent 3 × 3 determinant. If it is zero the lines are coplanar, and the plane containing them has normal b1 × b2. If it is non-zero, the lines are skew.

Q: Is the plane part of the JEE Main 2026 syllabus?

After the 2024 revision, the JEE Main syllabus lists lines, skew lines and shortest distance but not planes as a separate topic. Line problems still use plane ideas such as normals, the line of intersection and points meeting a plane, so learning the core formulas on this page remains useful for JEE Main.

Q: Which plane topics matter most for JEE Advanced?

The equation of a plane and the distance of a point from a plane are listed explicitly. Advanced problems combine them with family of planes, image of a point, projection of a line on a plane, coplanar lines and bisector planes, often in multi-correct questions, so practise switching quickly between Cartesian and vector forms.

MathsThree Dimensional GeometryFor JEE aspirants

A plane is a flat surface that extends without end, and the equation of a plane in 3D is always linear: $a x + b y + cz + d = 0$ , where $(a, b, c)$ is a normal to the plane. This page covers every standard form of the equation of a plane, special planes, coplanarity of points, the angle between planes, distance, foot and image of a point, the family of planes, angle bisector planes, and how a plane relates to a line, including coplanar lines and the projection of a line on a plane.

Key Formulas: Quick Reference

General form: $a x + b y + cz + d = 0$ , normal has direction ratios $a, b, c$
Normal form: $l x + m y + n z = p$ ; vector form: $r \cdot \overset{n}{^} = p$
Point-normal form: $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$
Intercept form: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
Angle between planes: $cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣}$
Distance of $(x_{1}, y_{1}, z_{1})$ from the plane: $\frac{∣ a x _{1} + b y _{1} + c z _{1} + d ∣}{a ^{2} + b ^{2} + c ^{2}}$
Distance between parallel planes: $\frac{∣ d _{1} - d _{2} ∣}{a ^{2} + b ^{2} + c ^{2}}$
Foot: $(x_{1} + ak, y_{1} + bk, z_{1} + c k)$ with $k = - \frac{a x _{1} + b y _{1} + c z _{1} + d}{a ^{2} + b ^{2} + c ^{2}}$ ; image: use $2 k$
Family of planes through the line of intersection: $P_{1} + λ P_{2} = 0$
Bisector planes: $\frac{P _{1}}{∣ n _{1} ∣} = \pm \frac{P _{2}}{∣ n _{2} ∣}$
Angle between a line and a plane: $sin ϕ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$

1. Definition of a Plane

Consider the locus of a point $P (x, y, z)$ . If $x$ , $y$ , $z$ are allowed to vary in different combinations, we get a set of points like $P$ , and the surface on which these points lie is the locus of $P$ . It may be a plane or a curved surface. If $Q$ is any other point on the locus and every point of the straight line $P Q$ lies on it, however small $P Q$ is and in whatever direction, the locus is a plane; otherwise it is a curved surface.

A plane is a surface such that the line joining any two of its points lies completely in it. Equivalently, it is a surface in which the line joining any two points is perpendicular to a fixed straight line, called the normal to the plane.

2. Equation of a Plane in Different Forms

General form

Every first-degree equation in $x$ , $y$ , $z$ represents a plane:

a x + b y + cz + d = 0

Here $a$ , $b$ , $c$ (not all zero) are the direction ratios of the normal to the plane.

Normal form

If $p$ is the length of the perpendicular from the origin to the plane and $l$ , $m$ , $n$ are the direction cosines of that perpendicular, the plane is

l x + m y + n z = p

In vector form, a plane at distance $p$ from the origin with unit normal $\overset{n}{^}$ is $r \cdot \overset{n}{^} = p$ .

Figure 1: Normal form

l x + m y + n z = p

O N

is the perpendicular from the origin to the plane,

p = O N

, and

l, m, n

are the direction cosines of

O N

Reducing the general form to normal form. Write the equation as $a x + b y + cz = q$ with $q \geq 0$ (multiply by $- 1$ if needed), then divide by $a^{2} + b^{2} + c^{2}$ :

\frac{a}{a ^{2} + b ^{2} + c ^{2}} x + \frac{b}{a ^{2} + b ^{2} + c ^{2}} y + \frac{c}{a ^{2} + b ^{2} + c ^{2}} z = \frac{q}{a ^{2} + b ^{2} + c ^{2}}

The coefficients of $x$ , $y$ , $z$ are now the direction cosines of the normal and the right-hand side is $p$ .

Point-normal form

The plane passing through $P (x_{1}, y_{1}, z_{1})$ whose normal has direction ratios $a, b, c$ is

a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0

In vector form, the plane through the point with position vector $a$ and normal to $n$ is $(r - a) \cdot n = 0$ , or $r \cdot n = a \cdot n$ .

Figure 2: Plane through

A (a)

with normal

n

. Every

A R = r - a

in the plane is perpendicular to

n

, so

(r - a) \cdot n = 0

Plane through three points

The plane through three non-collinear points $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ and $(x_{3}, y_{3}, z_{3})$ is

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} = 0

An equivalent form is

x y z 1 x_{1} y_{1} z_{1} 1 x_{2} y_{2} z_{2} 1 x_{3} y_{3} z_{3} 1 = 0

Intercept form

The plane whose intercepts on the x, y and z axes are $a$ , $b$ and $c$ respectively is

\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 (ab c \neq = 0)

Figure 3: Intercept form. A plane cutting intercepts

a

b

c

on the axes has equation

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

Plane through a point and parallel to two vectors

The plane through the point with position vector $a$ and parallel to $b$ and $c$ is

r = a + λ b + μ c (parametric form), r \cdot (b \times c) = a \cdot (b \times c) (non-parametric form)

Solved Example 1

Find the equation of the plane on which the length of the normal from the origin is

10

and the direction ratios of this normal are

3, 2, 6

Solution:

In normal form the plane is $l x + m y + n z = p$ with $p = 10$ .

Since $3^{2} + 2^{2} + 6^{2} = 7$ , the direction cosines of the normal are $l = \frac{3}{7}$ , $m = \frac{2}{7}$ , $n = \frac{6}{7}$ .

So the plane is $\frac{3}{7} x + \frac{2}{7} y + \frac{6}{7} z = 10$ , that is $3 x + 2 y + 6 z = 70$ .

Solved Example 2

Find the equation of the plane passing through the point

(2, - 1, 3)

which is the foot of the perpendicular drawn from the origin to the plane.

Solution:

The perpendicular from the origin passes through $(2, - 1, 3)$ , so the direction ratios of the normal are $2, - 1, 3$ .

The plane through $(2, - 1, 3)$ with this normal is $2 (x - 2) - 1 (y + 1) + 3 (z - 3) = 0$ , that is $2 x - y + 3 z - 14 = 0$ .

Solved Example 3

Find the intercepts of the plane

3 x + 4 y - 7 z = 84

on the axes. Also find the length of the perpendicular from the origin to this plane and the direction cosines of this normal.

Solution:

Dividing by $84$ : $\frac{x}{28} + \frac{y}{21} + \frac{z}{- 12} = 1$ . The intercepts are $28$ , $21$ and $- 12$ .

$3^{2} + 4^{2} + (- 7)^{2} = 74$ . Dividing $3 x + 4 y - 7 z = 84$ by $74$ gives the normal form

\frac{3}{74} x + \frac{4}{74} y - \frac{7}{74} z = \frac{84}{74}

So $p = \frac{84}{74}$ and the direction cosines of the normal are $\frac{3}{74}, \frac{4}{74}, - \frac{7}{74}$ .

Solved Example 4

Find the equation of the plane passing through

(1, 1, 2)

and parallel to the vectors

\hat{i} + \hat{j} + \hat{k}

and

\hat{i} - \hat{j}

Solution:

The normal is perpendicular to both vectors, so take $n = (\hat{i} + \hat{j} + \hat{k}) \times (\hat{i} - \hat{j})$ .

n = \hat{i} \hat{j} \hat{k} 111 1 - 10 = \hat{i} + \hat{j} - 2 \hat{k}

The plane is $1 (x - 1) + 1 (y - 1) - 2 (z - 2) = 0$ , that is $x + y - 2 z + 2 = 0$ .

3. Special Planes

Plane	Equation
YZ-plane / parallel to YZ-plane	$x = 0$ / $x = d$
ZX-plane / parallel to ZX-plane	$y = 0$ / $y = d$
XY-plane / parallel to XY-plane	$z = 0$ / $z = d$
Parallel to the x-axis	$b y + cz + d = 0$
Parallel to the y-axis	$a x + cz + d = 0$
Parallel to the z-axis	$a x + b y + d = 0$
Through the origin	$a x + b y + cz = 0$
Parallel to $a x + b y + cz + d = 0$	$a x + b y + cz + λ = 0$

If the coefficient of $x$ is zero, the normal $(0, b, c)$ is perpendicular to the x-axis, so the plane is parallel to the x-axis. The same reasoning gives the other rows.

Figure 4: A missing variable means a parallel plane.

z = k

is parallel to the XY-plane;

b y + cz + d = 0

has no

x

-term and is parallel to the x-axis.

Solved Example 5

Find the equation of the plane passing through

(2, 3, - 4)

and

(1, - 1, 3)

and parallel to the x-axis.

Solution:

The plane through $(2, 3, - 4)$ is $a (x - 2) + b (y - 3) + c (z + 4) = 0$ ...(1)

Since $(1, - 1, 3)$ lies on it: $- a - 4 b + 7 c = 0$ , that is $a + 4 b - 7 c = 0$ ...(2)

Since the plane is parallel to the x-axis (perpendicular to the YZ-plane), its normal is perpendicular to the x-axis: $1 \cdot a + 0 \cdot b + 0 \cdot c = 0$ , so $a = 0$ .

Then (2) gives $4 b = 7 c$ , so $\frac{b}{7} = \frac{c}{4}$ . Putting $a = 0$ , $b = 7$ , $c = 4$ in (1): $7 (y - 3) + 4 (z + 4) = 0$ , that is $7 y + 4 z = 5$ .

Solved Example 6

Find the equation of the plane parallel to the plane

x + 5 y - 4 z + 5 = 0

whose intercepts on the axes have sum

150

Solution:

Any plane parallel to the given plane is $x + 5 y - 4 z = k$ . Its intercepts are $k$ , $\frac{k}{5}$ and $- \frac{k}{4}$ .

$k + \frac{k}{5} - \frac{k}{4} = 150 \Rightarrow \frac{19 k}{20} = 150 \Rightarrow k = \frac{3000}{19}$

The required plane is $x + 5 y - 4 z = \frac{3000}{19}$ .

4. Coplanarity of Four Points

Four points are coplanar if one of them lies on the plane passing through the other three. For $A (x_{1}, y_{1}, z_{1})$ , $B (x_{2}, y_{2}, z_{2})$ , $C (x_{3}, y_{3}, z_{3})$ and $D (x_{4}, y_{4}, z_{4})$ the condition is

x_{2} - x_{1} y_{2} - y_{1} z_{2} - z_{1} x_{3} - x_{1} y_{3} - y_{1} z_{3} - z_{1} x_{4} - x_{1} y_{4} - y_{1} z_{4} - z_{1} = 0

In vector form, $A$ , $B$ , $C$ , $D$ are coplanar if $[A B A C A D] = 0$ , which means the tetrahedron $A B C D$ has zero volume.

Solved Example 7

Show that the points

(0, - 1, 0)

(2, 1, - 1)

(1, 1, 1)

and

(3, 3, 0)

are coplanar, and find the plane containing them.

Solution:

Let $A (0, - 1, 0)$ , $B (2, 1, - 1)$ , $C (1, 1, 1)$ , $D (3, 3, 0)$ . The plane through $A$ is $a x + b (y + 1) + cz = 0$ ...(1)

It passes through $B$ and $C$ : $2 a + 2 b - c = 0$ ...(2) and $a + 2 b + c = 0$ ...(3)

From (2) and (3) by cross-multiplication, $\frac{a}{2 + 2} = \frac{b}{- 1 - 2} = \frac{c}{4 - 2}$ , so $\frac{a}{4} = \frac{b}{- 3} = \frac{c}{2}$ .

Plane (1) becomes $4 x - 3 (y + 1) + 2 z = 0$ , that is $4 x - 3 y + 2 z - 3 = 0$ .

For $D (3, 3, 0)$ : $12 - 9 + 0 - 3 = 0$ , so $D$ lies on this plane. Hence $A$ , $B$ , $C$ , $D$ are coplanar.

Check with the determinant: $A B = (2, 2, - 1)$ , $A C = (1, 2, 1)$ , $A D = (3, 4, 0)$ and $2 (0 - 4) - 2 (0 - 3) - 1 (4 - 6) = - 8 + 6 + 2 = 0$ .

5. Angle Between Two Planes

The angle between two planes is defined as the angle between their normals drawn from any point. For the planes $a_{1} x + b_{1} y + c_{1} z + d_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$ ,

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}}

In vector form, for $r \cdot n_{1} = d_{1}$ and $r \cdot n_{2} = d_{2}$ , $cos θ = \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣∣ n _{2} ∣}$ . Take the modulus for the acute angle.

Figure 5: Viewed along their line of intersection, the angle

θ

between two planes equals the angle between their normals

n_{1}

and

n_{2}

Planes are	Cartesian condition	Vector condition
Perpendicular	$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$	$n_{1} \cdot n_{2} = 0$
Parallel	$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$	$n_{1} = λ n_{2}$

Solved Example 8

Find the angle between the planes

2 x - y + z = 11

and

x + y + 2 z = 3

Solution:

cos θ = \frac{2 \cdot 1 + ( - 1 ) \cdot 1 + 1 \cdot 2}{2 ^{2} + ( - 1 ) ^{2} + 1 ^{2} 1 ^{2} + 1 ^{2} + 2 ^{2}} = \frac{3}{6} = \frac{1}{2}

Therefore $θ = \frac{π}{3}$ .

Solved Example 9

Find the equation of the plane passing through

(2, 2, 1)

and

(9, 3, 6)

and perpendicular to the plane

x + 3 y + 3 z = 8

Solution:

The normal of the required plane is perpendicular to the segment direction $(9 - 2, 3 - 2, 6 - 1) = (7, 1, 5)$ and to the normal $(1, 3, 3)$ of the given plane.

$(7, 1, 5) \times (1, 3, 3) = (1 \cdot 3 - 5 \cdot 3, 5 \cdot 1 - 7 \cdot 3, 7 \cdot 3 - 1 \cdot 1) = (- 12, - 16, 20)$ , which is proportional to $(3, 4, - 5)$ .

The plane is $3 (x - 2) + 4 (y - 2) - 5 (z - 1) = 0$ , that is $3 x + 4 y - 5 z = 9$ .

Solved Example 10

A tetrahedron has vertices

O (0, 0, 0)

A (1, 2, 1)

B (2, 1, 3)

and

C (- 1, 1, 2)

. Prove that the angle between the faces

O A B

and

A B C

cos^{- 1} (\frac{19}{35})

Solution:

Normal to face $O A B$ : $O A \times O B = (1, 2, 1) \times (2, 1, 3) = (2 \cdot 3 - 1 \cdot 1, 1 \cdot 2 - 1 \cdot 3, 1 \cdot 1 - 2 \cdot 2) = (5, - 1, - 3)$ .

Normal to face $A B C$ : $A B \times A C = (1, - 1, 2) \times (- 2, - 1, 1) = (1, - 5, - 3)$ .

$cos θ = \frac{5 + 5 + 9}{35 35} = \frac{19}{35}$ , so the angle between the faces is $cos^{- 1} (\frac{19}{35})$ .

6. A Point and a Plane: Sides, Distance, Foot and Image

Sides of a plane

A plane divides space into two parts. The points $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ lie on the same side of $a x + b y + cz + d = 0$ if $a x_{1} + b y_{1} + c z_{1} + d$ and $a x_{2} + b y_{2} + c z_{2} + d$ have the same sign, and on opposite sides if they have opposite signs.

Figure 6: Sides of a plane. Points giving

a x + b y + cz + d

the same sign lie on the same side; opposite signs mean the segment joining them crosses the plane.

Ratio in which a plane divides a segment

The plane $a x + b y + cz + d = 0$ divides the join of $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ in the ratio

- \frac{a x _{1} + b y _{1} + c z _{1} + d}{a x _{2} + b y _{2} + c z _{2} + d}

A positive ratio means internal division (the points are on opposite sides); a negative ratio means external division. In particular, the XY-plane divides the join in the ratio $- \frac{z _{1}}{z _{2}}$ , the YZ-plane in $- \frac{x _{1}}{x _{2}}$ and the ZX-plane in $- \frac{y _{1}}{y _{2}}$ .

Perpendicular distance of a point from a plane

The length of the perpendicular from $P (x_{1}, y_{1}, z_{1})$ to the plane $a x + b y + cz + d = 0$ is

\frac{a x _{1} + b y _{1} + c z _{1} + d}{a ^{2} + b ^{2} + c ^{2}}

In vector form, the distance from the point with position vector $a$ to the plane $r \cdot n = d$ is $\frac{∣ a \cdot n - d ∣}{∣ n ∣}$ .

Figure 7: Foot

N

and image

P^{'}

of a point in a plane.

P N

is the perpendicular distance and

N

is the midpoint of

P P^{'}

Foot of the perpendicular and image of a point

The foot $(x, y, z)$ of the perpendicular from $(x_{1}, y_{1}, z_{1})$ to $a x + b y + cz + d = 0$ is given by

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = - \frac{a x _{1} + b y _{1} + c z _{1} + d}{a ^{2} + b ^{2} + c ^{2}}

The image $(x^{'}, y^{'}, z^{'})$ of $(x_{1}, y_{1}, z_{1})$ in the plane is given by

\frac{x ^{'} - x _{1}}{a} = \frac{y ^{'} - y _{1}}{b} = \frac{z ^{'} - z _{1}}{c} = - \frac{2 ( a x _{1} + b y _{1} + c z _{1} + d )}{a ^{2} + b ^{2} + c ^{2}}

These come from two facts: the join of the point and its image is parallel to the normal $(a, b, c)$ , so $x^{'} = x_{1} + aλ$ , $y^{'} = y_{1} + bλ$ , $z^{'} = z_{1} + c λ$ ; and the midpoint $(\frac{x ^{'} + x _{1}}{2}, \frac{y ^{'} + y _{1}}{2}, \frac{z ^{'} + z _{1}}{2})$ lies on the plane, which fixes $λ$ .

Distance between two parallel planes

For $a x + b y + cz + d_{1} = 0$ and $a x + b y + cz + d_{2} = 0$ (same $a, b, c$ ),

d = \frac{∣ d _{1} - d _{2} ∣}{a ^{2} + b ^{2} + c ^{2}}

Figure 8: Distance between parallel planes with the same

a, b, c

d = \frac{∣ d _{1} - d _{2} ∣}{a ^{2} + b ^{2} + c ^{2}}

Solved Example 11

Show that the points

(1, 2, 3)

and

(2, - 1, 4)

lie on opposite sides of the plane

x + 4 y + z - 3 = 0

Solution:

For $(1, 2, 3)$ : $1 + 4 (2) + 3 - 3 = 9$ . For $(2, - 1, 4)$ : $2 - 4 + 4 - 3 = - 1$ .

The values have opposite signs, so the points lie on opposite sides of the plane.

Solved Example 12

Find the ratio in which the XY-plane divides the line joining

A (1, 2, 3)

and

B (2, 3, 6)

Solution:

The XY-plane divides the join in the ratio $- \frac{z _{1}}{z _{2}} = - \frac{3}{6} = - \frac{1}{2}$ .

The ratio is negative, so the XY-plane divides $A B$ externally in the ratio $1 : 2$ . This is expected, since both points have positive z-coordinates and lie on the same side of the XY-plane.

Solved Example 13

Find the image of the point

P (3, 5, 7)

in the plane

2 x + y + z = 0

Solution:

Direction ratios of the normal are $2, 1, 1$ . Let $Q$ be the image and let $P Q$ meet the plane at $R$ . Then $P Q$ is along the normal, so $R = (2 r + 3, r + 5, r + 7)$ for some $r$ .

$R$ lies on the plane: $2 (2 r + 3) + (r + 5) + (r + 7) = 0 \Rightarrow 6 r + 18 = 0 \Rightarrow r = - 3$ , so $R = (- 3, 2, 4)$ .

$R$ is the midpoint of $P Q$ . If $Q = (α, β, γ)$ , then $\frac{α + 3}{2} = - 3$ , $\frac{β + 5}{2} = 2$ , $\frac{γ + 7}{2} = 4$ .

So $α = - 9$ , $β = - 1$ , $γ = 1$ , and the image is $Q (- 9, - 1, 1)$ .

Solved Example 14

For the point

(1, 0, 2)

and the plane

2 x + y + z = 5

, find (i) the perpendicular distance, (ii) the foot of the perpendicular, (iii) the image of the point.

Solution:

Here $a x_{1} + b y_{1} + c z_{1} + d = 2 + 0 + 2 - 5 = - 1$ and $a^{2} + b^{2} + c^{2} = 6$ .

(i) Distance $= \frac{∣ - 1∣}{6} = \frac{1}{6}$ .

(ii) $k = - \frac{- 1}{6} = \frac{1}{6}$ , so the foot is $(1 + \frac{2}{6}, 0 + \frac{1}{6}, 2 + \frac{1}{6}) = (\frac{4}{3}, \frac{1}{6}, \frac{13}{6})$ .

(iii) Using $2 k = \frac{1}{3}$ , the image is $(1 + \frac{2}{3}, \frac{1}{3}, 2 + \frac{1}{3}) = (\frac{5}{3}, \frac{1}{3}, \frac{7}{3})$ .

Solved Example 15

Find the distance between the planes

2 x - y + 2 z = 4

and

6 x - 3 y + 6 z = 2

Solution:

The planes are $2 x - y + 2 z - 4 = 0$ ...(1) and $6 x - 3 y + 6 z - 2 = 0$ ...(2). Since $\frac{2}{6} = \frac{- 1}{- 3} = \frac{2}{6}$ , they are parallel.

Make the coefficients equal: (2) becomes $2 x - y + 2 z - \frac{2}{3} = 0$ .

Distance $= \frac{- 4 + \frac{2}{3}}{2 ^{2} + ( - 1 ) ^{2} + 2 ^{2}} = \frac{10/3}{3} = \frac{10}{9}$ units.

7. Family of Planes

If $u = 0$ and $v = 0$ are two non-parallel planes, the equation of any plane passing through their line of intersection is $u + λ v = 0$ . For $a_{1} x + b_{1} y + c_{1} z + d_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$ :

a_{1} x + b_{1} y + c_{1} z + d_{1} + λ (a_{2} x + b_{2} y + c_{2} z + d_{2}) = 0, λ \in R

This is also the equation of a plane through a line given in non-symmetric form. In vector form, the plane through the intersection of $r \cdot n_{1} = d_{1}$ and $r \cdot n_{2} = d_{2}$ is $r \cdot (n_{1} + λ n_{2}) = d_{1} + λ d_{2}$ .

Figure 9: Family of planes. Every plane through the line of intersection of

P_{1} = 0

and

P_{2} = 0

has the form

P_{1} + λ P_{2} = 0

No value of

λ

gives the plane

v = 0

itself. If a condition seems to have no solution for

λ

, check whether

v = 0

is the answer.

Solved Example 16

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

2 x - 4 y + 3 z + 5 = 0

and

x + y + z = 6

and parallel to the straight line having direction ratios

1, - 1, - 1

Solution:

The required plane is $(2 x - 4 y + 3 z + 5) + λ (x + y + z - 6) = 0$ , that is

(2 + λ) x + (- 4 + λ) y + (3 + λ) z + (5 - 6 λ) = 0

The plane is parallel to the line, so its normal is perpendicular to the line: $a l + bm + c n = 0$ .

$1 (2 + λ) - 1 (- 4 + λ) - 1 (3 + λ) = 0 \Rightarrow 3 - λ = 0 \Rightarrow λ = 3$

The required plane is $5 x - y + 6 z - 13 = 0$ .

Solved Example 17

The plane

x - y - z = 4

is rotated through

9 0^{\circ}

about its line of intersection with the plane

x + y + 2 z = 4

. Find its equation in the new position.

Solution:

The new plane passes through the line of intersection, so it is

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

It is perpendicular to $x - y - z = 4$ : $(1 + k) - (1 - k) - (2 - k) = 0 \Rightarrow 3 k - 2 = 0 \Rightarrow k = \frac{2}{3}$ .

Putting $k = \frac{2}{3}$ and multiplying by $3$ : $5 x + y + 4 z = 20$ .

Solved Example 18

Find the equation of the plane through the point

(1, 1, 1)

which passes through the line of intersection of the planes

x + y + z = 6

and

2 x + 3 y + 4 z + 5 = 0

Solution:

Any plane through the line of intersection is $x + y + z - 6 + k (2 x + 3 y + 4 z + 5) = 0$ .

It passes through $(1, 1, 1)$ : $1 + 1 + 1 - 6 + k (2 + 3 + 4 + 5) = 0 \Rightarrow 14 k = 3 \Rightarrow k = \frac{3}{14}$ .

Multiplying by $14$ : $14 (x + y + z - 6) + 3 (2 x + 3 y + 4 z + 5) = 0$ , that is $20 x + 23 y + 26 z - 69 = 0$ .

Solved Example 19

If the planes

x - cy - b z = 0

c x - y + a z = 0

and

b x + a y - z = 0

pass through a straight line, find the value of

a^{2} + b^{2} + c^{2} + 2 ab c

Solution:

All three planes pass through the origin. They have a common line exactly when their normals are all perpendicular to that line, that is when the three normals are coplanar:

1 - c - b c - 1 a ba - 1 = 0

Expanding: $1 (1 - a^{2}) + c (- c - ab) - b (a c + b) = 0 \Rightarrow 1 - a^{2} - c^{2} - 2 ab c - b^{2} = 0$ .

Hence $a^{2} + b^{2} + c^{2} + 2 ab c = 1$ . The same result follows by writing the third plane as a member of the family of the first two and comparing coefficients.

8. Angle Bisector Planes

The equations of the planes bisecting the angles between $a_{1} x + b_{1} y + c_{1} z + d_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$ are

\frac{a _{1} x + b _{1} y + c _{1} z + d _{1}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2}} = \pm \frac{a _{2} x + b _{2} y + c _{2} z + d _{2}}{a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}

Every point on a bisector plane is at equal distance from both planes.

Figure 10: Bisector planes of two planes, seen edge-on along their line of intersection. Points on a bisector are equidistant from both planes, giving

\frac{P _{1}}{∣ n _{1} ∣} = \pm \frac{P _{2}}{∣ n _{2} ∣}

. With

d_{1}, d_{2} > 0

, the

+

sign gives the bisector of the angle containing

O

; here

O

lies in the acute angle.

Which bisector is which

Rewrite both equations so that the constant terms $d_{1}$ and $d_{2}$ are positive.
The $+$ sign then gives the bisector of the angle that contains the origin.
If $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} > 0$ , the origin lies in the obtuse angle, so the $+$ sign gives the obtuse-angle bisector.
If $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} < 0$ , the origin lies in the acute angle, so the $+$ sign gives the acute-angle bisector.

Solved Example 20

Find the planes bisecting the angles between the planes

2 x + y + 2 z = 9

and

3 x - 4 y + 12 z + 13 = 0

. Which of them bisects the acute angle? Does the origin lie in the acute angle or the obtuse angle?

Solution:

With positive constant terms the planes are $- 2 x - y - 2 z + 9 = 0$ ...(1) and $3 x - 4 y + 12 z + 13 = 0$ ...(2). Their normals have lengths $3$ and $13$ .

\frac{- 2 x - y - 2 z + 9}{3} = \pm \frac{3 x - 4 y + 12 z + 13}{13}

Taking the $+$ sign: $13 (- 2 x - y - 2 z + 9) = 3 (3 x - 4 y + 12 z + 13)$ , which gives $35 x + y + 62 z = 78$ ...(3)

Taking the $-$ sign: $13 (- 2 x - y - 2 z + 9) = - 3 (3 x - 4 y + 12 z + 13)$ , which gives $17 x + 25 y - 10 z = 156$ ...(4)

Now $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = (- 2) (3) + (- 1) (- 4) + (- 2) (12) = - 26 < 0$ .

So the origin lies in the acute angle, and the bisector of the acute angle is (3): $35 x + y + 62 z = 78$ .

9. A Line and a Plane

Angle between a line and a plane

For the plane $a x + b y + cz + d = 0$ , the normal has direction ratios $a, b, c$ . If a straight line has direction cosines $l, m, n$ , the angle $θ$ between the line and the normal is given by $cos θ = \frac{a l + bm + c n}{a ^{2} + b ^{2} + c ^{2}}$ . The angle between the line and the plane is $ϕ = \frac{π}{2} - θ$ , so

sin ϕ = \frac{∣ a l + bm + c n ∣}{a ^{2} + b ^{2} + c ^{2} l ^{2} + m ^{2} + n ^{2}}

In vector form, for the line $r = a + λ b$ and the plane $r \cdot n = d$ , $sin ϕ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$ .

Figure 11: The angle

ϕ

between a line and a plane is the complement of the angle

θ

between the line and the normal, so

sin ϕ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}

Conditions

Line and plane are	Cartesian condition	Vector condition
Parallel	$a l + bm + c n = 0$	$b \cdot n = 0$
Perpendicular	$\frac{a}{l} = \frac{b}{m} = \frac{c}{n}$	$b \times n = 0$
Line lies in the plane	$a l + bm + c n = 0$ and $a x_{1} + b y_{1} + c z_{1} + d = 0$	$b \cdot n = 0$ and $a \cdot n = d$

In the last row, $(x_{1}, y_{1}, z_{1})$ is any point of the line $\frac{x - x _{1}}{l} = \frac{y - y _{1}}{m} = \frac{z - z _{1}}{n}$ .

Standard problems

Point of intersection: put the general point of the line into the equation of the plane and solve for the parameter.
Distance of a point from a plane measured parallel to a line: draw the line through the point parallel to the given line, find where it meets the plane, and measure the distance to that point.
Plane containing a given line: take a point of the line on the plane and make the normal perpendicular to the direction of the line.
Projection of a line on a plane: find the plane through the line perpendicular to the given plane; the projection is the line of intersection of these two planes.

Figure 12: Projection of a line on a plane.

A^{'} B^{'}

is where the given plane meets the plane that contains

A B

and is perpendicular to the given plane.

Solved Example 21

Find the values of

a

and

b

for which the line

\frac{x - 2}{a} = \frac{y + 3}{4} = \frac{z - 6}{- 2}

is perpendicular to the plane

3 x - 2 y + b z + 10 = 0

Solution:

The line is perpendicular to the plane when its direction ratios are proportional to the normal $(3, - 2, b)$ :

$\frac{a}{3} = \frac{4}{- 2} = \frac{- 2}{b}$

From $\frac{a}{3} = - 2$ , $a = - 6$ . From $\frac{- 2}{b} = - 2$ , $b = 1$ .

Solved Example 22

Find the distance of the point

(1, 0, - 3)

from the plane

x - y - z = 9

measured parallel to the line

\frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{- 6}

Solution:

The line through $Q (1, 0, - 3)$ parallel to the given line is $\frac{x - 1}{2} = \frac{y}{3} = \frac{z + 3}{- 6} = r$ .

Any point on it is $P (2 r + 1, 3 r, - 6 r - 3)$ . If $P$ lies on the plane: $(2 r + 1) - 3 r - (- 6 r - 3) = 9 \Rightarrow 5 r + 4 = 9 \Rightarrow r = 1$ .

So $P = (3, 3, - 9)$ and $P Q = (3 - 1)^{2} + (3 - 0)^{2} + (- 9 + 3)^{2} = 4 + 9 + 36 = 7$ units.

Solved Example 23

Find the equation of the plane passing through

(1, 2, 0)

which contains the line

\frac{x + 3}{3} = \frac{y - 1}{4} = \frac{z - 2}{- 2}

Solution:

Any plane through $(1, 2, 0)$ is $a (x - 1) + b (y - 2) + cz = 0$ ...(1)

It contains the line, so its normal is perpendicular to the line: $3 a + 4 b - 2 c = 0$ ...(2)

The point $(- 3, 1, 2)$ of the line lies on (1): $- 4 a - b + 2 c = 0$ ...(3)

From (2) and (3), $\frac{a}{8 - 2} = \frac{b}{8 - 6} = \frac{c}{- 3 + 16}$ , so $\frac{a}{6} = \frac{b}{2} = \frac{c}{13}$ .

The plane is $6 (x - 1) + 2 (y - 2) + 13 z = 0$ , that is $6 x + 2 y + 13 z - 10 = 0$ .

Solved Example 24

Find the equation of the projection of the line

\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 3}{4}

on the plane

x + 2 y + z = 9

Solution:

Let $A B$ be the given line and $C D$ its projection. The plane $A B C D$ contains $A B$ and is perpendicular to the given plane. It passes through $(1, - 1, 3)$ :

$a (x - 1) + b (y + 1) + c (z - 3) = 0$ with $2 a - b + 4 c = 0$ (contains $A B$ ) and $a + 2 b + c = 0$ (perpendicular to the given plane).

By cross-multiplication, $\frac{a}{- 1 - 8} = \frac{b}{4 - 2} = \frac{c}{4 + 1}$ , so $\frac{a}{9} = \frac{b}{- 2} = \frac{c}{- 5}$ .

Plane $A B C D$ : $9 (x - 1) - 2 (y + 1) - 5 (z - 3) = 0$ , that is $9 x - 2 y - 5 z + 4 = 0$ .

The projection is the intersection of $9 x - 2 y - 5 z + 4 = 0$ and $x + 2 y + z - 9 = 0$ . Its direction ratios satisfy $9 l - 2 m - 5 n = 0$ and $l + 2 m + n = 0$ , giving $\frac{l}{8} = \frac{m}{- 14} = \frac{n}{20}$ , or $4, - 7, 10$ .

For a point, put $z = 0$ : $9 x - 2 y + 4 = 0$ and $x + 2 y - 9 = 0$ , so $x = \frac{1}{2}$ , $y = \frac{17}{4}$ .

The projection is $\frac{x - \frac{1}{2}}{4} = \frac{y - \frac{17}{4}}{- 7} = \frac{z}{10}$ .

10. Coplanar Lines

Lines in symmetric form

The lines $\frac{x - α}{l} = \frac{y - β}{m} = \frac{z - γ}{n}$ and $\frac{x - α ^{'}}{l ^{'}} = \frac{y - β ^{'}}{m ^{'}} = \frac{z - γ ^{'}}{n ^{'}}$ are coplanar (they intersect or are parallel) if

α^{'} - α β^{'} - β γ^{'} - γ l mn l^{'} m^{'} n^{'} = 0

and then the plane containing them is

x - α y - β z - γ l mn l^{'} m^{'} n^{'} = 0

Lines in general form

The lines $a x + b y + cz + d = 0 = a^{'} x + b^{'} y + c^{'} z + d^{'}$ and $α x + β y + γ z + δ = 0 = α^{'} x + β^{'} y + γ^{'} z + δ^{'}$ are coplanar if

ab c d a^{'} b^{'} c^{'} d^{'} α β γ δ α^{'} β^{'} γ^{'} δ^{'} = 0

Vector form

The lines $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ are coplanar if $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = 0$ , and the plane containing them is $(r - a_{1}) \cdot (b_{1} \times b_{2}) = 0$ .

This is the same scalar triple product used for the shortest distance between two lines. Zero shortest distance with non-parallel directions means the lines intersect, which is exactly when they are coplanar.

Solved Example 25

Show that the lines

\frac{x - 3}{2} = \frac{y + 1}{- 3} = \frac{z + 2}{1}

and

\frac{x - 7}{- 3} = \frac{y}{1} = \frac{z + 7}{2}

are coplanar. Also find the equation of the plane containing them.

Solution:

Any point on the first line is $(2 r + 3, - 3 r - 1, r - 2)$ and on the second line is $(- 3 R + 7, R, 2 R - 7)$ . If the lines meet:

$2 r + 3 = - 3 R + 7 \Rightarrow 2 r + 3 R = 4$ ...(1)

$- 3 r - 1 = R \Rightarrow - 3 r - R = 1$ ...(2)

$r - 2 = 2 R - 7 \Rightarrow r - 2 R = - 5$ ...(3)

Solving (1) and (2), $r = - 1$ and $R = 2$ , which also satisfy (3). So the lines intersect (at $(1, 2, - 3)$ ) and are therefore coplanar.

The plane containing them is

x - 3 y + 1 z + 2 2 - 31 - 312 = 0

$(x - 3) (- 6 - 1) - (y + 1) (4 + 3) + (z + 2) (2 - 9) = 0 \Rightarrow - 7 (x - 3) - 7 (y + 1) - 7 (z + 2) = 0$

So the plane is $x + y + z = 0$ .

Common Mistakes to Avoid

Watch out

Treating $a, b, c$ in $a x + b y + cz + d = 0$ as direction cosines. They are only direction ratios of the normal; divide by $a^{2} + b^{2} + c^{2}$ before reading $p$ .
Using the parallel-planes distance formula before making the coefficients of $x$ , $y$ , $z$ identical in both equations.
Using $cos$ instead of $sin$ for the angle between a line and a plane. The dot product with the normal gives the complement of the required angle.
Mixing up the conditions: $a l + bm + c n = 0$ means the line is parallel to the plane, while $\frac{a}{l} = \frac{b}{m} = \frac{c}{n}$ means it is perpendicular.
Concluding that a line lies in a plane from $a l + bm + c n = 0$ alone. A point of the line must also satisfy the plane.
Choosing the bisector containing the origin without first making both constant terms positive.
Forgetting the factor $2$ when moving from the foot of the perpendicular to the image.
Using $P_{1} + λ P_{2} = 0$ and missing the case where the answer is the plane $P_{2} = 0$ itself.

Frequently Asked Questions

What does the general equation of a plane tell you?

In $a x + b y + cz + d = 0$ , the coefficients $a, b, c$ are direction ratios of the normal to the plane, and $\frac{∣ d ∣}{a ^{2} + b ^{2} + c ^{2}}$ is its distance from the origin. A missing variable means the plane is parallel to that axis, and $d = 0$ means it passes through the origin.

How do you find the equation of a plane through three points?

Take two vectors in the plane, such as $A B$ and $A C$ , and find the normal $A B \times A C$ . Then use the point-normal form with point $A$ . This gives the same result as setting the $3 \times 3$ determinant form of the three-point equation equal to zero.

How do you find the distance of a point from a plane?

Substitute the point into the left side of $a x + b y + cz + d = 0$ , take the modulus, and divide by $a^{2} + b^{2} + c^{2}$ . The sign of the substituted value, before taking the modulus, also tells you which side of the plane the point lies on.

How do you find the image of a point in a plane?

Move from the point along the normal $(a, b, c)$ by a parameter $λ$ , and require the midpoint of the point and its image to lie on the plane. This gives $λ = - \frac{2 ( a x _{1} + b y _{1} + c z _{1} + d )}{a ^{2} + b ^{2} + c ^{2}}$ . Half of that value gives the foot of the perpendicular.

What is the angle between a line and a plane?

It is the angle between the line and its projection on the plane. Because the dot product of the line direction with the normal gives the complementary angle, the formula uses sine: $sin ϕ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$ . If this is zero, the line is parallel to the plane.

How do you check whether two lines are coplanar?

Compute the scalar triple product $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2})$ , or the equivalent $3 \times 3$ determinant. If it is zero the lines are coplanar, and the plane containing them has normal $b_{1} \times b_{2}$ . If it is non-zero, the lines are skew.

Is the plane part of the JEE Main 2026 syllabus?

After the 2024 revision, the JEE Main syllabus lists lines, skew lines and shortest distance but not planes as a separate topic. Line problems still use plane ideas such as normals, the line of intersection and points meeting a plane, so learning the core formulas on this page remains useful for JEE Main.

Which plane topics matter most for JEE Advanced?

The equation of a plane and the distance of a point from a plane are listed explicitly. Advanced problems combine them with family of planes, image of a point, projection of a line on a plane, coplanar lines and bisector planes, often in multi-correct questions, so practise switching quickly between Cartesian and vector forms.

Previous year questions on The Plane

6 questions from past papers, each with a step-by-step solution.

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