The Equation of a Plane Revision Notes for AQA A-Level Further Maths

The Equation of a Plane

What is a plane?

A plane is a two-dimensional surface that extends infinitely in all directions. Think of it as a flat sheet that goes on forever in space. In three-dimensional geometry, we can describe planes using different mathematical forms.

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Understanding planes is fundamental to 3D geometry - they represent flat surfaces in space, similar to how lines represent one-dimensional objects in 2D geometry.

Vector equation of a plane

Defining a plane using position vectors

To define a plane, you need:

A point on the plane with position vector a
Two non-parallel vectors b and c that lie in the plane

Any other vector on the plane can be described as a combination of b and c. This means any point on the plane has position vector:

$\mathbf{r} = \mathbf{a} + s\mathbf{b} + t\mathbf{c}$

where $s$ and $t$ are scalar parameters that can take any real values.

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Key Definition: The vector equation of a plane containing point with position vector a and non-parallel vectors b and c is:

$\mathbf{r} = \mathbf{a} + s\mathbf{b} + t\mathbf{c}$

This equation describes every point on the plane by varying the parameters $s$ and $t$ .

Column vector form

You can also write the vector equation in column vector form:

$\mathbf{r} = \begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix} + s\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} + t\begin{pmatrix} x_3 \\ y_3 \\ z_3 \end{pmatrix}$

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This form is particularly useful when working with coordinates directly, as it makes substitution and calculation more straightforward.

Worked example 1: Finding the vector equation

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Worked Example: Finding the Vector Equation of a Plane

Question: Find the equation of the plane containing the points $(1, 4, -2)$ , $(0, 3, 2)$ , and $(-5, 0, 3)$ .

Solution:

First, choose one point as your reference. Let $\mathbf{a} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}$ .

Now find two vectors in the plane by subtracting position vectors:

$\mathbf{b} = (0\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) - (\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}) = -\mathbf{i} - \mathbf{j} + 4\mathbf{k}$
$\mathbf{c} = (-5\mathbf{i} + 0\mathbf{j} + 3\mathbf{k}) - (\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}) = -6\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}$

The vector equation is: $\mathbf{r} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k} + s(-\mathbf{i} - \mathbf{j} + 4\mathbf{k}) + t(-6\mathbf{i} - 4\mathbf{j} + 5\mathbf{k})$

In column form: $\mathbf{r} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} + s\begin{pmatrix} -1 \\ -1 \\ 4 \end{pmatrix} + t\begin{pmatrix} -6 \\ -4 \\ 5 \end{pmatrix}$

Note: You can use any of the three points as a, and any combination of position vectors to create b and c, as long as they are not parallel.

Scalar product form

The normal vector

Consider a plane containing point with position vector a and with a perpendicular vector n. This perpendicular vector is called the normal to the plane.

Any vector on the plane (such as $\mathbf{r} - \mathbf{a}$ ) will be perpendicular to the normal n.

Using the definition of scalar product: $(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0$

Expanding the brackets: $\mathbf{r} \cdot \mathbf{n} - \mathbf{a} \cdot \mathbf{n} = 0$

Rearranging: $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$

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Key Definition: The scalar product equation of a plane perpendicular to vector n and passing through point with position vector a is:

$\mathbf{r} \cdot \mathbf{n} = p$

where $p = \mathbf{a} \cdot \mathbf{n}$

This form is particularly powerful because it uses the normal vector, which is perpendicular to the plane.

Finding the normal vector

If you know two vectors a and b that lie in the plane, you can find the normal vector using the vector cross product:

$\mathbf{n} = \mathbf{a} \times \mathbf{b}$

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The cross product gives a vector perpendicular to both a and b, hence perpendicular to the plane. This is one of the most important applications of the cross product in 3D geometry.

Cartesian form

Converting to Cartesian form

You can convert the scalar product form into Cartesian form by writing $\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ and calculating the scalar product.

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Key Definition: The Cartesian equation of a plane is:

$ax + by + cz + d = 0$

where the vector $a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$ is the normal (perpendicular) to the plane.

This is often the most convenient form for calculations and checking if points lie on the plane.

Worked example 2: Finding scalar product and Cartesian forms

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Worked Example: Converting Between Plane Equation Forms

Question: The plane $\Pi$ contains vectors $\mathbf{a} = \mathbf{i} - \mathbf{j} + 3\mathbf{k}$ and $\mathbf{b} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ , and passes through point $(1, 0, -2)$ . Find the equation of $\Pi$ in:

a) Scalar product form
b) Cartesian form

Solution:

Part a) Scalar product form

First, find the normal vector using the cross product:

$\mathbf{n} = \mathbf{a} \times \mathbf{b}$

Calculate the determinant: $\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 3 \\ 3 & 1 & -2 \end{vmatrix}$

$= \mathbf{i}((-1)(-2) - (3)(1)) - \mathbf{j}((1)(-2) - (3)(3)) + \mathbf{k}((1)(1) - (-1)(3))$

$= \mathbf{i}(2 - 3) - \mathbf{j}(-2 - 9) + \mathbf{k}(1 + 3)$

$= -\mathbf{i} + 11\mathbf{j} + 4\mathbf{k}$

Now find the value of $p$ using point $(1, 0, -2)$ :

Position vector: $\mathbf{a} = \mathbf{i} + 0\mathbf{j} - 2\mathbf{k}$

$p = \mathbf{a} \cdot \mathbf{n} = (\mathbf{i} - 2\mathbf{k}) \cdot (-\mathbf{i} + 11\mathbf{j} + 4\mathbf{k})$

$= -1 + 0 - 8 = -9$

Therefore, the scalar product form is: $\mathbf{r} \cdot (-\mathbf{i} + 11\mathbf{j} + 4\mathbf{k}) = -9$

Part b) Cartesian form

Replace $\mathbf{r}$ with $x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ :

$(x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) \cdot (-\mathbf{i} + 11\mathbf{j} + 4\mathbf{k}) = -9$

Calculate the scalar product: $-x + 11y + 4z = -9$

Rearranging: $-x + 11y + 4z + 9 = 0$

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Exam Tip: Be careful with negative signs when calculating determinants and scalar products. Check your working carefully, especially the middle term in the cross product calculation.

Checking if two planes are the same

The vector equation of a plane is not unique because you can choose any two vectors and any point on the plane. However, the scalar product and Cartesian forms will simplify to a unique equation.

Strategy for checking if planes are identical

To check whether two equations represent the same plane:

Check if the normal vectors are parallel - if they're not parallel, the planes are different
Choose a point that you know is on one plane and check if it also lies on the other plane
Conclude whether the planes are the same, parallel, or intersecting

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This two-step verification process is essential: parallel normals tell you the planes might be the same or parallel, and checking a common point confirms they are identical.

Worked example 3: Proving planes are identical

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Worked Example: Verifying Two Planes are Identical

Question: The planes $\Pi_1$ and $\Pi_2$ have equations $\mathbf{r} \cdot (\mathbf{i} - 6\mathbf{j} + 2\mathbf{k}) = 1$ and $\mathbf{r} = \mathbf{i} + \mathbf{j} + 3\mathbf{k} + s(\mathbf{j} + 3\mathbf{k}) + t(2\mathbf{i} - \mathbf{k})$ respectively. Show that $\Pi_1$ and $\Pi_2$ are the same plane.

Solution:

Step 1: Calculate the normal vector to $\Pi_2$ using the cross product:

$\mathbf{n} = (\mathbf{j} + 3\mathbf{k}) \times (2\mathbf{i} - \mathbf{k})$

$= -\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}$

$= -(\mathbf{i} - 6\mathbf{j} + 2\mathbf{k})$

The normal vectors are parallel (one is $-1$ times the other), so the planes are either the same or parallel.

Step 2: Check if point $(1, 1, 3)$ (from $\Pi_2$ when $s = t = 0$ ) lies on $\Pi_1$ :

$(\mathbf{i} + \mathbf{j} + 3\mathbf{k}) \cdot (\mathbf{i} - 6\mathbf{j} + 2\mathbf{k}) = 1 - 6 + 6 = 1$ ✓

Step 3: Since the normals are parallel and they share a point, $\Pi_1$ and $\Pi_2$ represent the same plane.

Angles between planes and lines

Angle between a plane and a line

Consider the intersection of plane $\mathbf{r} \cdot \mathbf{n} = p$ and line $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$ .

The acute angle $\alpha$ between the vectors n and b is given by:

$\cos \alpha = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$

Since $\theta = 90° - \alpha$ , we have $\sin \theta = \cos \alpha$

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Key Formula: The acute angle θ between plane $\mathbf{r} \cdot \mathbf{n} = p$ and line $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$ is:

$\sin \theta = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$

Notice that we use sin for line-plane angles, not cos!

Angle between two planes

For two planes $\mathbf{r} \cdot \mathbf{n}_1 = p_1$ and $\mathbf{r} \cdot \mathbf{n}_2 = p_2$ , the angle between the planes is the same as the angle between their normal vectors.

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Key Formula: The acute angle θ between planes with normal vectors n₁ and n₂ is:

$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$

Notice that we use cos for plane-plane angles, not sin!

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Use the absolute value to ensure you get the acute angle. If the question asks for the obtuse angle, calculate $180° - \theta$ .

Worked example 4: Obtuse angle between two planes

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Worked Example: Finding the Obtuse Angle Between Planes

Question: Calculate the obtuse angle between planes $\mathbf{r} \cdot (13\mathbf{i} - 9\mathbf{j} + 5\mathbf{k}) = 8$ and $\mathbf{r} \cdot (2\mathbf{j} + 3\mathbf{k}) = 4$ .

Solution:

Step 1: Identify the normal vectors:

$\mathbf{n}_1 = 13\mathbf{i} - 9\mathbf{j} + 5\mathbf{k}$
$\mathbf{n}_2 = 2\mathbf{j} + 3\mathbf{k}$

Step 2: Calculate the dot product: $\mathbf{n}_1 \cdot \mathbf{n}_2 = (13)(0) + (-9)(2) + (5)(3) = 0 - 18 + 15 = -3$

Step 3: Calculate the magnitudes: $|\mathbf{n}_1| = \sqrt{13^2 + (-9)^2 + 5^2} = \sqrt{169 + 81 + 25} = \sqrt{275}$ $|\mathbf{n}_2| = \sqrt{0^2 + 2^2 + 3^2} = \sqrt{13}$

Step 4: Apply the formula: $\cos \theta = \frac{|-3|}{\sqrt{275}\sqrt{13}} = \frac{3}{\sqrt{275 \times 13}}$

$\theta = \arccos\left(\frac{3}{\sqrt{3575}}\right) = 87.1°$

Step 5: Since we need the obtuse angle: $180° - 87.1° = 92.9°$

Worked example 5: Angle between line and plane

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Worked Example: Finding the Angle Between a Line and Plane

Question: Calculate the acute angle between plane $2x + 3y - z = 1$ and line $\frac{x}{4} = \frac{y+1}{-2} = \frac{z-2}{7}$ .

Solution:

Step 1: Identify the normal vector and direction vector:

Normal to plane: $\mathbf{n} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}$
Direction vector of line: $\mathbf{b} = 4\mathbf{i} - 2\mathbf{j} + 7\mathbf{k}$

Step 2: Calculate the dot product: $\mathbf{b} \cdot \mathbf{n} = (4)(2) + (-2)(3) + (7)(-1) = 8 - 6 - 7 = -5$

Step 3: Calculate the magnitudes: $|\mathbf{b}| = \sqrt{4^2 + (-2)^2 + 7^2} = \sqrt{16 + 4 + 49} = \sqrt{69}$ $|\mathbf{n}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$

Step 4: Apply the formula: $\sin \theta = \frac{|-5|}{\sqrt{69}\sqrt{14}} = \frac{5}{\sqrt{966}}$

$\theta = \arcsin\left(\frac{5}{\sqrt{966}}\right) = 9.26°$

Strategy summary

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Key Strategy: Calculating Angles Between Planes or Between a Line and Plane

Identify the normal vector(s) - from the plane equation(s)
Identify the direction vector - if working with a line
Use $\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$ to find the acute angle between two planes
Use $\sin \theta = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$ to find the acute angle between line and plane
Subtract from 180° if an obtuse angle is required

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Critical Distinction: Remember that for a line-plane angle you use sin, but for a plane-plane angle you use cos. This is because of the geometry involved - the angle between the line and plane is complementary to the angle between the line's direction vector and the plane's normal.

Common exam traps

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Watch Out for These Common Mistakes:

Forgetting absolute values - angles should be positive, so use absolute values in the formulas
Confusing sin and cos - line-plane uses sin, plane-plane uses cos
Sign errors in cross products - be especially careful with the middle term when calculating determinants
Using wrong point - when converting to scalar product form, make sure you use a point that actually lies on the plane
Not checking if vectors are parallel - when finding the vector equation, ensure b and c are not parallel

Remember!

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Key Points to Remember:

A plane is defined by a point and two non-parallel vectors: $\mathbf{r} = \mathbf{a} + s\mathbf{b} + t\mathbf{c}$
The normal vector is perpendicular to the plane and can be found using the cross product: $\mathbf{n} = \mathbf{a} \times \mathbf{b}$
Scalar product form: $\mathbf{r} \cdot \mathbf{n} = p$ where $p = \mathbf{a} \cdot \mathbf{n}$
Cartesian form: $ax + by + cz + d = 0$ where $a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$ is the normal
For angles: use sin for line-plane, cos for plane-plane, and always use absolute values to get positive angles

The Equation of a Plane (AQA A-Level Further Maths): Revision Notes