Vectors are fundamental in electronics engineering, representing quantities that have both magnitude and direction such as current, voltage, and electromagnetic fields. Mastery of vector theory enables technicians to analyze circuit behaviors, signal directions, and spatial orientations of components with precision. This chapter introduces vector concepts in two and three dimensions and explores position vectors, essential for spatial analysis in electronics system design and troubleshooting.

9.1 Vectors and Scalar in Two and Three Dimensions

Vectors describe quantities with magnitude and direction, while scalars have only magnitude. In electronics engineering, vector quantities include current and voltage represented in phasor form, while scalar quantities include resistance and power magnitude. Understanding vector operations in two and three dimensions is crucial for analyzing complex circuits and electromagnetic fields in practical Kenyan engineering contexts like communication towers and control systems.

9.1.1 Vector and Scalar Quantities in Electronics Engineering

Vectors have both magnitude and direction, whereas scalars have only magnitude. In electronics:

• Voltage (phasor form) is a vector representing magnitude and phase angle.
• Current (phasor form) is also a vector with magnitude and phase.
• Resistance is a scalar quantity.
• Power (real power) is scalar, but complex power is vectorial.
• Electric field strength is a vector in spatial analysis.

$$\text{Vector} = \text{Magnitude} + \text{Direction}$$

$$\text{Scalar} = \text{Magnitude only}$$

Worked Examples

Example 1: A voltage phasor has magnitude 230 V and phase angle 30°. Express it as a vector.

Given: Magnitude \( V = 230 \, V \), Angle \( \theta = 30^\circ \)

$$\vec{V} = V \angle \theta$$

$$\vec{V} = 230 \angle 30^\circ$$

Answer: Voltage vector \( \vec{V} = 230 \angle 30^\circ \) V

Example 2: Current magnitude is 5 A with phase angle -45°. Express the current vector.

Given: \( I = 5 \, A \), \( \theta = -45^\circ \)

$$\vec{I} = I \angle \theta$$

$$\vec{I} = 5 \angle -45^\circ$$

Answer: Current vector \( \vec{I} = 5 \angle -45^\circ \) A

Example 3: A resistance of 10 Ω is given. Is this a vector or scalar?

Given: \( R = 10 \, \Omega \)

Resistance has magnitude only.

Answer: Resistance is scalar with value 10 Ω

9.1.2 Vector Representation in Two Dimensions

Vectors in two dimensions are represented by components along the x- and y-axes. In electronics, two-dimensional vectors represent quantities like voltage and current phasors on the complex plane.

A vector \( \vec{A} \) is expressed as:

$$\vec{A} = A_x \hat{i} + A_y \hat{j}$$

where \( A_x = A \cos \theta \), \( A_y = A \sin \theta \), and \( \theta \) is the angle with the x-axis.

Worked Examples

Example 1: Find components of a voltage vector of 120 V at 60°.

Given: \( A = 120 \, V \), \( \theta = 60^\circ \)

$$A_x = A \cos \theta = 120 \cos 60^\circ$$

$$A_x = 120 \times 0.5 = 60 \, V$$

$$A_y = A \sin \theta = 120 \sin 60^\circ$$

$$A_y = 120 \times 0.866 = 103.92 \, V$$

Answer: \( \vec{A} = 60 \hat{i} + 103.92 \hat{j} \, V \)

Example 2: Calculate components of current vector 10 A at -30°.

Given: \( I = 10 \, A \), \( \theta = -30^\circ \)

$$I_x = 10 \cos (-30^\circ) = 10 \times 0.866 = 8.66 \, A$$

$$I_y = 10 \sin (-30^\circ) = 10 \times (-0.5) = -5 \, A$$

Answer: \( \vec{I} = 8.66 \hat{i} - 5 \hat{j} \, A \)

Example 3: A vector has \( \vec{V} = 40 \hat{i} - 30 \hat{j} \) V. Find its magnitude and angle.

Given: \( V_x = 40 \, V \), \( V_y = -30 \, V \)

$$|\vec{V}| = \sqrt{V_x^2 + V_y^2} = \sqrt{40^2 + (-30)^2}$$

$$|\vec{V}| = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, V$$

$$\theta = \tan^{-1} (\frac{V_y}{V_x}) = \tan^{-1} (\frac{-30}{40}) = -36.87^\circ$$

Answer: Magnitude 50 V, Angle -36.87°

9.1.3 Vector Representation in Three Dimensions

In electronics engineering, three-dimensional vectors are used to analyze electromagnetic fields and antenna orientations. A vector in 3D has components along x, y, and z axes:

$$\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$$

The magnitude is:

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

Worked Examples

Example 1: Find the magnitude of vector \( \vec{A} = 3 \hat{i} + 4 \hat{j} + 12 \hat{k} \) A/m.

Given: \( A_x = 3 \), \( A_y = 4 \), \( A_z = 12 \)

$$|\vec{A}| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144}$$

$$|\vec{A}| = \sqrt{169} = 13 \, A/m$$

Answer: Magnitude is 13 A/m

Example 2: Given \( \vec{B} = 5 \hat{i} - 2 \hat{j} + 6 \hat{k} \) T, find magnitude.

Given: \( B_x = 5 \), \( B_y = -2 \), \( B_z = 6 \)

$$|\vec{B}| = \sqrt{5^2 + (-2)^2 + 6^2} = \sqrt{25 + 4 + 36}$$

$$|\vec{B}| = \sqrt{65} = 8.062 \, T$$

Answer: Magnitude is 8.062 T

Example 3: A magnetic field vector has components \( 1 \hat{i} + 2 \hat{j} + 2 \hat{k} \) T. Calculate the magnitude.

Given: \( B_x = 1 \), \( B_y = 2 \), \( B_z = 2 \)

$$|\vec{B}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4}$$

$$|\vec{B}| = \sqrt{9} = 3 \, T$$

Answer: Magnitude is 3 T

9.1.4 Operations on Vectors: Addition and Subtraction

Vector addition and subtraction combine vectors representing quantities such as voltages or currents in circuits. Addition is done component-wise, similarly subtraction.

If

$$\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$$

and

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

then

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

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

Worked Examples

Example 1: Add vectors \( \vec{A} = 3 \hat{i} + 4 \hat{j} \) V and \( \vec{B} = 1 \hat{i} + 2 \hat{j} \) V.

Given:

\( \vec{A} = 3 \hat{i} + 4 \hat{j} \)

\( \vec{B} = 1 \hat{i} + 2 \hat{j} \)

$$\vec{A} + \vec{B} = (3 + 1) \hat{i} + (4 + 2) \hat{j}$$

$$= 4 \hat{i} + 6 \hat{j}$$

Answer: \( \vec{A} + \vec{B} = 4 \hat{i} + 6 \hat{j} \) V

Example 2: Subtract \( \vec{B} = 2 \hat{i} + 5 \hat{j} \) A from \( \vec{A} = 7 \hat{i} + 3 \hat{j} \) A.

Given:

\( \vec{A} = 7 \hat{i} + 3 \hat{j} \)

\( \vec{B} = 2 \hat{i} + 5 \hat{j} \)

$$\vec{A} - \vec{B} = (7 - 2) \hat{i} + (3 - 5) \hat{j}$$

$$= 5 \hat{i} - 2 \hat{j}$$

Answer: \( \vec{A} - \vec{B} = 5 \hat{i} - 2 \hat{j} \) A

Example 3: Add 3D vectors \( \vec{A} = 1 \hat{i} + 2 \hat{j} + 3 \hat{k} \) and \( \vec{B} = 4 \hat{i} + 0 \hat{j} + 5 \hat{k} \).

Given:

\( \vec{A} = 1 \hat{i} + 2 \hat{j} + 3 \hat{k} \)

\( \vec{B} = 4 \hat{i} + 0 \hat{j} + 5 \hat{k} \)

$$\vec{A} + \vec{B} = (1+4) \hat{i} + (2+0) \hat{j} + (3+5) \hat{k}$$

$$= 5 \hat{i} + 2 \hat{j} + 8 \hat{k}$$

Answer: \( \vec{A} + \vec{B} = 5 \hat{i} + 2 \hat{j} + 8 \hat{k} \)

Example 4: Subtract \( \vec{B} = 6 \hat{i} + 3 \hat{j} + 1 \hat{k} \) from \( \vec{A} = 10 \hat{i} + 8 \hat{j} + 2 \hat{k} \).

Given:

\( \vec{A} = 10 \hat{i} + 8 \hat{j} + 2 \hat{k} \)

\( \vec{B} = 6 \hat{i} + 3 \hat{j} + 1 \hat{k} \)

$$\vec{A} - \vec{B} = (10 - 6) \hat{i} + (8 - 3) \hat{j} + (2 - 1) \hat{k}$$

$$= 4 \hat{i} + 5 \hat{j} + 1 \hat{k}$$

Answer: \( \vec{A} - \vec{B} = 4 \hat{i} + 5 \hat{j} + 1 \hat{k} \)

Example 5: Two voltage vectors \( \vec{V_1} = 50 \hat{i} + 30 \hat{j} \) V and \( \vec{V_2} = -20 \hat{i} + 40 \hat{j} \) V are applied. Find the resultant vector.

Given:

\( \vec{V_1} = 50 \hat{i} + 30 \hat{j} \)

\( \vec{V_2} = -20 \hat{i} + 40 \hat{j} \)

$$\vec{V} = \vec{V_1} + \vec{V_2} = (50 - 20) \hat{i} + (30 + 40) \hat{j}$$

$$= 30 \hat{i} + 70 \hat{j}$$

Answer: Resultant voltage vector is \( 30 \hat{i} + 70 \hat{j} \) V

Chapter Summary

This chapter introduced the fundamental concepts of vectors and scalars in two and three dimensions, emphasizing their differences and applications. It covered vector operations including addition and subtraction, which are essential for combining or comparing vector quantities. The concept of position vectors was explained to determine the location of points in space relative to an origin. The resolution of vectors into components was presented to simplify complex vector problems. The chapter then explored scalar and vector products, highlighting their significance in calculating magnitudes and directions. Advanced vector calculus topics such as gradient, curl, and divergence were discussed to describe spatial variations in scalar and vector fields. Together, these concepts provide a comprehensive foundation for applying vector theory in engineering mathematics.