By the end of this chapter, you will be able to:
Mastering these skills will help you tackle real engineering problems with confidence and precision in your future trade.
Vectors are fundamental in quantity surveying calculations involving forces, displacements, and directions in two and three-dimensional spaces. Mastery of vector theory enables quantity surveyors to accurately determine quantities, resolve structural forces, and interpret site measurements, particularly when working with complex building geometries or site layouts. This chapter focuses on understanding vectors and scalars, and applying vector operations essential for advanced engineering computations in the Kenyan construction context.
Vectors and scalar quantities are the foundation of engineering mathematics applied in quantity surveying. Scalars have magnitude only, while vectors have both magnitude and direction. Quantity surveyors encounter vectors when calculating forces on beams, displacements in structures, or gradients on site plans. Understanding vector representation in two and three dimensions is vital for solving real-world surveying problems.
Vectors are quantities described by both magnitude and direction, represented as arrows in space. Scalars have magnitude only and no direction. Examples include:
The vector \(\vec{v}\) in two dimensions is represented as \(\vec{v} = v_x \hat{i} + v_y \hat{j}\), where \(v_x\) and \(v_y\) are components along the x- and y-axes respectively.
$$\vec{v} = v_x \hat{i} + v_y \hat{j}$$
In three dimensions, a vector has an additional z-component:
$$\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$$
Example 1: A site engineer measures a displacement vector on a flat plane as 5 m east and 3 m north. Represent this displacement vector.
Given: \(v_x = 5\) m, \(v_y = 3\) m
$$\vec{v} = 5 \hat{i} + 3 \hat{j}$$
Answer: \(\vec{v} = 5 \hat{i} + 3 \hat{j}\) m
Example 2: A load force acts with components 2000 N east, 1500 N north, and 1000 N vertically upwards. Express this force vector.
Given: \(F_x = 2000\) N, \(F_y = 1500\) N, \(F_z = 1000\) N
$$\vec{F} = 2000 \hat{i} + 1500 \hat{j} + 1000 \hat{k}$$
Answer: \(\vec{F} = 2000 \hat{i} + 1500 \hat{j} + 1000 \hat{k}\) N
Example 3: Find the magnitude of the displacement vector \(\vec{v} = 5 \hat{i} + 3 \hat{j}\).
Given: \(v_x = 5\) m, \(v_y = 3\) m
$$|\vec{v}| = \sqrt{v_x^2 + v_y^2}$$
$$= \sqrt{5^2 + 3^2}$$
$$= \sqrt{25 + 9}$$
$$= \sqrt{34}$$
$$= 5.83 \text{ m}$$
Answer: 5.83 m
Example 4: Calculate the magnitude of the force vector \(\vec{F} = 2000 \hat{i} + 1500 \hat{j} + 1000 \hat{k}\).
Given: \(F_x = 2000\) N, \(F_y = 1500\) N, \(F_z = 1000\) N
$$|\vec{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2}$$
$$= \sqrt{2000^2 + 1500^2 + 1000^2}$$
$$= \sqrt{4,000,000 + 2,250,000 + 1,000,000}$$
$$= \sqrt{7,250,000}$$
$$= 2694.44 \text{ N}$$
Answer: 2694.44 N
Example 5: A displacement vector in 3D has components \(v_x = 4\) m, \(v_y = -3\) m, \(v_z = 12\) m. Find its magnitude.
Given: \(v_x = 4\) m, \(v_y = -3\) m, \(v_z = 12\) m
$$|\vec{v}| = \sqrt{4^2 + (-3)^2 + 12^2}$$
$$= \sqrt{16 + 9 + 144}$$
$$= \sqrt{169}$$
$$= 13 \text{ m}$$
Answer: 13 m
Vectors in two-dimensional space are resolved into orthogonal components along the x- and y-axes. This allows quantity surveyors to analyze forces and displacements in planar structures such as walls and beams.
The magnitude of a vector \(\vec{v}\) is:
$$|\vec{v}| = \sqrt{v_x^2 + v_y^2}$$
The direction \(\theta\) of \(\vec{v}\) relative to the x-axis is:
$$\theta = \tan^{-1} \frac{v_y}{v_x}$$
Example 1: A force vector has components \(F_x = 3000\) N and \(F_y = 4000\) N. Find the magnitude and direction.
Given: \(F_x = 3000\) N, \(F_y = 4000\) N
$$|\vec{F}| = \sqrt{3000^2 + 4000^2}$$
$$= \sqrt{9,000,000 + 16,000,000}$$
$$= \sqrt{25,000,000}$$
$$= 5000 \text{ N}$$
$$\theta = \tan^{-1} \frac{4000}{3000}$$
$$= \tan^{-1} (1.333)$$
$$= 53.13^\circ$$
Answer: Magnitude = 5000 N, Direction = 53.13° from x-axis
Example 2: A displacement vector has magnitude 10 m and makes an angle of 30° with the x-axis. Find the components \(v_x\) and \(v_y\).
Given: \(|\vec{v}| = 10\) m, \(\theta = 30^\circ\)
$$v_x = |\vec{v}| \cos \theta$$
$$= 10 \times \cos 30^\circ$$
$$= 10 \times 0.866$$
$$= 8.66 \text{ m}$$
$$v_y = |\vec{v}| \sin \theta$$
$$= 10 \times \sin 30^\circ$$
$$= 10 \times 0.5$$
$$= 5 \text{ m}$$
Answer: \(v_x = 8.66\) m, \(v_y = 5\) m
Example 3: A force vector of magnitude 2500 N acts at 120° from the positive x-axis. Find its components.
Given: \(|\vec{F}| = 2500\) N, \(\theta = 120^\circ\)
$$F_x = 2500 \cos 120^\circ$$
$$= 2500 \times (-0.5)$$
$$= -1250 \text{ N}$$
$$F_y = 2500 \sin 120^\circ$$
$$= 2500 \times 0.866$$
$$= 2165 \text{ N}$$
Answer: \(F_x = -1250\) N, \(F_y = 2165\) N
Example 4: Find the magnitude and direction of a displacement vector with components \(v_x = -6\) m and \(v_y = 8\) m.
Given: \(v_x = -6\) m, \(v_y = 8\) m
$$|\vec{v}| = \sqrt{(-6)^2 + 8^2}$$
$$= \sqrt{36 + 64}$$
$$= \sqrt{100}$$
$$= 10 \text{ m}$$
$$\theta = \tan^{-1} \frac{8}{-6}$$
$$= \tan^{-1} (-1.333)$$
Since \(v_x\) is negative and \(v_y\) positive, vector lies in second quadrant:
$$\theta = 180^\circ - 53.13^\circ = 126.87^\circ$$
Answer: Magnitude = 10 m, Direction = 126.87°
In three-dimensional space, vectors have components along x-, y-, and z-axes. Quantity surveyors use 3D vectors when working with multi-storey buildings or complex site topographies.
The magnitude of a vector \(\vec{v}\) is:
$$|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$$
Direction cosines \(\alpha, \beta, \gamma\) represent the angles between the vector and the coordinate axes:
$$\cos \alpha = \frac{v_x}{|\vec{v}|}, \quad \cos \beta = \frac{v_y}{|\vec{v}|}, \quad \cos \gamma = \frac{v_z}{|\vec{v}|}$$
Example 1: A force vector has components \(F_x = 1000\) N, \(F_y = 2000\) N, \(F_z = 1500\) N. Find the magnitude.
Given: \(F_x = 1000\) N, \(F_y = 2000\) N, \(F_z = 1500\) N
$$|\vec{F}| = \sqrt{1000^2 + 2000^2 + 1500^2}$$
$$= \sqrt{1,000,000 + 4,000,000 + 2,250,000}$$
$$= \sqrt{7,250,000}$$
$$= 2694.44 \text{ N}$$
Answer: 2694.44 N
Example 2: Calculate the direction cosines of the vector \(\vec{v} = 3 \hat{i} + 4 \hat{j} + 12 \hat{k}\).
Given: \(v_x=3\), \(v_y=4\), \(v_z=12\)
$$|\vec{v}| = \sqrt{3^2 + 4^2 + 12^2}$$
$$= \sqrt{9 + 16 + 144}$$
$$= \sqrt{169} = 13$$
$$\cos \alpha = \frac{3}{13} = 0.231$$
$$\cos \beta = \frac{4}{13} = 0.308$$
$$\cos \gamma = \frac{12}{13} = 0.923$$
Answer: \(\cos \alpha = 0.231\), \(\cos \beta = 0.308\), \(\cos \gamma = 0.923\)
Example 3: A displacement vector \(\vec{v}\) has magnitude 20 m and direction cosines 0.6, 0.8, and 0. Calculate its components.
Given: \(|\vec{v}| = 20\) m, \(\cos \alpha = 0.6\), \(\cos \beta = 0.8\), \(\cos \gamma = 0\)
$$v_x = |\vec{v}| \cos \alpha = 20 \times 0.6 = 12 \text{ m}$$
$$v_y = 20 \times 0.8 = 16 \text{ m}$$
$$v_z = 20 \times 0 = 0 \text{ m}$$
Answer: \(v_x = 12\) m, \(v_y = 16\) m, \(v_z = 0\) m
Example 4: Find the magnitude and direction cosines of a vector with components \(v_x = -5\), \(v_y = 12\), \(v_z = -13\).
Given: \(v_x = -5\), \(v_y = 12\), \(v_z = -13\)
$$|\vec{v}| = \sqrt{(-5)^2 + 12^2 + (-13)^2}$$
$$= \sqrt{25 + 144 + 169}$$
$$= \sqrt{338} = 18.38$$
$$\cos \alpha = \frac{-5}{18.38} = -0.272$$
$$\cos \beta = \frac{12}{18.38} = 0.653$$
$$\cos \gamma = \frac{-13}{18.38} = -0.707$$
Answer: Magnitude = 18.38, \(\cos \alpha = -0.272\), \(\cos \beta = 0.653\), \(\cos \gamma = -0.707\)
Create a free account to open more of this chapter.
Free: practical guides, quick cards, workplace scenarios and more.
Create a free accountThis chapter introduced the fundamental concepts of vectors and scalar quantities in both two and three dimensions, highlighting their differences and applications in engineering contexts. It explored various operations on vectors, including addition, subtraction, and multiplication, demonstrating how these operations are vital for solving spatial problems. The chapter also examined the position of vectors, explaining how vectors can represent points in space relative to an origin or other reference points. Additionally, the resolution of vectors into components was covered, showing how complex vectors can be broken down into simpler parts along coordinate axes for easier analysis. Through these topics, students gained a comprehensive understanding of vector theory principles necessary for practical engineering calculations. The chapter emphasized the importance of precise vector manipulation to accurately model forces, directions, and positions in civil engineering projects.
A force vector in two dimensions is given by \( \mathbf{F} = 3\mathbf{i} + 4\mathbf{j} \) kN. Calculate the magnitude of the force. (2 marks)
A displacement vector in three dimensions is \( \mathbf{D} = 5\mathbf{i} + 2\mathbf{j} + 6\mathbf{k} \) m. Find the scalar magnitude of \( \mathbf{D} \). (2 marks)
At the start of this chapter we promised you would be able to:
Tick each one you can genuinely do.
So, are you there yet?
You're competent when you can confidently do 50% or more of what this chapter promised.
Sign in to record how you're doing.