Vectors are fundamental in electronics engineering for representing quantities that have both magnitude and direction, such as currents, voltages, and electromagnetic fields. Understanding vector operations in two and three dimensions is essential for analyzing circuit behaviors, signal propagation, and antenna orientations. This chapter develops the mathematical tools of vector theory and applies them to practical electronics engineering problems encountered in Kenya and beyond.

9.1 Vectors and scalar in two and three dimensions

Vectors describe quantities with magnitude and direction, while scalars have magnitude only. In electronics engineering, voltages and currents can be represented as vectors to analyze their phase relationships and magnitudes in AC circuits and electromagnetic wave propagation. This section focuses on the representation of vectors in 2D and 3D space and how to perform addition and subtraction operations on them.

9.1.1 Vectors and scalars in two dimensions

Vectors in two dimensions are represented by components along the x and y axes. A vector \(\mathbf{A}\) is written as \(\mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j}\), where \(A_x\) and \(A_y\) are the scalar components, and \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors along the x and y axes respectively.

$$ \mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j} $$

The magnitude \(|\mathbf{A}|\) is given by

$$ |\mathbf{A}| = \sqrt{A_x^2 + A_y^2} $$

Worked Examples

Example 1: An AC voltage vector has components \(V_x = 120\,V\) and \(V_y = 90\,V\). Find the magnitude of the voltage vector.

Given: \(V_x = 120\,V\), \(V_y = 90\,V\)

$$ |\mathbf{V}| = \sqrt{V_x^2 + V_y^2} $$

$$ = \sqrt{(120)^2 + (90)^2} $$

$$ = \sqrt{14400 + 8100} $$

$$ = \sqrt{22500} = 150\,V $$

Answer: 150 V

Example 2: A current vector has components \(I_x = 3\,A\) and \(I_y = 4\,A\). Calculate the magnitude and direction angle \(\theta\) relative to the x-axis.

Given: \(I_x = 3\,A\), \(I_y = 4\,A\)

$$ |\mathbf{I}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\,A $$

$$ \theta = \tan^{-1} (\frac{I_y}{I_x}) = \tan^{-1} (\frac{4}{3}) = 53.13^\circ $$

Answer: Magnitude = 5 A, Angle = 53.13°

Example 3: Find the components of a voltage vector of magnitude 230 V making an angle of 30° with the x-axis.

Given: \(|\mathbf{V}| = 230\,V\), \(\theta = 30^\circ\)

$$ V_x = |\mathbf{V}| \cos \theta = 230 \times \cos 30^\circ = 230 \times 0.866 = 198.18\,V $$

$$ V_y = |\mathbf{V}| \sin \theta = 230 \times \sin 30^\circ = 230 \times 0.5 = 115\,V $$

Answer: \(V_x = 198.18\,V\), \(V_y = 115\,V\)

9.1.2 Vectors and scalars in three dimensions

In three dimensions, vectors have components along x, y, and z axes. A vector \(\mathbf{B}\) is expressed as

$$ \mathbf{B} = B_x \mathbf{i} + B_y \mathbf{j} + B_z \mathbf{k} $$

where \(\mathbf{k}\) is the unit vector along the z-axis. The magnitude is

$$ |\mathbf{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2} $$

Three-dimensional vectors are important in electronics for analyzing magnetic fields and antenna radiation patterns.

Worked Examples

Example 1: A magnetic field vector has components \(B_x = 0.5\,T\), \(B_y = 0.3\,T\), and \(B_z = 0.4\,T\). Find the magnitude.

Given: \(B_x = 0.5\,T\), \(B_y = 0.3\,T\), \(B_z = 0.4\,T\)

$$ |\mathbf{B}| = \sqrt{0.5^2 + 0.3^2 + 0.4^2} $$

$$ = \sqrt{0.25 + 0.09 + 0.16} = \sqrt{0.5} = 0.707\,T $$

Answer: 0.707 T

Example 2: Find the components of an electric field vector of magnitude 100 V/m making angles 45°, 60°, and 30° with x, y, and z axes respectively.

Given: \(|\mathbf{E}|=100\,V/m\), \(\theta_x=45^\circ\), \(\theta_y=60^\circ\), \(\theta_z=30^\circ\)

$$ E_x = |\mathbf{E}| \cos \theta_x = 100 \times \cos 45^\circ = 100 \times 0.707 = 70.7\,V/m $$

$$ E_y = 100 \times \cos 60^\circ = 100 \times 0.5 = 50\,V/m $$

$$ E_z = 100 \times \cos 30^\circ = 100 \times 0.866 = 86.6\,V/m $$

Answer: \(E_x = 70.7\,V/m\), \(E_y = 50\,V/m\), \(E_z = 86.6\,V/m\)

Example 3: Given \(\mathbf{C} = 2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\), find the magnitude and the direction cosines.

Given: \(C_x=2\), \(C_y=3\), \(C_z=6\)

$$ |\mathbf{C}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 $$

Direction cosines are

$$ \cos \alpha = \frac{C_x}{|\mathbf{C}|} = \frac{2}{7} = 0.286 $$

$$ \cos \beta = \frac{C_y}{|\mathbf{C}|} = \frac{3}{7} = 0.429 $$

$$ \cos \gamma = \frac{C_z}{|\mathbf{C}|} = \frac{6}{7} = 0.857 $$

Answer: Magnitude = 7, Direction cosines = 0.286, 0.429, 0.857

9.1.3 Addition of vectors

Vector addition combines two or more vectors to produce a resultant vector, useful in electronics for summing currents or voltages with direction. The components of the resultant vector are the sum of the respective components of the individual vectors.

$$ \mathbf{R} = \mathbf{A} + \mathbf{B} = (A_x + B_x) \mathbf{i} + (A_y + B_y) \mathbf{j} + (A_z + B_z) \mathbf{k} $$

Worked Examples

Example 1: Add two 2D voltage vectors \(\mathbf{V_1} = 100 \mathbf{i} + 50 \mathbf{j}\) V and \(\mathbf{V_2} = 60 \mathbf{i} + 80 \mathbf{j}\) V.

Given: \(\mathbf{V_1} = 100 \mathbf{i} + 50 \mathbf{j}\), \(\mathbf{V_2} = 60 \mathbf{i} + 80 \mathbf{j}\)

$$ \mathbf{V_R} = (100 + 60) \mathbf{i} + (50 + 80) \mathbf{j} = 160 \mathbf{i} + 130 \mathbf{j} $$

Magnitude:

$$ |\mathbf{V_R}| = \sqrt{160^2 + 130^2} = \sqrt{25600 + 16900} = \sqrt{42500} = 206.16\,V $$

Answer: \(\mathbf{V_R} = 160 \mathbf{i} + 130 \mathbf{j}\), Magnitude = 206.16 V

Example 2: Add two 3D current vectors \(\mathbf{I_1} = 2 \mathbf{i} + 3 \mathbf{j} + 1 \mathbf{k}\) A and \(\mathbf{I_2} = 4 \mathbf{i} + 1 \mathbf{j} + 5 \mathbf{k}\) A.

Given: \(\mathbf{I_1} = 2 \mathbf{i} + 3 \mathbf{j} + 1 \mathbf{k}\), \(\mathbf{I_2} = 4 \mathbf{i} + 1 \mathbf{j} + 5 \mathbf{k}\)

$$ \mathbf{I_R} = (2 + 4) \mathbf{i} + (3 + 1) \mathbf{j} + (1 + 5) \mathbf{k} = 6 \mathbf{i} + 4 \mathbf{j} + 6 \mathbf{k} $$

Magnitude:

$$ |\mathbf{I_R}| = \sqrt{6^2 + 4^2 + 6^2} = \sqrt{36 + 16 + 36} = \sqrt{88} = 9.38\,A $$

Answer: \(\mathbf{I_R} = 6 \mathbf{i} + 4 \mathbf{j} + 6 \mathbf{k}\), Magnitude = 9.38 A

Example 3: Two electromagnetic wave vectors \(\mathbf{E_1} = 5 \mathbf{i} + 7 \mathbf{j} + 2 \mathbf{k}\) V/m and \(\mathbf{E_2} = -3 \mathbf{i} + 4 \mathbf{j} + 6 \mathbf{k}\) V/m interfere. Find the resultant vector.

Given: \(\mathbf{E_1} = 5 \mathbf{i} + 7 \mathbf{j} + 2 \mathbf{k}\), \(\mathbf{E_2} = -3 \mathbf{i} + 4 \mathbf{j} + 6 \mathbf{k}\)

$$ \mathbf{E_R} = (5 - 3) \mathbf{i} + (7 + 4) \mathbf{j} + (2 + 6) \mathbf{k} = 2 \mathbf{i} + 11 \mathbf{j} + 8 \mathbf{k} $$

Magnitude:

$$ |\mathbf{E_R}| = \sqrt{2^2 + 11^2 + 8^2} = \sqrt{4 + 121 + 64} = \sqrt{189} = 13.75\,V/m $$

Answer: \(\mathbf{E_R} = 2 \mathbf{i} + 11 \mathbf{j} + 8 \mathbf{k}\), Magnitude = 13.75 V/m

9.1.4 Subtraction of vectors

Vector subtraction finds the difference between two vectors and is used in electronics to determine relative phase or difference between signals. The components of the difference vector are obtained by subtracting the corresponding components.

$$ \mathbf{D} = \mathbf{A} - \mathbf{B} = (A_x - B_x) \mathbf{i} + (A_y - B_y) \mathbf{j} + (A_z - B_z) \mathbf{k} $$

Worked Examples

Example 1: Subtract two 2D vectors \(\mathbf{V_1} = 150 \mathbf{i} + 80 \mathbf{j}\) V and \(\mathbf{V_2} = 100 \mathbf{i} + 50 \mathbf{j}\) V.

Given: \(\mathbf{V_1} = 150 \mathbf{i} + 80 \mathbf{j}\), \(\mathbf{V_2} = 100 \mathbf{i} + 50 \mathbf{j}\)

$$ \mathbf{V_D} = (150 - 100) \mathbf{i} + (80 - 50) \mathbf{j} = 50 \mathbf{i} + 30 \mathbf{j} $$

Magnitude:

$$ |\mathbf{V_D}| = \sqrt{50^2 + 30^2} = \sqrt{2500 + 900} = \sqrt{3400} = 58.31\,V $$

Answer: \(\mathbf{V_D} = 50 \mathbf{i} + 30 \mathbf{j}\), Magnitude = 58.31 V

Example 2: Subtract two 3D current vectors \(\mathbf{I_1} = 8 \mathbf{i} + 5 \mathbf{j} + 3 \mathbf{k}\) A and \(\mathbf{I_2} = 3 \mathbf{i} + 7 \mathbf{j} + 1 \mathbf{k}\) A.

Given: \(\mathbf{I_1} = 8 \mathbf{i} + 5 \mathbf{j} + 3 \mathbf{k}\), \(\mathbf{I_2} = 3 \mathbf{i} + 7 \mathbf{j} + 1 \mathbf{k}\)

$$ \mathbf{I_D} = (8 - 3) \mathbf{i} + (5 - 7) \mathbf{j} + (3 - 1) \mathbf{k} = 5 \mathbf{i} - 2 \mathbf{j} + 2 \mathbf{k} $$

Magnitude:

$$ |\mathbf{I_D}| = \sqrt{5^2 + (-2)^2 + 2^2} = \sqrt{25 + 4 + 4} = \sqrt{33} = 5.74\,A $$

Answer: \(\mathbf{I_D} = 5 \mathbf{i} - 2 \mathbf{j} + 2 \mathbf{k}\), Magnitude = 5.74 A

Example 3: Find the difference between two electromagnetic vectors \(\mathbf{E_1} = 7 \mathbf{i} + 2 \mathbf{j} + 4 \mathbf{k}\) V/m and \(\mathbf{E_2} = 3 \mathbf{i} + 6 \mathbf{j} + 5 \mathbf{k}\) V/m.

Given: \(\mathbf{E_1} = 7 \mathbf{i} + 2 \mathbf{j} + 4 \mathbf{k}\), \(\mathbf{E_2} = 3 \mathbf{i} + 6 \mathbf{j} + 5 \mathbf{k}\)

$$ \mathbf{E_D} = (7 - 3) \mathbf{i} + (2 - 6) \mathbf{j} + (4 - 5) \mathbf{k} = 4 \mathbf{i} - 4 \mathbf{j} - 1 \mathbf{k} $$

Magnitude:

$$ |\mathbf{E_D}| = \sqrt{4^2 + (-4)^2 + (-1)^2} = \sqrt{16 + 16 + 1} = \sqrt{33} = 5.74\,V/m $$

Answer: \(\mathbf{E_D} = 4 \mathbf{i} - 4 \mathbf{j} - 1 \mathbf{k}\), Magnitude = 5.74 V/m

Chapter Summary

This chapter introduced vectors and scalars in two and three dimensions, emphasizing their fundamental differences and how vectors represent quantities with both magnitude and direction. Operations on vectors, including addition and subtraction, were demonstrated to combine or compare vector quantities effectively. Position vectors were defined to locate points in space relative to an origin, providing a basis for further vector analysis. The resolution of vectors into components was explained to simplify calculations by breaking vectors into perpendicular directions. The chapter then explored scalar and vector products, showing how these operations produce scalar or vector results useful in engineering contexts. Advanced vector calculus concepts such as gradient, curl, and divergence were introduced to describe how scalar and vector fields change in space, which are essential for understanding physical phenomena in electronics engineering. These topics collectively build a strong foundation for applying vector theory in practical engineering problems.