Three-Dimensional Geometry: Planes notes: ISC Class 12 Mathematics

1. Vector and Cartesian equations of a plane

Through the point $\vec{a}$ with normal $\vec{n}$ : $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$ , i.e. $\vec{r} \cdot \vec{n} = d$ with $d = \vec{a} \cdot \vec{n}$ .
Cartesian form $ax + by + cz = d$ : the coefficients $a, b, c$ are direction ratios of the normal.
A point lies on the plane exactly when its coordinates satisfy the equation.
Parallel planes have proportional normals; a plane through the origin has $d = 0$ .

Every line in the plane through $Q$ is perpendicular to the normal $\vec{n}$ (sketch, not to scale).

One plane, several ways to write it.
Form	Equation	What it tells you
Vector	$\vec{r} \cdot \vec{n} = d$	$\vec{n}$ is a normal
Through $\vec{a}$ with normal $\vec{n}$	$(\vec{r} - \vec{a}) \cdot \vec{n} = 0$	$d = \vec{a} \cdot \vec{n}$
Cartesian	$ax + by + cz = d$	$a, b, c$ are direction ratios of the normal
One-point	$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$	through $(x_1, y_1, z_1)$

Point and normal give the whole plane.
Step	Working
Constant	$\vec{a} \cdot \vec{n} = 2 - 2 - 6 = -6$
Vector form	$\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = -6$
Cartesian form	$2x + y - 2z = -6$
Check the point	$2 - 2 - 6 = -6$

Formulas

Plane through a point
$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$

Key terms

Normal: A vector perpendicular to every line lying in the plane.

2. Plane through three points

Through non-collinear $A$ , $B$ , $C$ : $\vec{n} = \vec{AB} \times \vec{AC}$ , and $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$ .
Equivalently $(\vec{r} - \vec{a}) \cdot [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})] = 0$ .
Always substitute all three points into the answer.

Three points that are not collinear fix a plane; $\vec{AB} \times \vec{AC}$ is a normal (sketch, not to scale).

Cross two edges, then use one point.
Step	Working
Two edges	$\vec{AB} = \hat{i} + \hat{j} - \hat{k},\ \vec{AC} = -\hat{i} + 3\hat{j} + \hat{k}$
Normal	$\vec{AB} \times \vec{AC} = 4\hat{i} + 4\hat{k} \parallel \hat{i} + \hat{k}$
Plane	$x + z = 2$
Check $B$ and $C$	$2 + 0 = 2,\ 0 + 2 = 2$

Formulas

Normal from three points
$\vec{n} = \vec{AB} \times \vec{AC}$

Key terms

Non-collinear points: Points not on one line; three such points fix exactly one plane.

3. One-point, intercept and normal forms

One-point form: $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$ .
Intercept form: $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$ for intercepts $a$ , $b$ , $c$ (divide so the right side is $1$ ).
Normal form: $lx + my + nz = p$ (vector form $\vec{r} \cdot \hat{n} = p$ ), where $l, m, n$ are direction cosines of the normal and $p \ge 0$ is the distance from the origin. Divide by $|\vec{n}|$ , or by $-|\vec{n}|$ if the constant is negative.

$4x + 3y + 6z = 12$ is $\dfrac{x}{3} + \dfrac{y}{4} + \dfrac{z}{2} = 1$ (sketch, not to scale).

Make the right side $1$, then read the denominators.
Step	Working
Divide by $20$	$\dfrac{x}{4} + \dfrac{y}{-10} + \dfrac{z}{2} = 1$
Intercepts	$4$ , $-10$ , $2$
Points	$(4, 0, 0)$ , $(0, -10, 0)$ , $(0, 0, 2)$

Normal form: $p$ is never negative.
Step	Rule
Length of the normal	$\|\vec{n}\| = \sqrt{a^2 + b^2 + c^2}$
Divide	divide the equation by $\|\vec{n}\|$ ; use $-\|\vec{n}\|$ if $d < 0$
Result	$lx + my + nz = p$ with $p \ge 0$
Meaning	$l, m, n$ : direction cosines of the normal; $p$ : distance from the origin

Divide by $-7$ because the constant is negative.
Step	Working
Rewrite	$2x + 6y - 3z = -21$
Length	$\sqrt{4 + 36 + 9} = 7$
Divide by $-7$	$-\dfrac27x - \dfrac67y + \dfrac37z = 3$
Read	direction cosines $-\dfrac27, -\dfrac67, \dfrac37$ ; $p = 3$

Spot these at a glance.
Plane	Equation	Feature
Through the origin	$ax + by + cz = 0$	$d = 0$
Parallel to the $yz$ -plane	$x = k$	normal along the $x$ -axis
Parallel to the $x$ -axis	$by + cz = d$	no $x$ -term
Parallel to $ax + by + cz = d$	$ax + by + cz = k$	same normal

Formulas

Intercept form
$\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$

Key terms

Intercept: The coordinate at which a plane cuts an axis.
Normal form: An equation of the plane with unit normal and non-negative constant.

4. Distance of a point from a plane

From $(x_1, y_1, z_1)$ to $ax + by + cz = d$ : $\dfrac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}$ ; vector form $\dfrac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}$ .
Parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ (same coefficients): $\dfrac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$ .
Foot of the perpendicular from $P$ : where the line through $P$ along $\vec{n}$ meets the plane. Image: $2F - P$ .

$PF$ runs along the normal; $PF = \dfrac{|\vec{n} \cdot \vec{p} - d|}{|\vec{n}|}$ (sketch, not to scale).

Substitute, subtract, divide by $|\vec{n}|$.
Distance	Formula
Point $(x_1, y_1, z_1)$ from $ax + by + cz = d$	$\dfrac{\|ax_1 + by_1 + cz_1 - d\|}{\sqrt{a^2 + b^2 + c^2}}$
Point $\vec{a}$ from $\vec{r} \cdot \vec{n} = d$	$\dfrac{\|\vec{a} \cdot \vec{n} - d\|}{\|\vec{n}\|}$
Origin from $ax + by + cz = d$	$\dfrac{\|d\|}{\sqrt{a^2 + b^2 + c^2}}$
Parallel planes $ax + by + cz = d_1, d_2$	$\dfrac{\|d_1 - d_2\|}{\sqrt{a^2 + b^2 + c^2}}$ (same $a, b, c$ )

The modulus makes the distance positive.
Step	Working
Substitute	$6(2) + 2(-3) - 3(4) - 8 = -14$
Length	$\sqrt{36 + 4 + 9} = 7$
Distance	$\dfrac{\|-14\|}{7} = 2$

Make the left sides identical first.
Step	Working
Same coefficients	$2x - 4y + 4z = -10 \Rightarrow x - 2y + 2z = -5$
Length	$\sqrt{1 + 4 + 4} = 3$
Distance	$\dfrac{\|4 - (-5)\|}{3} = 3$
Check with a point	$(4, 0, 0) \text{ from the second plane: } \dfrac{\|8 + 10\|}{6} = 3$

Image of $P$: $P' = 2F - P$ (sketch, not to scale).

Formulas

Point to plane
$d = \dfrac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}$

Key terms

Foot of the perpendicular: The point of the plane nearest to the given point.

5. A line and a plane: meeting point, parallel, lying in

To find where $\vec{r} = \vec{a} + \lambda\vec{b}$ meets $\vec{r} \cdot \vec{n} = d$ : substitute the general point, solve for $\lambda$ , put it back, and check.
If $\vec{b} \cdot \vec{n} = 0$ the line is parallel to the plane: it lies in the plane if one of its points satisfies the equation, and never meets it otherwise.
If $\vec{b} \parallel \vec{n}$ the line is perpendicular to the plane.

A line that is not parallel to a plane meets it at exactly one point (sketch, not to scale).

Five steps to the meeting point.

General point, substitute, solve, check.
Step	Working
General point	$(1 + 2\lambda, -1 + \lambda, 2 - \lambda)$
Substitute	$(1 + 2\lambda) + (-1 + \lambda) + (2 - \lambda) = 2 + 2\lambda = 8$
Solve	$\lambda = 3$
Point	$(7, 2, -1)$ ; check $7 + 2 - 1 = 8$

Meets, parallel, or lies in?

Formulas

Meeting parameter
$\lambda = \dfrac{d - \vec{a} \cdot \vec{n}}{\vec{b} \cdot \vec{n}}$

Key terms

Line parallel to a plane: A line whose direction is perpendicular to the normal: $\vec{b} \cdot \vec{n} = 0$ .

6. Angles

Between two planes: $\cos\theta = \dfrac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1|\,|\vec{n}_2|}$ . Perpendicular planes: $a_1a_2 + b_1b_2 + c_1c_2 = 0$ ; parallel planes: proportional normals.
Between a line and a plane: $\sin\phi = \dfrac{|\vec{b} \cdot \vec{n}|}{|\vec{b}|\,|\vec{n}|}$ (the complement of the angle with the normal).

Planes: cosine. Line and plane: sine.
Angle	Formula
Between planes with normals $\vec{n}_1$ , $\vec{n}_2$	$\cos\theta = \dfrac{\|\vec{n}_1 \cdot \vec{n}_2\|}{\|\vec{n}_1\|\,\|\vec{n}_2\|}$
Line with direction $\vec{b}$ and plane with normal $\vec{n}$	$\sin\phi = \dfrac{\|\vec{b} \cdot \vec{n}\|}{\|\vec{b}\|\,\|\vec{n}\|}$
Perpendicular planes	$\vec{n}_1 \cdot \vec{n}_2 = 0$
Line parallel to plane	$\vec{b} \cdot \vec{n} = 0$
Line perpendicular to plane	$\vec{b} \parallel \vec{n}$

The angle between planes is the angle between normals.
Step	Working
Normals	$2, -1, 1$ and $1, 1, 2$
Dot product	$2 - 1 + 2 = 3$
Lengths	$\sqrt6 \text{ and } \sqrt6$
Angle	$\cos\theta = \dfrac36 = \dfrac12,\ \theta = 60^\circ$

Sine, not cosine, for a line and a plane.
Step	Working
Direction and normal	$1, 1, 2$ and $2, -1, 1$
Dot product	$2 - 1 + 2 = 3$
Lengths	$\sqrt6 \text{ and } \sqrt6$
Angle	$\sin\phi = \dfrac36 = \dfrac12,\ \phi = 30^\circ$

Formulas

Line and plane
$\sin\phi = \dfrac{|\vec{b} \cdot \vec{n}|}{|\vec{b}|\,|\vec{n}|}$

Key terms

Angle between planes: The angle between their normals.

Exam tips

How ISC marks are awarded

Plane through a point (2 marks): normal and constant, then the equation.
Plane through three points (3 marks): two edge vectors, normal by cross product, equation with a check.
Distance (2–3 marks): substitution, length of the normal, answer with modulus.
Meeting point (3 marks): general point, $\lambda$ , point and check.
Angle (2 marks): dot product and lengths, then the angle.

Common mistakes

Using the parallel-planes formula before making the coefficients equal.
Dropping the modulus, so that a distance comes out negative.
Using cosine for the angle between a line and a plane.
A negative $p$ in normal form.
Reading the constant $d$ as part of the normal.

1. Vector and Cartesian equations of a plane

Formulas

Key terms

2. Plane through three points

Formulas

Key terms

3. One-point, intercept and normal forms

Formulas

Key terms

4. Distance of a point from a plane

Formulas

Key terms

5. A line and a plane: meeting point, parallel, lying in

Formulas

Key terms

6. Angles

Formulas

Key terms

Exam tips

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