JEE Mains Maths · Three Dimensional Geometry

Equation of a Plane

Writing a plane from a point and a normal, three points or its intercepts; finding the normal as a cross product when the plane contains lines or is perpendicular to other planes; the family of planes through a line of intersection; and the angle between planes.

All Three Dimensional Geometry notes JEE Mains Maths strategy

Why this matters

Fifty-two PYQs. A plane needs a point and a normal, and in most questions the work is finding the normal: a cross product, or one number in the family P₁ + λP₂. Four ideas cover the page.

Concept 1 of 4: A plane from a point and a normal, three points, or its intercepts

A plane is fixed by one point on it and a normal vector

\vec n=(a,b,c)

: every vector lying in the plane is perpendicular to

\vec n

. That gives

a(x-x_1)+b(y-y_1)+c(z-z_1)=0

, and in

ax+by+cz+d=0

the coefficients ARE the normal. Three points give two in-plane vectors, whose cross product is the normal. A plane cutting the axes at

a

b

c

\frac xa+\frac yb+\frac zc=1

Definition

Point and normal: $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ .
Three points: normal $\vec{AB}\times\vec{AC}$ .
Intercepts: $\frac xa+\frac yb+\frac zc=1$ .
Foot of the perpendicular from $O$ is $F$ : normal $\vec{OF}$ , plane $\vec r\cdot\vec{OF}=|\vec{OF}|^2$ .
Perpendicular bisector of $AB$ : normal $\vec{AB}$ , through the midpoint.
Four points coplanar: scalar triple product of the three edges from one of them is $0$ .

Point-normal form

a(x-x_1)+b(y-y_1)+c(z-z_1)=0

Worked example

Find the plane through

(1,0,0)

(0,2,0)

and

(0,0,3)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 9 · Q82Moderate

Example 1 · Three Dimensional Geometry · Equation of a Plane

Let the plane

ax+by+cz+d=0

bisect the line joining the points

(4,-3,1)

and

(2,3,-5)

at right angles. If

a,b,c,d

are integers, then the minimum value of

\left(a^{2}+b^{2}+c^{2}+d^{2}\right)

Intercepts are not the normal

\frac x2+\frac y3+\frac z6=1

has normal

\left(\frac12,\frac13,\frac16\right)

, i.e.

(3,2,1)

, not

(2,3,6)

. Clear the fractions before reading the normal.

Drill 11 more on a plane from a point and a normal, three points, or its intercepts

Concept 2 of 4: The normal as a cross product: planes containing lines or perpendicular to planes

Whenever two directions lie IN the plane, their cross product is its normal. A plane containing a line and a point uses the line's direction and the vector from a point of the line to the given point. A plane containing two lines uses both directions. A plane perpendicular to two other planes contains both of their normals, so its normal is their cross product.

Definition

Contains a line (point $A$ , direction $\vec d$ ) and a point $P$ : $\vec n=\vec d\times\vec{AP}$ .
Contains two lines (or parallel to two directions): $\vec n=\vec d_1\times\vec d_2$ .
Perpendicular to two planes: $\vec n=\vec n_1\times\vec n_2$ .
Contains a line and is perpendicular to a plane: $\vec n=\vec d\times\vec n_1$ .
Contains an axis: that axis's unit vector is one in-plane direction.

Normal from two in-plane directions

\vec n=\vec u\times\vec v

Worked example

Find the plane containing the line

x=y=z

and the point

(1,2,3)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 11 April 2023 · Q62Moderate

Example 2 · Three Dimensional Geometry · Equation of a Plane

If equation of the plane that contains the point

( - 2,3,5)

and is perpendicular to each of the planes

2x + 4y + 5z = 8

and

3x - 2y + 3z = 5

\alpha x + \beta y + \gamma z + 97 = 0

then

\alpha + \beta + \gamma =

Perpendicular to a plane means CONTAINING its normal

A plane perpendicular to

x+y+z=1

has

(1,1,1)

lying in it, so

(1,1,1)

goes into the cross product. Using

(1,1,1)

as the new normal gives a PARALLEL plane instead.

Drill 17 more on the normal as a cross product: planes containing lines or perpendicular to planes

Concept 3 of 4: The family of planes through a line of intersection

Every plane through the line where

P_1=0

and

P_2=0

meet can be written

P_1+\lambda P_2=0

. The whole question reduces to finding one number

\lambda

from the extra condition: a point on the plane, perpendicular to another plane, parallel to a line or an axis, a distance from a point, or a rotation through a right angle.

Definition

Family: $P_1+\lambda P_2=0$ , normal $\vec n_1+\lambda\vec n_2$ .
Through a point: substitute it.
Perpendicular to a plane with normal $\vec m$ : $(\vec n_1+\lambda\vec n_2)\cdot\vec m=0$ .
Parallel to a line or axis with direction $\vec d$ : $(\vec n_1+\lambda\vec n_2)\cdot\vec d=0$ .
$P_1$ rotated through $90^\circ$ about the line: the new normal is perpendicular to $\vec n_1$ .
The form never produces $P_2$ itself; check it separately if needed.

Planes through a line of intersection

P_1+\lambda P_2=0

Worked example

Find the plane through the line of intersection of

x+y+z=1

and

2x+3y+4z=5

that passes through

(1,1,1)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q163Moderate

Example 3 · Three Dimensional Geometry · Equation of a Plane

A plane

P

contains the line of intersection of the plane

\overrightarrow{r}\cdot (\widehat{i}+\widehat{j}+\widehat{k}) = 6

and

\overrightarrow{r}\cdot (2\widehat{i}+ 3\widehat{j}+ 4\widehat{k}) = - 5

. If

P

passes through the point

(0,2, - 2)

, then the square of distance of the point

(12,12,18)

from the plane

P

Parallel to a line: dot with its DIRECTION

A plane parallel to a line has the line's direction perpendicular to its normal:

\vec n\cdot\vec d=0

. Setting

\vec n

parallel to

\vec d

instead makes the plane perpendicular to the line.

Drill 18 more on the family of planes through a line of intersection

Concept 4 of 4: The angle between two planes; parallel and perpendicular planes

Two planes meet at the same angle as their normals. So the angle comes from the dot product of the normals, parallel planes have proportional normals, and perpendicular planes have normals with zero dot product.

Definition

$\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}$ .
Parallel: $\vec n_1\parallel\vec n_2$ . Perpendicular: $\vec n_1\cdot\vec n_2=0$ .
The points equidistant from $A$ and $B$ form a plane with normal $\vec{AB}$ .

Angle between planes

\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1|\,|\vec n_2|}

Worked example

Find the angle between

x+y=0

and

y+z=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 14 June 2022 · Q165Moderate

Example 4 · Three Dimensional Geometry · Equation of a Plane

Let the points on the plane

P

be equidistant from the points

( - 4,2,1)

and

(2, - 2,3)

. Then the acute angle between the plane

P

and the plane

2x + y +

3z = 1

Take the absolute value for the acute angle

A negative dot product gives the obtuse angle between the normals. The angle between planes is conventionally the acute one, so use

|\vec n_1\cdot\vec n_2|

Drill 2 more on the angle between two planes; parallel and perpendicular 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 (4)

A plane from a point and a normal, three points, or its intercepts
Point-normal form
$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
The normal as a cross product: planes containing lines or perpendicular to planes
Normal from two in-plane directions
$\vec n=\vec u\times\vec v$
The family of planes through a line of intersection
Planes through a line of intersection
$P_1+\lambda P_2=0$
The angle between two planes; parallel and perpendicular planes
Angle between planes
$\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1|\,|\vec n_2|}$

Watch out for (4)

Intercepts are not the normal
→ A plane from a point and a normal, three points, or its intercepts
Perpendicular to a plane means CONTAINING its normal
→ The normal as a cross product: planes containing lines or perpendicular to planes
Parallel to a line: dot with its DIRECTION
→ The family of planes through a line of intersection
Take the absolute value for the acute angle
→ The angle between two planes; parallel and perpendicular planes

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