6.1.3 Angle between Lines

Introduction

The scalar (dot) product of two vectors is a fundamental tool in 3D geometry. It is used to:

Find the angle between two vectors.
Express the equation of a plane.
Calculate angles between lines, between planes, and between a line and a plane. The scalar product formula is:

\mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta

where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$ .

Key Applications of the Scalar Product

Angle Between Two Lines

The angle between two lines is the angle between their direction vectors.

Let $\mathbf{b}_1$ and $\mathbf{b}_2$ be the direction vectors of two lines:

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

Equation of a Plane

The equation of a plane can be expressed as:

\mathbf{r} \times \mathbf{n} = k

where:

$\mathbf{r}$ is the position vector of any point on the plane.
$\mathbf{n}$ is the normal vector to the plane.
$k$ is a constant determined by substituting a point $\mathbf{r}_0$ into $\mathbf{n} \times \mathbf{r}_0 = k$

Angle Between Two Planes

The angle between two planes is the angle between their normal vectors.

Let $\mathbf{n}_1$ and $\mathbf{n}_2$ be the normal vectors of two planes:

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

Angle Between a Line and a Plane

The angle between a line and a plane is the complement of the angle between the line's direction vector and the plane's normal vector.

If $\mathbf{b}$ is the direction vector of the line and $\mathbf{n}$ is the plane's normal vector:

\sin \theta = \frac{\mathbf{b} \times \mathbf{n}}{|\mathbf{b}||\mathbf{n}|}

where $\theta$ is the angle between the line and the plane.

Worked Examples

lightbulbExample

Example 1: Find the Angle Between Two Lines

Find the angle between the lines:

\mathbf{r}_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix}

\mathbf{r}_2 = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 3 \\ 4 \end{pmatrix}

Step 1: Extract direction vectors:

\mathbf{b}_1 = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix}

\mathbf{b}_2 = \begin{pmatrix} -1 \\ 3 \\ 4 \end{pmatrix}

Step 2: Compute the dot product:

\mathbf{b}_1 \times \mathbf{b}_2 = (4)(-1) + (-2)(3) + (5)(4)

= -4 - 6 + 20 = :highlight[10]

Step 3: Find magnitudes:

|\mathbf{b}_1| = \sqrt{4^2 + (-2)^2 + 5^2} = \sqrt{16 + 4 + 25} = :highlight[\sqrt{45}]

|\mathbf{b}_2| = \sqrt{(-1)^2 + 3^2 + 4^2} = \sqrt{1 + 9 + 16} = :highlight[\sqrt{26}]

Step 4: Calculate $\cos \theta$ :

\cos \theta = \frac{\mathbf{b}_1 \times \mathbf{b}_2}{|\mathbf{b}_1||\mathbf{b}_2|} = \frac{10}{\sqrt{45} \times \sqrt{26}} = \frac{10}{\sqrt{1170}}

Step 5: Find $\theta$ :

\theta = \arccos\left(\frac{10}{\sqrt{1170}}\right)

lightbulbExample

Example 2: Find the Equation of a Plane

Find the equation of the plane passing through (1, 2, 3) with normal vector

\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}.

Step 1: Substitute into $\mathbf{r} \times \mathbf{n} = k$

Let $\mathbf{r}_0 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$

k = \mathbf{n} \times \mathbf{r}_0 = (2)(1) + (-1)(2) + (4)(3)

= 2 - 2 + 12 = :highlight[12]

Step 2: Write the equation:

\mathbf{r} \times \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = 12

or in Cartesian form:

:success[2x - y + 4z = 12]

lightbulbExample

Example 3: Angle Between Two Planes

Find the angle between the planes:

x - y + 3z = 5, \quad 4x + y - z = 7

Step 1: Extract normal vectors:

\mathbf{n}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}

\mathbf{n}_2 = \begin{pmatrix} 4 \\ 1 \\ -1 \end{pmatrix}

Step 2: Compute the dot product:

\mathbf{n}_1 \times \mathbf{n}_2 = (2)(4) + (-1)(1) + (3)(-1)

= 8 - 1 - 3 = :highlight[4]

Step 3: Find magnitudes:

|\mathbf{n}_1| = \sqrt{2^2 + (-1)^2 + 3^2}

= \sqrt{4 + 1 + 9} = :highlight[\sqrt{14}]

|\mathbf{n}_2| = \sqrt{4^2 + 1^2 + (-1)^2}

= \sqrt{16 + 1 + 1} = :highlight[\sqrt{18}]

Step 4: Calculate $\cos \theta$ :

\cos \theta = \frac{\mathbf{n}_1 \times \mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{4}{\sqrt{14} \times \sqrt{18}}

= \frac{4}{\sqrt{252}} = \frac{4}{6\sqrt{7}} = :highlight[\frac{2}{3\sqrt{7}}]

Step 5: Find $\theta$ :

\theta = \arccos\left(\frac{2}{3\sqrt{7}}\right)

Note Summary

infoNote

Common Mistakes:

Incorrectly applying the scalar product: Always ensure the vectors are correctly substituted into $\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos \theta$
Mixing up line and plane vectors: Use direction vectors for lines and normal vectors for planes.
Neglecting to normalize vectors: Forgetting to divide by magnitudes leads to incorrect $cos⁡θ$ values.
Confusing sine and cosine relationships: For line-plane angles, ensure the use of $\sin \theta$ instead of $\cos \theta$ .
Forgetting absolute values in $\cos \theta$ : Negative dot products can occur, but the magnitude of $\cos \theta$ should always be between 0 and 1.

infoNote

Key Formulas:

Scalar Product:

\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos \theta

Angle Between Two Lines:

\cos \theta = \frac{\mathbf{b}_1 \times \mathbf{b}_2}{|\mathbf{b}_1||\mathbf{b}_2|}

Equation of a Plane:

\mathbf{r} \times \mathbf{n} = k, \quad \text{or} \quad ax + by + cz = d

Angle Between Two Planes:

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

Angle Between a Line and a Plane:

\sin \theta = \frac{\mathbf{b} \times \mathbf{n}}{|\mathbf{b}||\mathbf{n}|}

Angle between Lines Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

6.1.3 Angle between Lines

Introduction

Key Applications of the Scalar Product

Angle Between Two Lines

Equation of a Plane

Angle Between Two Planes

Angle Between a Line and a Plane

Worked Examples

Note Summary

Common Mistakes:

Key Formulas:

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