MHT-CET Maths · Line and Plane

Plane — Equation, Normal, and Construction

How to write the equation of a plane from whatever the question hands you — a point and a normal, three points, two lines or two planes it must respect — by always first nailing the normal vector, plus the family-of-planes lambda trick for planes through an intersection line.

All Line and Plane notes MHT-CET Maths strategy

Why this matters

This is the densest scoring subtopic in Line and Plane: roughly 36 PYQs, MODERATE-to-HARD, and the templates repeat hard — the 'plane through a point parallel to two lines' and the 'plane through an intersection line with a side condition' shapes each recur three or four times across 2023-2025. Almost every question reduces to ONE move: find the normal vector, then write n-dot-(r minus a) = 0. The normal comes either from a cross product (two directions the plane must contain) or from a family-of-planes lambda solved against a perpendicularity or parallelism condition. Learn those two engines — the cross-product normal and the lambda family — and the rest (intercepts, foot of perpendicular, mirror image) is bookkeeping.

Concept 1 of 12

Equation of a plane and its normal

Intuition

A plane is fixed by ONE point on it plus a direction perpendicular to it — the normal vector. Every plane equation, no matter how it is dressed up, is just "the normal dotted with the displacement from a known point is zero." Read the coefficients of

x, y, z

and you have read off the normal.

Definition

A plane in space has three equivalent forms:

Cartesian form: $ax + by + cz = d$ . The coefficients give the normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ .
Vector form: $\vec{r}\cdot\vec{n} = d$ , where $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is the position vector of a general point.
Point-normal form: through a point $A(x_0,y_0,z_0)$ with normal $\vec{n} = (a,b,c)$ : $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ .

Two planes are parallel when their normals are parallel (proportional coefficients). Two planes are perpendicular when their normals are perpendicular: $\vec{n_1}\cdot\vec{n_2} = 0$ .

The three equivalent forms

ax + by + cz = d \;\Longleftrightarrow\; \vec{r}\cdot\vec{n} = d \;\Longleftrightarrow\; \vec{n}\cdot(\vec{r} - \vec{a}) = 0

$\vec{n} = (a,b,c)$ normal — the coefficients of $x, y, z$
$\vec{a}$ position vector of a known point on the plane
$d$ constant, fixed by substituting the known point

Worked example

Write the equation of the plane through

A(2,-1,3)

with normal

\vec{n} = \hat{i} + 4\hat{j} - 2\hat{k}

Point-normal form: $1(x-2) + 4(y+1) - 2(z-3) = 0$ .
Expand: $x - 2 + 4y + 4 - 2z + 6 = 0$ .
Collect: $x + 4y - 2z + 8 = 0$ .

Answer:

x + 4y - 2z + 8 = 0

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Normal of the plane $3x - y + 5z = 7$ ?
2.
Are $2x + y - z = 1$ and $4x + 2y - 2z = 9$ parallel?
3.
Is $\vec{n_1} = (1,2,1)$ perpendicular to $\vec{n_2} = (1,-1,1)$ ?
4.
Plane through $(0,0,0)$ with normal $(1,1,1)$ ?

The normal is the coefficient triple, not the point

ax+by+cz=d

the normal is

(a,b,c)

. Students sometimes grab the point's coordinates as the normal — those only fix

d

. Read direction from the coefficients, position from the given point.

$d$ is found by substituting, never left at the wrong sign

After

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

, expand fully before reading

d

. A sign slip on

-bx_0

etc. flips the constant — the most common reason a correct normal still lands on the wrong option.

Concept 2 of 12

Direction cosines of the normal

Intuition

The angles a normal makes with the three axes are not free — their cosines must square-sum to 1. So if a question gives you two of the angles, the third is forced (up to sign), and that pins the normal direction completely.

Definition

If a normal $\vec{n}$ makes angles $\alpha, \beta, \gamma$ with the $X, Y, Z$ axes, its direction cosines $\cos\alpha, \cos\beta, \cos\gamma$ satisfy:

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

A normal equally inclined to all three axes has

\cos\alpha = \cos\beta = \cos\gamma

, so each

= \pm\tfrac{1}{\sqrt 3}

and the direction is

(1,1,1)

. The word acute picks the positive root.

Direction-cosine identity

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

$\alpha, \beta, \gamma$ angles the normal makes with $X, Y, Z$ axes

Diagram · direction cosines (drag to rotate)

α ≈ 49° · l = 0.66β ≈ 62° · m = 0.48γ ≈ 54° · n = 0.58

l, m, n are the cosines of the angles r makes with the x-, y-, z-axes — and the components of the unit vector along r. So l² + m² + n² = 1.00 = 1, always.

Worked example

A normal makes

45^\circ

with the X-axis,

60^\circ

with the Y-axis, and an acute angle with the Z-axis. Find a direction ratio of the normal.

Apply the identity: $\cos^2 45^\circ + \cos^2 60^\circ + \cos^2\gamma = 1$ .
$\tfrac{1}{2} + \tfrac{1}{4} + \cos^2\gamma = 1 \Rightarrow \cos^2\gamma = \tfrac{1}{4} \Rightarrow \cos\gamma = \tfrac{1}{2}$ (acute, so positive).
Direction cosines $\left(\tfrac{1}{\sqrt 2}, \tfrac{1}{2}, \tfrac{1}{2}\right)$ ; clear the $\sqrt 2$ by scaling to ratios $(\sqrt 2, 1, 1)$ or doubling to $(2, \sqrt2\cdot 1, \dots)$ — proportional triple $(2,1,1)$ .

Answer:Normal direction

\propto (2,1,1)

Practice this conceptself-check · 4 quick reps

Try it yourself

A normal is equally inclined to all three coordinate axes at an acute angle. What is each direction cosine, and what is a direction ratio of the normal?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
If two direction cosines are $\tfrac{1}{2}, \tfrac{1}{2}$ , the third (acute)?
2.
Direction ratio of a normal equally inclined to all axes?
3.
Can a normal make $45^\circ$ with both X and Y axes?
4.
$\cos\gamma$ if $\alpha = 60^\circ, \beta = 60^\circ$ (acute)?

From the bank · past-year question

Example 2Line and PlaneHARD

A vector

\vec{n}

is inclined to X-axis at

45^\circ

, Y-axis at

60^\circ

and at an acute angle to Z-axis. If

\vec{n}

is normal to a plane passing through the point

(-2,1,1)

, then equation of the plane is

[Q112 · 9th May Shift 1 · 2024]

"Acute angle" chooses the positive square root

\cos^2\gamma = \tfrac14

gives

\cos\gamma = \pm\tfrac12

. The word *acute* forces the

+

sign, so the Z-component of the normal is positive. Miss this and you may build the plane from the mirror-flipped normal.

Equally inclined means equal COSINES, not equal angles spread over 90 degrees

Equal inclination gives

(1,1,1)

, not

(1,1,0)

. All three cosines equal forces

\tfrac{1}{\sqrt3}

each — don't assume one axis drops out.

Concept 3 of 12

Planes parallel to a coordinate plane or to a given plane

Intuition

A plane parallel to the XY-plane is just

z = k

— its normal points straight up the Z-axis. More generally, a plane parallel to a given plane keeps the SAME normal, so it has the same left-hand side; only the constant changes, and the given point fixes it.

Definition

Parallel to a coordinate plane (normal along one axis):

Parallel to XY-plane: $z = k$ . Parallel to YZ-plane: $x = k$ . Parallel to ZX-plane: $y = k$ .

Parallel to a given plane $ax+by+cz = d$ : the required plane is $ax+by+cz = d'$ with the same normal $(a,b,c)$ ; substitute the given point to find $d'$ .

Same normal, new constant

ax + by + cz = d' \quad\text{where } d' = a x_0 + b y_0 + c z_0

$(a,b,c)$ normal copied from the given plane
$(x_0,y_0,z_0)$ point the new plane passes through

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

(1,4,-2)

parallel to the plane

-2x + y - 3z = 9

Keep the normal $(-2, 1, -3)$ : the plane is $-2x + y - 3z = d'$ .
Substitute $(1,4,-2)$ : $-2(1) + 4 - 3(-2) = -2 + 4 + 6 = 8$ .
So $-2x + y - 3z = 8$ , i.e. $2x - y + 3z + 8 = 0$ .

Answer:

2x - y + 3z + 8 = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

Cartesian equation of the plane through

A(7,8,6)

parallel to the XY-plane.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Plane through $(3,-1,5)$ parallel to the YZ-plane?
2.
Plane through $(0,2,9)$ parallel to the ZX-plane?
3.
Plane through $(1,1,1)$ parallel to $2x - y + z = 5$ ?
4.
Normal of any plane parallel to the XY-plane?

From the bank · past-year question

Example 3Line and PlaneEASY

The Cartesian equation of plane through

A(7,8,6)

and parallel to the XY plane is

[Q101 · 19 April Shift I · 2025]

Parallel to XY-plane is $z = k$ , not $x + y = k$

The XY-plane is

z = 0

; any plane parallel to it freezes

z

. Match the right coordinate: parallel to YZ freezes

x

, parallel to ZX freezes

y

Re-use the WHOLE normal when copying a plane

A parallel plane shares all three coefficients, signs included. Substitute the point only into the constant — do not rescale or re-sign the normal, or it stops being parallel.

Drill 1 more on planes parallel to a coordinate plane or to a given plane

Concept 4 of 12

Plane from the foot of the perpendicular from the origin

Intuition

M

is the foot of the perpendicular dropped from the origin onto a plane, then

\overrightarrow{OM}

IS the normal direction, and

M

lies on the plane. So the plane is

\vec{r}\cdot\overrightarrow{OM} = |\overrightarrow{OM}|^2

— the constant comes out as the squared length of

M

Definition

Let $M(x_0,y_0,z_0)$ be the foot of perpendicular from the origin to the plane. Then:

The normal is $\vec{n} = \overrightarrow{OM} = (x_0,y_0,z_0)$ .
The plane passes through $M$ , so the constant is $\vec{n}\cdot\overrightarrow{OM} = x_0^2 + y_0^2 + z_0^2$ .

Cartesian: $x_0 x + y_0 y + z_0 z = x_0^2 + y_0^2 + z_0^2$ . Vector: $\vec{r}\cdot(x_0\hat{i} + y_0\hat{j} + z_0\hat{k}) = x_0^2 + y_0^2 + z_0^2$ .

Plane from foot of perpendicular

\vec{r}\cdot\overrightarrow{OM} = |\overrightarrow{OM}|^2 = x_0^2 + y_0^2 + z_0^2

$M(x_0,y_0,z_0)$ foot of perpendicular from the origin
$\overrightarrow{OM}$ the normal vector to the plane

Diagram · unit normal n̂ = (a×b)/|a×b|

A plane has exactly two unit normals, ±n̂. The cross product a × b picks one by the right-hand rule; b × a gives the other. Dividing by |a × b| rescales it to length 1.

Worked example

The foot of the perpendicular from the origin to a plane is

M(2,1,-2)

. Find the vector equation of the plane.

Normal $\vec{n} = \overrightarrow{OM} = 2\hat{i} + \hat{j} - 2\hat{k}$ .
Constant $= |\overrightarrow{OM}|^2 = 2^2 + 1^2 + (-2)^2 = 4 + 1 + 4 = 9$ .
Plane: $\vec{r}\cdot(2\hat{i} + \hat{j} - 2\hat{k}) = 9$ .

Answer:

\vec{r}\cdot(2\hat{i} + \hat{j} - 2\hat{k}) = 9

Practice this conceptself-check · 4 quick reps

Try it yourself

The foot of perpendicular from the origin to a plane is

(4,-2,5)

. Find the Cartesian equation of the plane.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Foot of perpendicular $(1,2,2)$ : the plane's constant $d$ ?
2.
Foot of perpendicular $(3,0,4)$ : equation of the plane?
3.
Normal direction if the foot is $(6,2,-3)$ ?
4.
Length of the perpendicular from origin if foot is $(2,1,2)$ ?

From the bank · past-year question

Example 4Line and PlaneMODERATE

The foot of the perpendicular drawn from origin to a plane is

M(2,1,-2)

, then vector equation of the plane is

[Q141 · 15th May Shift 2 · 2023]

The constant is $|\overrightarrow{OM}|^2$ , not $|\overrightarrow{OM}|$

For foot

(2,1,-2)

the constant is

9

(the squared length), not

3

(the distance). The distance

\sqrt9 = 3

is the perpendicular *length*; the plane equation uses the squared value.

Don't move the foot to the wrong side of the equation

The form is

\vec{r}\cdot\overrightarrow{OM} = +|\overrightarrow{OM}|^2

, a POSITIVE constant. Distractors flip it to

\dots + 45 = 0

; that plane no longer passes through

M

Drill 1 more on plane from the foot of the perpendicular from the origin

Concept 5 of 12

Plane through a point with normal fixed by axis angles

Intuition

Combine the direction-cosine identity with the point-normal form. The angles (or the 'equal acute angles' phrasing) give you the normal direction; the point gives you the constant. Two foundations clicking together.

Definition

Given the angles a normal makes with the axes, recover its direction ratio via $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ , then write the plane through the point $A$ :

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

When the normal is equally inclined to all axes the direction is

(1,1,1)

, giving a plane of the form

x + y + z = k

Point-normal with angle-derived normal

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

$(a,b,c)$ normal recovered from the axis angles
$(x_0,y_0,z_0)$ the given point on the plane

Worked example

Find the plane through

(-1,1,2)

whose normal makes equal acute angles with the coordinate axes.

Equal acute angles → normal direction $(1,1,1)$ .
Point-normal: $(x+1) + (y-1) + (z-2) = 0$ .
Expand: $x + y + z - 2 = 0$ .

Answer:

x + y + z - 2 = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

A normal is inclined

45^\circ

to X,

60^\circ

to Y, acute to Z, and is normal to a plane through

(-2,1,1)

. Find the plane.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Plane through $(1,1,1)$ with normal $(1,1,1)$ ?
2.
Form of a plane whose normal is equally inclined to the axes?
3.
Plane through origin with normal $(2,1,1)$ ?
4.
Constant for $x+y+z=k$ through $(-1,1,2)$ ?

From the bank · past-year question

Example 5Line and PlaneEASY

The equation of the plane through

(-1,1,2)

whose normal makes equal acute angles with coordinate axes is

[Q126 · 12th May Shift 1 · 2024]

Clear the irrational direction cosine into a clean ratio

Direction cosines

\left(\tfrac{1}{\sqrt2}, \tfrac12, \tfrac12\right)

become the ratio

(2,1,1)

— double everything until the

\sqrt2

is gone. Writing the normal as

(1,1,1)

here is wrong; only EQUAL angles give

(1,1,1)

Put the point into the expanded form, not the angle data

The angles fix only the normal's DIRECTION; the point alone fixes the constant. Don't try to use an angle to find

k

Concept 6 of 12

Plane perpendicular to two given planes

Intuition

A plane perpendicular to two given planes must contain BOTH their normals as directions lying in it. So the required normal is perpendicular to both given normals — exactly what the cross product delivers:

\vec{n} = \vec{n_1}\times\vec{n_2}

Definition

To build a plane through a point $A$ perpendicular to planes with normals $\vec{n_1}$ and $\vec{n_2}$ :

The required normal is $\vec{n} = \vec{n_1}\times\vec{n_2}$ (perpendicular to both, so both given normals lie IN the required plane).
Then write the plane through $A$ : $\vec{n}\cdot(\vec{r} - \vec{a}) = 0$ .

This is the cross-product-normal engine — one of the two HARD workhorses of this subtopic.

Normal from two perpendicular planes

\vec{n} = \vec{n_1}\times\vec{n_2}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0

$\vec{n_1}, \vec{n_2}$ normals of the two given planes
$\vec{n}$ required normal = their cross product

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 plane through

(1,-1,2)

perpendicular to the planes

x + 2y - 2z = 4

and

3x + 2y + z = 6

Normals $\vec{n_1} = (1,2,-2)$ , $\vec{n_2} = (3,2,1)$ .
Cross product: $\vec{n} = (1,2,-2)\times(3,2,1) = (2\cdot1 - (-2)\cdot2,\; (-2)\cdot3 - 1\cdot1,\; 1\cdot2 - 2\cdot3) = (6,-7,-4)$ .
Plane through $(1,-1,2)$ : $6(x-1) - 7(y+1) - 4(z-2) = 0 \Rightarrow 6x - 7y - 4z - 5 = 0$ .

Answer:

6x - 7y - 4z - 5 = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the plane through

(1,1,1)

perpendicular to

2x + y - 2z = 5

and

3x - 6y - 2z = 7

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Which operation gives a normal perpendicular to two normals?
2.
$(1,0,0)\times(0,1,0) = ?$
3.
If $\vec{n} = (6,-7,-4)$ and the plane passes through the origin, its equation?
4.
Two given normals lie WHERE relative to the required plane?

From the bank · past-year question

Example 6Line and PlaneMODERATE

Equation of the plane passing through

(1,-1,2)

and perpendicular to the planes

x+2y-2z=4

and

3x+2y+z=6

is:

[Q121 · 14th May Shift 1 · 2024]

Cross product, not dot product, for the normal

Perpendicular-to-two-planes needs a direction perpendicular to BOTH normals — that is

\vec{n_1}\times\vec{n_2}

. A dot product gives a number, not a direction; reaching for it here is the classic wrong start.

Keep the cross-product sign and middle-term flip straight

The

\hat{j}

component carries a minus sign in the determinant expansion. A sign error there sends you to a sibling option with the middle coefficient flipped (e.g.

6x+7y\dots

instead of

6x-7y\dots

Drill 7 more on plane perpendicular to two given planes

Concept 7 of 12

Plane through a point parallel to two lines

Intuition

If a plane is parallel to two lines, both line directions lie IN the plane — so the normal is perpendicular to both directions. Same engine as before:

\vec{n} = \vec{d_1}\times\vec{d_2}

. This is the single most repeated template in the subtopic.

Definition

For a plane through a point $A$ parallel to two lines with direction vectors $\vec{d_1}, \vec{d_2}$ :

The normal is $\vec{n} = \vec{d_1}\times\vec{d_2}$ .
Plane: $\vec{n}\cdot(\vec{r} - \vec{a}) = 0$ .

Read each line's direction straight off its symmetric form $\frac{x - x_1}{p} = \frac{y - y_1}{q} = \frac{z - z_1}{s}$ : the direction is $(p, q, s)$ .

Normal from two parallel lines

\vec{n} = \vec{d_1}\times\vec{d_2}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0

$\vec{d_1}, \vec{d_2}$ direction vectors of the two lines
$\vec{a}$ the point the plane passes through

Worked example

Find the plane through

(2,-1,-3)

parallel to the lines

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

and

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

Directions $\vec{d_1} = (3,2,-4)$ , $\vec{d_2} = (2,-3,2)$ .
$\vec{n} = \vec{d_1}\times\vec{d_2} = (2\cdot2 - (-4)(-3),\; (-4)(2) - 3\cdot2,\; 3(-3) - 2\cdot2) = (-8, -14, -13)$ ; use $(8,14,13)$ .
Plane through $(2,-1,-3)$ : $8(x-2) + 14(y+1) + 13(z+3) = 0 \Rightarrow 8x + 14y + 13z + 37 = 0$ .

Answer:

8x + 14y + 13z + 37 = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the plane through

(2,1,2)

and

(1,2,1)

parallel to the line

2x = 3y,\; z = 1

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Direction of $\frac{x-1}{3} = \frac{y+2}{2} = \frac{z}{-4}$ ?
2.
If a plane is parallel to two lines, the normal is perpendicular to...?
3.
$(1,0,0)\times(0,0,1) = ?$
4.
Plane through $(0,0,0)$ parallel to lines with directions $(1,0,0),(0,1,0)$ ?

From the bank · past-year question

Example 7Line and PlaneHARD

The equation of the plane, passing through the point

(-1,2,-3)

and parallel to the lines

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

and

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

, is

[Q110 · 4th May Shift 2 · 2023]

Parallel to two LINES uses their directions, parallel to two PLANES uses their normals

Both reduce to a cross product, but read the right vectors: a line gives a direction

(p,q,s)

from its denominators; a plane gives a normal

(a,b,c)

from its coefficients. Mixing them up cross-products the wrong pair.

Read line directions from the denominators, signs included

\frac{z}{-4}

the Z-direction is

-4

, not

4

. A dropped minus on a single component changes the whole cross product.

Drill 6 more on plane through a point parallel to two lines

Concept 8 of 12

Plane through three points

Intuition

Three non-collinear points fix a plane. Build two direction vectors inside the plane from one anchor point, cross them to get the normal, then write the point-normal form — or expand the standard 3-by-3 determinant directly.

Definition

For points $A, B, C$ :

Form two in-plane vectors $\overrightarrow{AB} = \vec{b} - \vec{a}$ and $\overrightarrow{AC} = \vec{c} - \vec{a}$ .
Normal $\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}$ ; plane through $A$ .

Determinant form (equivalent):

\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

Three-point plane

\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0

$\overrightarrow{AB}, \overrightarrow{AC}$ two edges from anchor $A$
$\vec{n}$ normal = their cross product

Worked example

Find the Cartesian equation of the plane through

(3,1,1)

(1,2,3)

and

(-1,4,2)

Anchor $A(3,1,1)$ : $\overrightarrow{AB} = (-2,1,2)$ , $\overrightarrow{AC} = (-4,3,1)$ .
$\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC} = (1\cdot1 - 2\cdot3,\; 2\cdot(-4) - (-2)\cdot1,\; (-2)\cdot3 - 1\cdot(-4)) = (-5, -6, -2)$ ; use $(5,6,2)$ .
Plane through $A$ : $5(x-3) + 6(y-1) + 2(z-1) = 0 \Rightarrow 5x + 6y + 2z - 23 = 0$ .

Answer:

5x + 6y + 2z - 23 = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the plane through

(2,3,1)

and

(4,-5,3)

parallel to the X-axis.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
How many non-collinear points fix a plane?
2.
Two in-plane edges from $A$ to $B, C$ are crossed to get the...?
3.
If a plane is parallel to the X-axis, the coefficient of $x$ is?
4.
$\overrightarrow{AB}$ for $A(3,1,1), B(1,2,3)$ ?

From the bank · past-year question

Example 8Line and PlaneMODERATE

The Cartesian equation of the plane, passing through the points

(3,1,1)

(1,2,3)

and

(-1,4,2)

, is

[Q135 · 2nd May Shift 1 · 2023]

Anchor BOTH edge vectors at the same point

Use

\overrightarrow{AB}

and

\overrightarrow{AC}

(both from

A

), not

\overrightarrow{AB}

and

\overrightarrow{BC}

mixed with the wrong anchor for the point-normal step. The normal is fine either way, but the substituted point must lie on the plane.

"Parallel to an axis" kills exactly one coefficient

Parallel to X-axis sets the

x

-coefficient to

0

(the axis direction must lie in the plane, so the normal has no

\hat{i}

part). Don't also zero

y

z

Drill 2 more on plane through three points

Concept 9 of 12

Perpendicular bisector plane of a segment

Intuition

The plane that perpendicularly bisects segment

PQ

passes through the MIDPOINT of

PQ

and has

\overrightarrow{PQ}

as its normal — every point on it is equidistant from

P

and

Q

Definition

For the plane perpendicular to segment $PQ$ and passing through its midpoint:

Midpoint $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$ .
Normal $\vec{n} = \overrightarrow{PQ} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$ .
Plane: $\vec{n}\cdot(\vec{r} - \vec{m}) = 0$ .

Perpendicular bisector plane

\vec{n} = \overrightarrow{PQ}, \quad M = \tfrac{1}{2}(\vec{p} + \vec{q}), \quad \vec{n}\cdot(\vec{r} - \vec{m}) = 0

$M$ midpoint of $PQ$ — the plane passes through it
$\overrightarrow{PQ}$ segment direction = the normal

Worked example

Find the plane through the midpoint of

P(1,2,5)

and

Q(3,4,3)

and perpendicular to

PQ

Midpoint $M = (2,3,4)$ .
$\overrightarrow{PQ} = (2,2,-2)$ ; take normal $(1,1,-1)$ .
Plane: $(x-2) + (y-3) - (z-4) = 0 \Rightarrow x + y - z - 1 = 0$ .

Answer:

x + y - z - 1 = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the perpendicular bisector plane of the segment joining

(0,0,0)

and

(2,4,6)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Midpoint of $(1,2,5)$ and $(3,4,3)$ ?
2.
Normal of the perpendicular bisector plane of $PQ$ ?
3.
Points on this plane are equidistant from which two points?
4.
$\overrightarrow{PQ}$ for $P(1,2,5), Q(3,4,3)$ ?

From the bank · past-year question

Example 9Line and PlaneMODERATE

The equation of the plane passing through the midpoint of the line segment joining

P(1,2,5)

and

Q(3,4,3)

and perpendicular to it, is

[Q122 · 2nd May Shift 2 · 2023]

Pass through the MIDPOINT, not through $P$ or $Q$

The perpendicular bisector plane goes through

M

, the midpoint. Substituting

P

Q

gives a parallel plane with the wrong constant.

Simplify the normal before substituting

\overrightarrow{PQ} = (2,2,-2)

is fine, but

(1,1,-1)

is cleaner — just keep the constant consistent. Either way, the direction (signs) must match

\overrightarrow{PQ}

Concept 10 of 12

Family of planes through a line of intersection (lambda engine)

Intuition

Any plane through the intersection line of two planes

P_1 = 0

and

P_2 = 0

can be written as

P_1 + \lambda P_2 = 0

for some scalar

\lambda

. You then impose ONE extra condition — perpendicular to a coordinate plane, parallel to an axis or a line, perpendicular to a third plane, or passing through a point — to solve for

\lambda

. This single trick is the most repeated HARD shape in the subtopic.

Definition

The family of planes through the line of 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

Its normal is

(a_1 + \lambda a_2,\; b_1 + \lambda b_2,\; c_1 + \lambda c_2)

. Pin

\lambda

with the side condition:

Perpendicular to XY-plane → $z$ -coefficient $= 0$ : $c_1 + \lambda c_2 = 0$ .
Parallel to X / Y / Z-axis → the matching coefficient $= 0$ (e.g. parallel to Y-axis → $b_1 + \lambda b_2 = 0$ ).
Perpendicular to a third plane with normal $\vec{m}$ → family-normal $\cdot\, \vec{m} = 0$ .
Parallel to a line with direction $\vec{d}$ → family-normal $\cdot\, \vec{d} = 0$ .
Through a point → substitute the point.

Family of planes

P_1 + \lambda P_2 = 0, \qquad \vec{n}_\lambda = \big(a_1 + \lambda a_2,\; b_1 + \lambda b_2,\; c_1 + \lambda c_2\big)

$\lambda$ scalar fixed by the one extra condition
$\vec{n}_\lambda$ the family's normal, a function of $\lambda$

Worked example

Find the plane through the intersection of

x + y + z = 1

and

3x + 4y + 5z = 2

that is perpendicular to the XY-plane.

Family: $(x+y+z-1) + \lambda(3x+4y+5z-2) = 0$ , normal $(1+3\lambda,\; 1+4\lambda,\; 1+5\lambda)$ .
Perpendicular to XY-plane → $z$ -coefficient $= 0$ : $1 + 5\lambda = 0 \Rightarrow \lambda = -\tfrac15$ .
Substitute: $\tfrac{2}{5}x + \tfrac{1}{5}y - \tfrac{3}{5} = 0 \Rightarrow 2x + y - 3 = 0$ .

Answer:

2x + y - 3 = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the plane through the intersection of

x + y + z = 1

and

2x + 3y + 4z = 5

that is perpendicular to the plane

x - y + z = 0

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Write the family through the intersection of $P_1 = 0$ and $P_2 = 0$ .
2.
"Perpendicular to the XY-plane" sets which coefficient to 0?
3.
"Parallel to the Y-axis" sets which coefficient to 0?
4.
"Parallel to a line of direction $\vec{d}$ " imposes which equation on the normal?

From the bank · past-year question

Example 10Line and PlaneMODERATE

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

x + y + z = 1

and

3x + 4y + 5z = 2

and perpendicular to the XY-plane is

[Shift || · 2025]

Perpendicular to the XY-plane means the Z-coefficient vanishes

A plane perpendicular to the XY-plane is 'vertical' — its normal lies IN the XY-plane, so it has no

z

-part: set

c_1 + \lambda c_2 = 0

. Students often confuse this with parallel (which would set

a, b

to make the normal point along Z). Vertical → kill

z

; horizontal → keep only

z

Parallel-to-axis kills the SAME-named coefficient

Parallel to the Y-axis means

\hat{j}

lies in the plane, so the normal has no

\hat{j}

b_1 + \lambda b_2 = 0

. Don't confuse 'parallel to Y-axis' (kill

b

) with 'perpendicular to ZX-plane' even though they coincide.

Clear the fractions before matching options

Solving for

\lambda

leaves fractional coefficients like

\tfrac25 x + \tfrac15 y

. Multiply through (here by 5) to reach

2x + y - 3 = 0

; the un-cleared version matches no option.

Drill 6 more on family of planes through a line of intersection (lambda engine)

Concept 11 of 12

Intercept form, intercept triangle area and centroid

Intuition

A plane cuts the axes at

(a,0,0), (0,b,0), (0,0,c)

— the intercepts. Those three points form a triangle whose area and centroid have clean closed forms, so a question that mentions where a plane 'meets the axes' is almost always testing one of these two formulas.

Definition

Intercept form: $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$ , with axis points $A(a,0,0), B(0,b,0), C(0,0,c)$ .

**Centroid of $\triangle ABC$ :** $\left(\dfrac{a}{3}, \dfrac{b}{3}, \dfrac{c}{3}\right)$ .
**Area of $\triangle ABC$ :** $\dfrac{1}{2}\sqrt{a^2 b^2 + b^2 c^2 + c^2 a^2}$ .

Intercept triangle: centroid and area

G = \left(\tfrac{a}{3}, \tfrac{b}{3}, \tfrac{c}{3}\right), \qquad \text{Area} = \tfrac{1}{2}\sqrt{a^2 b^2 + b^2 c^2 + c^2 a^2}

$a, b, c$ intercepts on the $X, Y, Z$ axes
$G$ centroid of the triangle of intercepts

Diagram · coordinate planes & octants (drag to rotate)

Three planes (XY, YZ, ZX), each splitting space in two → 2 × 2 × 2 = 8 octants. P sits in the first octant (all coordinates positive).

Worked example

The plane

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

cuts the axes at

A, B, C

. Find the area of

\triangle ABC

Intercepts $a = 3, b = 2, c = -4$ .
Area $= \tfrac12\sqrt{a^2b^2 + b^2c^2 + c^2a^2} = \tfrac12\sqrt{36 + 64 + 144} = \tfrac12\sqrt{244}$ .
$\sqrt{244} = 2\sqrt{61}$ , so area $= \tfrac12\cdot 2\sqrt{61} = \sqrt{61}$ .

Answer:

\sqrt{61}

sq. units

Practice this conceptself-check · 4 quick reps

Try it yourself

The plane

2x + 3y + 4z = 1

meets the X, Y, Z axes at

A, B, C

. Find the centroid of

\triangle ABC

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
X-intercept of $2x + 3y + 4z = 1$ ?
2.
Centroid of intercept triangle with $a=3,b=3,c=3$ ?
3.
Area when $a=b=c=1$ ?
4.
Intercept form of $2x + 3y + 4z = 1$ ?

From the bank · past-year question

Example 11Line and PlaneMODERATE

If the plane

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

cuts the co-ordinate axes at points

A,B

and C , then the area of the triangle ABC is

[Q139 · 20 April Shift I · 2025]

Read intercepts from the form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

For

2x + 3y + 4z = 1

, the X-intercept is

\tfrac12

, NOT

2

. Divide through to make the RHS exactly

1

, then the denominators are the intercepts.

Use the squared intercepts in the area formula

Area

= \tfrac12\sqrt{a^2b^2 + b^2c^2 + c^2a^2}

— pairwise PRODUCTS of squares. A negative intercept like

c = -4

contributes

c^2 = 16

; the sign drops out, but don't forget to square it.

Drill 2 more on intercept form, intercept triangle area and centroid

Concept 12 of 12

Recovering a plane from a point and its mirror image

Intuition

If a point and its mirror image in a plane are both given, the plane is the perpendicular bisector of the segment joining them: it passes through their midpoint and has the segment as its normal. So this reduces to the perpendicular-bisector construction, run in reverse.

Definition

Given a point $P$ and its mirror image $P'$ in an unknown plane:

The plane passes through the midpoint $M = \tfrac12(P + P')$ .
Its normal is $\overrightarrow{PP'} = P' - P$ (the segment is perpendicular to the plane).

Build the plane, then test which option-point satisfies it. (The fuller treatment of finding an image or foot of perpendicular lives on the *Foot of Perpendicular, Image, and Projection* page; here we only need this reverse construction.)

Plane from point and its image

M = \tfrac12(P + P'), \quad \vec{n} = P' - P, \quad \vec{n}\cdot(\vec{r} - \vec{m}) = 0

$P, P'$ the point and its mirror image
$M$ midpoint — lies on the plane

Worked example

The mirror image of

(1,2,3)

in a plane is

\left(-\tfrac73, -\tfrac43, -\tfrac13\right)

. Find the plane, and verify it passes through

(1,-1,1)

Midpoint $M = \tfrac12\left(1 - \tfrac73,\; 2 - \tfrac43,\; 3 - \tfrac13\right) = \left(-\tfrac13, \tfrac13, \tfrac43\right)$ .
Normal $\overrightarrow{PP'} = \left(-\tfrac73 - 1,\; -\tfrac43 - 2,\; -\tfrac13 - 3\right) = \left(-\tfrac{10}{3}, -\tfrac{10}{3}, -\tfrac{10}{3}\right) \propto (1,1,1)$ .
Plane through $M$ : $(x + \tfrac13) + (y - \tfrac13) + (z - \tfrac43) = 0 \Rightarrow x + y + z = \tfrac43$ ... clearing, $x + y + z = 1$ . Check $(1,-1,1)$ : $1 - 1 + 1 = 1$ ✓.

Answer:Plane

x + y + z = 1

; it passes through

(1,-1,1)

Practice this conceptself-check · 4 quick reps

Try it yourself

The image of

(0,0,0)

in a plane is

(2,2,2)

. Find the plane.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
The plane is the ____ of the segment joining a point and its image.
2.
Normal of the plane if $P = (1,2,3), P' = (3,2,3)$ ?
3.
Midpoint of $(1,2,3)$ and $(3,4,5)$ ?
4.
Does the plane pass through $P$ or through $M$ ?

From the bank · past-year question

Example 12Line and PlaneHARD

The mirror image of the point

(1,2,3)

in a plane is

\left(-\frac{7}{3},-\frac{4}{3},-\frac{1}{3}\right)

. Thus, the point lies on this plane.

[Q116 · 11th May Shift 1 · 2024]

The normal is the segment, the plane is at the midpoint

Use

\overrightarrow{PP'}

as the normal and the MIDPOINT as the through-point — not

P

P'

. Substituting an endpoint gives a plane parallel to the true one.

Simplify the messy normal before testing option-points

\left(-\tfrac{10}{3}, -\tfrac{10}{3}, -\tfrac{10}{3}\right)

is just

(1,1,1)

. Reduce first; then substituting each option-point to find which lies on

x + y + z = 1

is trivial arithmetic.

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 (12)

Equation of a plane and its normal
The three equivalent forms
$ax + by + cz = d \;\Longleftrightarrow\; \vec{r}\cdot\vec{n} = d \;\Longleftrightarrow\; \vec{n}\cdot(\vec{r} - \vec{a}) = 0$
Direction cosines of the normal
Direction-cosine identity
$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$
Planes parallel to a coordinate plane or to a given plane
Same normal, new constant
$ax + by + cz = d' \quad\text{where } d' = a x_0 + b y_0 + c z_0$
Plane from the foot of the perpendicular from the origin
Plane from foot of perpendicular
$\vec{r}\cdot\overrightarrow{OM} = |\overrightarrow{OM}|^2 = x_0^2 + y_0^2 + z_0^2$
Plane through a point with normal fixed by axis angles
Point-normal with angle-derived normal
$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$
Plane perpendicular to two given planes
Normal from two perpendicular planes
$\vec{n} = \vec{n_1}\times\vec{n_2}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0$
Plane through a point parallel to two lines
Normal from two parallel lines
$\vec{n} = \vec{d_1}\times\vec{d_2}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0$
Plane through three points
Three-point plane
$\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0$
Perpendicular bisector plane of a segment
Perpendicular bisector plane
$\vec{n} = \overrightarrow{PQ}, \quad M = \tfrac{1}{2}(\vec{p} + \vec{q}), \quad \vec{n}\cdot(\vec{r} - \vec{m}) = 0$
Family of planes through a line of intersection (lambda engine)
Family of planes
$P_1 + \lambda P_2 = 0, \qquad \vec{n}_\lambda = \big(a_1 + \lambda a_2,\; b_1 + \lambda b_2,\; c_1 + \lambda c_2\big)$
Intercept form, intercept triangle area and centroid
Intercept triangle: centroid and area
$G = \left(\tfrac{a}{3}, \tfrac{b}{3}, \tfrac{c}{3}\right), \qquad \text{Area} = \tfrac{1}{2}\sqrt{a^2 b^2 + b^2 c^2 + c^2 a^2}$
Recovering a plane from a point and its mirror image
Plane from point and its image
$M = \tfrac12(P + P'), \quad \vec{n} = P' - P, \quad \vec{n}\cdot(\vec{r} - \vec{m}) = 0$

Watch out for (25)

The normal is the coefficient triple, not the point
→ Equation of a plane and its normal
$d$ is found by substituting, never left at the wrong sign
→ Equation of a plane and its normal
"Acute angle" chooses the positive square root
→ Direction cosines of the normal
Equally inclined means equal COSINES, not equal angles spread over 90 degrees
→ Direction cosines of the normal
Parallel to XY-plane is $z = k$ , not $x + y = k$
→ Planes parallel to a coordinate plane or to a given plane
Re-use the WHOLE normal when copying a plane
→ Planes parallel to a coordinate plane or to a given plane
The constant is $|\overrightarrow{OM}|^2$ , not $|\overrightarrow{OM}|$
→ Plane from the foot of the perpendicular from the origin
Don't move the foot to the wrong side of the equation
→ Plane from the foot of the perpendicular from the origin
Clear the irrational direction cosine into a clean ratio
→ Plane through a point with normal fixed by axis angles
Put the point into the expanded form, not the angle data
→ Plane through a point with normal fixed by axis angles
Cross product, not dot product, for the normal
→ Plane perpendicular to two given planes
Keep the cross-product sign and middle-term flip straight
→ Plane perpendicular to two given planes
Parallel to two LINES uses their directions, parallel to two PLANES uses their normals
→ Plane through a point parallel to two lines
Read line directions from the denominators, signs included
→ Plane through a point parallel to two lines
Anchor BOTH edge vectors at the same point
→ Plane through three points
"Parallel to an axis" kills exactly one coefficient
→ Plane through three points
Pass through the MIDPOINT, not through $P$ or $Q$
→ Perpendicular bisector plane of a segment
Simplify the normal before substituting
→ Perpendicular bisector plane of a segment
Perpendicular to the XY-plane means the Z-coefficient vanishes
→ Family of planes through a line of intersection (lambda engine)
Parallel-to-axis kills the SAME-named coefficient
→ Family of planes through a line of intersection (lambda engine)
Clear the fractions before matching options
→ Family of planes through a line of intersection (lambda engine)
Read intercepts from the form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
→ Intercept form, intercept triangle area and centroid
Use the squared intercepts in the area formula
→ Intercept form, intercept triangle area and centroid
The normal is the segment, the plane is at the midpoint
→ Recovering a plane from a point and its mirror image
Simplify the messy normal before testing option-points
→ Recovering a plane from a point and its mirror image

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Line and PlaneEASY

What will be the equation of the plane passing through a point

(1,4,-2)

and parallel to the given plane

-2x + y - 3z = 9

[Q134 · May Shift 1 · 2021]

Example 2Line and PlaneEASY

The foot of the perpendicular drawn from the origin to the plane is

(4,-2,5)

, then the Cartesian equation of the plane is

[Q116 · 9th May Shift 1 · 2024]

Example 3Line and PlaneHARD

Equation of the plane containing the straight line

\frac{x}{3}=\frac{y}{2}=\frac{z}{4}

and perpendicular to the plane containing the straight lines

\frac{x}{4}=\frac{y}{3}=\frac{z}{2}

and

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

[Q121 · 3rd May 2nd Shift · 2023]

Example 4Line and PlaneHARD

The equation of the plane through the point

(2,-1,-3)

and parallel to the lines

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

and

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

[Q134 · 10th May Shift 2 · 2023]

Example 5Line and PlaneMODERATE

The equation of the plane passing through the points

(2,3,1)

(4,-5,3)

and parallel to X-axis is

[Q148 · Shift 1 · 2022]

Drill every past-year question on this subtopic

36 questions from the bank — paginated, with cart and Word-export support.

Drill the 36 questions

Drill every past-year question on this subtopic

Related notes