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Vectors are fundamental in land surveying for representing directions, displacements, and forces in three-dimensional space. Understanding vectors and scalars enables surveyors to accurately calculate positions, bearings, and distances essential for site layout, boundary determination, and construction planning. This chapter focuses on applying vector theory in surveying contexts commonly encountered in Kenya, emphasizing practical calculations and problem-solving.
Vectors and scalar quantities are the building blocks of mathematical descriptions in land surveying. Scalars have magnitude only, while vectors have both magnitude and direction, making them indispensable for representing physical quantities like displacement and force. Mastery of vector operations allows surveyors to solve coordinate transformations, compute resultant directions, and analyze field measurements efficiently.
Scalar quantities represent physical measurements with magnitude only. They do not have direction, which distinguishes them from vectors. Common scalar quantities in surveying include distance, speed, time, temperature, and mass.
The magnitude of a scalar is always a positive real number or zero. Scalars are combined using ordinary arithmetic operations such as addition, subtraction, multiplication, and division.
The formula for scalar addition is straightforward:
$$S = S_1 + S_2$$
where \(S\) is the resultant scalar quantity.
Example 1: A surveyor measures a chain length of 30 m and then adds another 50 m of tape. What is the total length?
Given: \(S_1 = 30\, m\), \(S_2 = 50\, m\)
$$S = S_1 + S_2$$
$$S = 30 + 50$$
$$S = 80\, m$$
Answer: 80 m
Example 2: A surveyor records a time interval of 2.5 hours and then another 1.75 hours. Find the total time.
Given: \(S_1 = 2.5\, h\), \(S_2 = 1.75\, h\)
$$S = S_1 + S_2$$
$$S = 2.5 + 1.75$$
$$S = 4.25\, h$$
Answer: 4.25 hours
Example 3: A temperature reading is 25°C in the morning and rises by 7.5°C in the afternoon. What is the final temperature?
Given: \(S_1 = 25^\circ C\), \(S_2 = 7.5^\circ C\)
$$S = S_1 + S_2$$
$$S = 25 + 7.5$$
$$S = 32.5^\circ C$$
Answer: 32.5 °C
Vectors have both magnitude and direction, making them suitable to represent quantities like displacement, velocity, and force in surveying. A vector is often written as \(\vec{A}\) or \(\mathbf{A}\) and can be expressed in component form relative to coordinate axes.
The general form of a vector in two dimensions is:
$$\vec{A} = A_x \hat{i} + A_y \hat{j}$$
where \(A_x\) and \(A_y\) are the components along the x- and y-axes respectively, and \(\hat{i}, \hat{j}\) are unit vectors.
The magnitude of \(\vec{A}\) is:
$$|\vec{A}| = \sqrt{A_x^2 + A_y^2}$$
Example 1: Find the magnitude of a displacement vector with components \(A_x = 40\, m\) and \(A_y = 30\, m\).
Given: \(A_x = 40\, m\), \(A_y = 30\, m\)
$$|\vec{A}| = \sqrt{A_x^2 + A_y^2}$$
$$|\vec{A}| = \sqrt{40^2 + 30^2}$$
$$|\vec{A}| = \sqrt{1600 + 900}$$
$$|\vec{A}| = \sqrt{2500}$$
$$|\vec{A}| = 50\, m$$
Answer: 50 m
Example 2: A surveyor records a vector with \(A_x = 15\, m\) east and \(A_y = 20\, m\) north. Calculate the magnitude.
Given: \(A_x = 15\, m\), \(A_y = 20\, m\)
$$|\vec{A}| = \sqrt{15^2 + 20^2}$$
$$|\vec{A}| = \sqrt{225 + 400}$$
$$|\vec{A}| = \sqrt{625}$$
$$|\vec{A}| = 25\, m$$
Answer: 25 m
Example 3: Determine the magnitude of vector \(\vec{A}\) with components \(A_x = -12\, m\) and \(A_y = 9\, m\).
Given: \(A_x = -12\, m\), \(A_y = 9\, m\)
$$|\vec{A}| = \sqrt{(-12)^2 + 9^2}$$
$$|\vec{A}| = \sqrt{144 + 81}$$
$$|\vec{A}| = \sqrt{225}$$
$$|\vec{A}| = 15\, m$$
Answer: 15 m
Vector addition combines two or more vectors to find a resultant vector, taking into account both magnitude and direction. Vector subtraction finds the difference between vectors, which is useful for calculating relative positions in surveying.
The component form for vector addition is:
$$\vec{R} = (A_x + B_x) \hat{i} + (A_y + B_y) \hat{j}$$
For subtraction:
$$\vec{D} = (A_x - B_x) \hat{i} + (A_y - B_y) \hat{j}$$
The magnitude of the resultant vector \(\vec{R}\) is:
$$|\vec{R}| = \sqrt{(A_x + B_x)^2 + (A_y + B_y)^2}$$
Example 1: A surveyor walks 50 m east (\(\vec{A}\)) then 30 m north (\(\vec{B}\)). Find the resultant displacement.
Given: \(A_x = 50\, m\), \(A_y = 0\, m\), \(B_x = 0\, m\), \(B_y = 30\, m\)
$$\vec{R} = (50 + 0) \hat{i} + (0 + 30) \hat{j}$$
$$\vec{R} = 50 \hat{i} + 30 \hat{j}$$
$$|\vec{R}| = \sqrt{50^2 + 30^2}$$
$$|\vec{R}| = \sqrt{2500 + 900}$$
$$|\vec{R}| = \sqrt{3400}$$
$$|\vec{R}| = 58.31\, m$$
Answer: 58.31 m
Example 2: Vector \(\vec{A}\) has components (40 m east, 20 m north), and vector \(\vec{B}\) is (10 m west, 15 m north). Find \(\vec{A} + \vec{B}\).
Given: \(A_x = 40\, m\), \(A_y = 20\, m\), \(B_x = -10\, m\), \(B_y = 15\, m\)
$$\vec{R} = (40 - 10) \hat{i} + (20 + 15) \hat{j}$$
$$\vec{R} = 30 \hat{i} + 35 \hat{j}$$
$$|\vec{R}| = \sqrt{30^2 + 35^2}$$
$$|\vec{R}| = \sqrt{900 + 1225}$$
$$|\vec{R}| = \sqrt{2125}$$
$$|\vec{R}| = 46.10\, m$$
Answer: 46.10 m
Example 3: A surveyor moves along vector \(\vec{A} = (60\, m, 40\, m)\) then returns along vector \(\vec{B} = (50\, m, 70\, m)\). Find the resultant displacement \(\vec{A} - \vec{B}\).
Given: \(A_x = 60\, m\), \(A_y = 40\, m\), \(B_x = 50\, m\), \(B_y = 70\, m\)
$$\vec{D} = (60 - 50) \hat{i} + (40 - 70) \hat{j}$$
$$\vec{D} = 10 \hat{i} - 30 \hat{j}$$
$$|\vec{D}| = \sqrt{10^2 + (-30)^2}$$
$$|\vec{D}| = \sqrt{100 + 900}$$
$$|\vec{D}| = \sqrt{1000}$$
$$|\vec{D}| = 31.62\, m$$
Answer: 31.62 m
The direction of a vector is expressed by the angle it makes with a reference axis, commonly the positive x-axis. Unit vectors have magnitude 1 and indicate direction only, often used to normalize vectors or express directions in surveying.
The direction angle \(\theta\) of a vector \(\vec{A} = A_x \hat{i} + A_y \hat{j}\) is:
$$\theta = \tan^{-1} (\frac{A_y}{A_x})$$
The unit vector \(\hat{u}\) in the direction of \(\vec{A}\) is:
$$\hat{u} = \frac{\vec{A}}{|\vec{A}|} = \frac{A_x}{|\vec{A}|} \hat{i} + \frac{A_y}{|\vec{A}|} \hat{j}$$
Example 1: Find the direction angle of vector \(\vec{A} = 30\, m \hat{i} + 40\, m \hat{j}\).
Given: \(A_x = 30\, m\), \(A_y = 40\, m\)
$$\theta = \tan^{-1} (\frac{40}{30})$$
$$\theta = \tan^{-1} (1.333)$$
$$\theta = 53.13^\circ$$
Answer: 53.13°
Example 2: Determine the unit vector for \(\vec{A} = 3 \hat{i} + 4 \hat{j}\).
Given: \(A_x = 3\), \(A_y = 4\)
$$|\vec{A}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
$$\hat{u} = \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$$
$$\hat{u} = 0.6 \hat{i} + 0.8 \hat{j}$$
Answer: \(0.6 \hat{i} + 0.8 \hat{j}\)
Example 3: A vector \(\vec{A} = (-5, 12)\). Find its direction angle and unit vector.
Given: \(A_x = -5\), \(A_y = 12\)
$$\theta = \tan^{-1} (\frac{12}{-5})$$
Calculate the angle in the correct quadrant:
$$\theta = \tan^{-1} (-2.4) = -67.38^\circ$$
Since \(A_x < 0\) and \(A_y > 0\), \(\theta = 180^\circ - 67.38^\circ = 112.62^\circ\)
Magnitude:
$$|\vec{A}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$
Unit vector:
$$\hat{u} = \frac{-5}{13} \hat{i} + \frac{12}{13} \hat{j} = -0.385 \hat{i} + 0.923 \hat{j}$$
Answer: Direction angle = 112.62°, Unit vector = \(-0.385 \hat{i} + 0.923 \hat{j}\)
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Create a free accountThis chapter introduced the fundamental concepts of vectors and scalar quantities, emphasizing the differences between these two types of physical quantities. It then explored vector algebra, including operations such as addition, subtraction, and multiplication, with particular focus on the dot product and cross product and their applications in engineering problems. The chapter also covered the concepts of gradient, divergence, and curl, which describe how vector fields change in space and are essential in analyzing fluid flow, electromagnetism, and other engineering phenomena. Through detailed explanations and calculations, students learned how to manipulate vectors algebraically and interpret their physical meanings. The integration of these topics provides a strong foundation for solving complex engineering problems involving multidimensional quantities. Mastery of vector theory is crucial for advanced studies and practical applications in various engineering disciplines.
Given two vectors \(\mathbf{A} = 3\mathbf{i} + 4\mathbf{j}\) and \(\mathbf{B} = 2\mathbf{i} + \mathbf{j}\), calculate the magnitude of \(\mathbf{A}\). (2 marks)
Calculate the scalar product (dot product) of vectors \(\mathbf{A} = 5\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and \(\mathbf{B} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k}\). (3 marks)
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