Angles in 3 Dimensions - Mathematics.lu

You have already seen the main formula back in S6. The scalar product can be defined by using the angle between the two vectors. From this, we can conclude a formula to find the angle between two vectors:

$\vec{u}\cdot\vec{v}=\|\vec{u}\|\cdot\|\vec{v}\|\cdot\cos(\theta)\implies \cos(\theta)=\frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\cdot\|\vec{v}\|}$

If we want the acute angle $\theta$ created by two vectors $\vec{u}$ and $\vec{v}$ we adjust the formula slightly by adding absolute value.

$\cos(\theta)=\frac{|\vec{u}\cdot\vec{v}|}{\|\vec{u}\|\cdot\|\vec{v}\|}$

Table of Contents

Angle Between Two Vectors

Given by the formula above.

Example: Given the vectors $\vec{u} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}$ in $\mathbb{R}^3$ , calculate the acute angle $\theta$ between them.

Angle Between Two Lines

The angle between two lines is simply the angle created by the direction vectors of the lines.

Example: Find the acute angle $\theta$ between the two lines $\ell_1$ and $\ell_2$ given by the equations:

$\ell_1: \begin{cases} x = 2 + t \\ y = -4 \\ z = 1 - t \end{cases} \quad \text{and} \quad \ell_2: \frac{x - 5}{0} = \frac{y + 1}{2} = \frac{z - 3}{2}$

In this case, we chose an easier example just for the purposes of simplicity. The method is literally the same as with vectors.

Angle Between Two Planes

The angle between two planes is calculated using the normal vectors. We can use simple geometry to understand that the formula remains exactly the same, though we will see that the case between a line and a plane changes our formula slightly.

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That is, if $\pi_1$ and $\pi_2$ are two planes with normal vectors $\vec{n}_1$ and $\vec{n}_2$ , the angle $\theta$ between them can be calculated as:

$\cos(\theta)=\frac{|\vec{n}_1\cdot\vec{n}_2|}{\|\vec{n}_1\|\cdot\|\vec{n_2}\|}$

Example: Calculate the acute angle $\theta$ between the two planes $\pi_1$ and $\pi_2$ given by their Cartesian equations:

$\pi_1: x - z + 4 = 0 \quad \text{and} \quad \pi_2: 2y + 2z - 7 = 0$

Angle Between a Plane and a Line

Example: Calculate the acute angle $\theta$ between the line $\ell$ and the plane $\pi$ given by their respective equations:

$\ell: \begin{cases} x = 2 + t \\ y = -4 \\ z = 1 - t \end{cases} \quad \text{and} \quad \pi: 2y + 2z - 7 = 0$

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Exercises

Extract the direction vectors for each pair of lines below and apply the absolute dot product formula to determine the exact acute angle $\theta$ between them.

1. $\ell_1: \begin{cases} x = t \\ y = -2 + t \\ z = 4 - 4t \end{cases} \quad \text{and} \quad \ell_2: \begin{cases} x = 3 - 2s \\ y = 1 + 2s \\ z = s \end{cases}$
2. $\ell_1: \frac{x - 2}{2} = \frac{y + 1}{-1} = \frac{z - 5}{2} \quad \text{and} \quad \ell_2: \frac{x + 4}{1} = \frac{y}{2} = \frac{z - 1}{2}$
3. $\ell_1(t) = \begin{pmatrix} 1 \\ 5 \\ 0 \end{pmatrix} + t\begin{pmatrix} 3 \\ 0 \\ -4 \end{pmatrix} \quad \text{and} \quad \ell_2(s) = \begin{pmatrix} -2 \\ 2 \\ 6 \end{pmatrix} + s\begin{pmatrix} 0 \\ 5 \\ 12 \end{pmatrix}$
4. $\ell_1: \begin{cases} x = -3 + 4t \\ y = 1 - t \\ z = 2 + 3t \end{cases} \quad \text{and} \quad \ell_2: \frac{x - 1}{1} = \frac{y + 2}{4} = \frac{z}{-1}$

Identify the components of the normal vectors from the lowercase Cartesian plane equations below to compute the acute angle of intersection $\theta$ separating the two plane structures.

1. $\pi_1: 2x - y + 2z - 1 = 0 \quad \text{and} \quad \pi_2: 3x + 4y - 12 = 0$
2. $\pi_1: x + 2y - z + 3 = 0 \quad \text{and} \quad \pi_2: 2x - y + 3z - 5 = 0$
3. $\pi_1: 4x - 4y + 2z + 7 = 0 \quad \text{and} \quad \pi_2: x + z - 9 = 0$
4. $\pi_1: 3x - y - 2z + 1 = 0 \quad \text{and} \quad \pi_2: x + 2y - z - 4 = 0$

Extract the line’s direction vector and the plane’s normal vector. Apply the complementary absolute sine product formula to calculate the acute inclination angle $\theta$ between the line path and the plane surface layer.

1. $\ell: \begin{cases} x = 1 + 2t \\ y = -3t \\ z = 4 + t \end{cases} \quad \text{and} \quad \pi: 2x - y - 2z + 6 = 0$
2. $\ell: \frac{x - 5}{3} = \frac{y + 2}{1} = \frac{z}{-4} \quad \text{and} \quad \pi: x + 4y - 3z - 2 = 0$
3. $\ell(t) = \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} + t\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \quad \text{and} \quad \pi: 3x - y + z + 5 = 0$
4. $\ell: \begin{cases} x = -t \\ y = 4 + 4t \\ z = 2 - 3t \end{cases} \quad \text{and} \quad \pi: 5x - z + 8 = 0$