By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems with confidence and precision.
Engineering Mathematics III for Land Survey professionals involves the application of vector theory to solve practical problems in surveying and mapping. Vectors provide a mathematical framework for representing quantities that have both magnitude and direction, essential for accurate land measurement, boundary delineation, and coordinate computations. This chapter introduces the fundamental concepts of vectors and scalar quantities, focusing on their definitions, properties, and applications within the context of land surveying in Kenya.
Land surveyors work extensively with both scalar and vector quantities. Scalars represent magnitude only, such as length or distance, while vectors incorporate direction, such as displacement or force. Understanding the distinction and manipulation of vectors is critical when calculating traverse coordinates, determining directions, and resolving field measurements.
Scalar quantities are physical measurements described by magnitude alone. Examples in land surveying include distance, area, temperature, and time. Vectors, however, require both magnitude and direction for complete description. Typical vector quantities in surveying include displacement, velocity, and force.
The magnitude of a vector is a non-negative scalar, and its direction is usually represented by an angle or unit vector. Vectors can be graphically represented as directed line segments, with length proportional to magnitude and orientation indicating direction.
The general representation 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.
Example 1: A surveyor measures a displacement of 50 m east. Represent this as a vector.
Given: Displacement magnitude = 50 m, direction = east (0° from x-axis)
$$\vec{D} = 50 \cos 0^\circ \hat{i} + 50 \sin 0^\circ \hat{j}$$
$$= 50 \times 1 \hat{i} + 50 \times 0 \hat{j}$$
$$= 50 \hat{i} + 0 \hat{j}$$
Answer: \(\vec{D} = 50 \hat{i} + 0 \hat{j}\) meters
Example 2: A boundary line runs 40 m north-east at 45°. Find the vector components.
Given: Magnitude = 40 m, angle = 45°
$$\vec{B} = 40 \cos 45^\circ \hat{i} + 40 \sin 45^\circ \hat{j}$$
$$= 40 \times \frac{\sqrt{2}}{2} \hat{i} + 40 \times \frac{\sqrt{2}}{2} \hat{j}$$
$$= 28.28 \hat{i} + 28.28 \hat{j}$$
Answer: \(\vec{B} = 28.28 \hat{i} + 28.28 \hat{j}\) meters
Example 3: A survey line has components \(A_x = -30\) m, \(A_y = 40\) m. Find the magnitude and direction.
Given: \(A_x = -30\) m, \(A_y = 40\) m
Magnitude:
$$|\vec{A}| = \sqrt{(-30)^2 + 40^2}$$
$$= \sqrt{900 + 1600}$$
$$= \sqrt{2500} = 50 \text{ m}$$
Direction:
$$\theta = \tan^{-1} (\frac{40}{-30})$$
$$= \tan^{-1}(-1.333) = -53.13^\circ$$
Since \(A_x\) is negative and \(A_y\) positive, vector lies in the second quadrant:
$$\theta = 180^\circ - 53.13^\circ = 126.87^\circ$$
Answer: Magnitude = 50 m, Direction = 126.87° from positive x-axis
Vector addition and subtraction are fundamental operations for combining or comparing displacements, traverse legs, or other vector quantities in surveying. These operations are performed component-wise.
If \(\vec{A} = A_x \hat{i} + A_y \hat{j}\) and \(\vec{B} = B_x \hat{i} + B_y \hat{j}\), then:
Addition:
$$\vec{R} = \vec{A} + \vec{B} = (A_x + B_x) \hat{i} + (A_y + B_y) \hat{j}$$
Subtraction:
$$\vec{D} = \vec{A} - \vec{B} = (A_x - B_x) \hat{i} + (A_y - B_y) \hat{j}$$
These operations help compute resultant displacements or relative positions between survey points.
Example 1: Two traverse legs are \(\vec{A} = 30 \hat{i} + 40 \hat{j}\) m and \(\vec{B} = -20 \hat{i} + 10 \hat{j}\) m. Find the resultant displacement.
Given:
\(\vec{A} = 30 \hat{i} + 40 \hat{j}\) m
\(\vec{B} = -20 \hat{i} + 10 \hat{j}\) m
$$\vec{R} = (30 + (-20)) \hat{i} + (40 + 10) \hat{j}$$
$$= 10 \hat{i} + 50 \hat{j}$$
Magnitude:
$$|\vec{R}| = \sqrt{10^2 + 50^2} = \sqrt{100 + 2500} = \sqrt{2600} = 50.99 \text{ m}$$
Direction:
$$\theta = \tan^{-1} (\frac{50}{10}) = \tan^{-1}(5) = 78.69^\circ$$
Answer: Resultant displacement = 50.99 m at 78.69°
Example 2: A surveyor needs the vector difference \(\vec{A} - \vec{B}\) where \(\vec{A} = 45 \hat{i} + 20 \hat{j}\) m and \(\vec{B} = 15 \hat{i} + 30 \hat{j}\) m.
Given:
\(\vec{A} = 45 \hat{i} + 20 \hat{j}\) m
\(\vec{B} = 15 \hat{i} + 30 \hat{j}\) m
$$\vec{D} = (45 - 15) \hat{i} + (20 - 30) \hat{j}$$
$$= 30 \hat{i} - 10 \hat{j}$$
Magnitude:
$$|\vec{D}| = \sqrt{30^2 + (-10)^2} = \sqrt{900 + 100} = \sqrt{1000} = 31.62 \text{ m}$$
Direction:
$$\theta = \tan^{-1} (\frac{-10}{30}) = \tan^{-1}(-0.333) = -18.43^\circ$$
Since \(A_x - B_x > 0\) and \(A_y - B_y < 0\), direction is in the fourth quadrant:
$$\theta = 360^\circ - 18.43^\circ = 341.57^\circ$$
Answer: Vector difference = 31.62 m at 341.57°
Example 3: Find the resultant vector of three legs:
\(\vec{A} = 20 \hat{i} + 15 \hat{j}\) m,
\(\vec{B} = -10 \hat{i} + 25 \hat{j}\) m,
\(\vec{C} = 5 \hat{i} - 10 \hat{j}\) m.
Given:
\(\vec{A} = 20 \hat{i} + 15 \hat{j}\) m
\(\vec{B} = -10 \hat{i} + 25 \hat{j}\) m
\(\vec{C} = 5 \hat{i} - 10 \hat{j}\) m
$$\vec{R} = (20 - 10 + 5) \hat{i} + (15 + 25 - 10) \hat{j}$$
$$= 15 \hat{i} + 30 \hat{j}$$
Magnitude:
$$|\vec{R}| = \sqrt{15^2 + 30^2} = \sqrt{225 + 900} = \sqrt{1125} = 33.54 \text{ m}$$
Direction:
$$\theta = \tan^{-1} (\frac{30}{15}) = \tan^{-1}(2) = 63.43^\circ$$
Answer: Resultant vector = 33.54 m at 63.43°
Scalar multiplication scales a vector by a real number, changing its magnitude but not its direction unless the scalar is negative, which reverses direction. Unit vectors have magnitude 1 and indicate direction only. They are useful in expressing directions of survey lines or bearings.
Given a vector \(\vec{A} = A_x \hat{i} + A_y \hat{j}\) and scalar \(k\), scalar multiplication is:
$$k \vec{A} = k A_x \hat{i} + k A_y \hat{j}$$
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: Multiply vector \(\vec{A} = 10 \hat{i} + 15 \hat{j}\) m by scalar \(k = 3\).
Given:
\(\vec{A} = 10 \hat{i} + 15 \hat{j}\) m, \(k = 3\)
$$k \vec{A} = 3 \times 10 \hat{i} + 3 \times 15 \hat{j}$$
$$= 30 \hat{i} + 45 \hat{j}$$
Answer: \(3 \vec{A} = 30 \hat{i} + 45 \hat{j}\) meters
Example 2: Find the unit vector of \(\vec{B} = 6 \hat{i} + 8 \hat{j}\) m.
Given:
\(\vec{B} = 6 \hat{i} + 8 \hat{j}\) m
Magnitude:
$$|\vec{B}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
Unit vector:
$$\hat{u} = \frac{6}{10} \hat{i} + \frac{8}{10} \hat{j} = 0.6 \hat{i} + 0.8 \hat{j}$$
Answer: Unit vector \(\hat{u} = 0.6 \hat{i} + 0.8 \hat{j}\)
Example 3: A vector \(\vec{C} = -12 \hat{i} + 5 \hat{j}\) m is multiplied by \(k = -2\). Find the resulting vector.
Given:
\(\vec{C} = -12 \hat{i} + 5 \hat{j}\) m, \(k = -2\)
$$k \vec{C} = -2 \times (-12) \hat{i} + -2 \times 5 \hat{j}$$
$$= 24 \hat{i} - 10 \hat{j}$$
Answer: \(-2 \vec{C} = 24 \hat{i} - 10 \hat{j}\) meters
Magnitude and direction define a vector completely and are crucial in fieldwork to translate vector components into practical bearings and distances. Magnitude is computed using the Pythagorean theorem, while direction is found using the inverse tangent function, adjusted for quadrant.
For vector \(\vec{A} = A_x \hat{i} + A_y \hat{j}\):
Magnitude:
$$|\vec{A}| = \sqrt{A_x^2 + A_y^2}$$
Direction (angle with the positive x-axis):
$$\theta = \tan^{-1} (\frac{A_y}{A_x})$$
Adjust \(\theta\) based on the signs of \(A_x\) and \(A_y\) to find the correct quadrant.
Example 1: Find the magnitude and direction of \(\vec{A} = 24 \hat{i} + 7 \hat{j}\) m.
Given:
\(A_x = 24\) m, \(A_y = 7\) m
Magnitude:
$$|\vec{A}| = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ m}$$
Direction:
$$\theta = \tan^{-1} (\frac{7}{24}) = \tan^{-1}(0.2917) = 16.26^\circ$$
Answer: Magnitude = 25 m, Direction = 16.26°
Example 2: A vector has components \(-15 \hat{i} + 20 \hat{j}\) m. Calculate magnitude and direction.
Given:
\(A_x = -15\) m, \(A_y = 20\) m
Magnitude:
$$|\vec{A}| = \sqrt{(-15)^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \text{ m}$$
Direction:
$$\theta = \tan^{-1} (\frac{20}{-15}) = \tan^{-1}(-1.333) = -53.13^\circ$$
Since \(A_x < 0\) and \(A_y > 0\), vector lies in second quadrant:
$$\theta = 180^\circ - 53.13^\circ = 126.87^\circ$$
Answer: Magnitude = 25 m, Direction = 126.87°
Example 3: Determine magnitude and direction of \(\vec{A} = -10 \hat{i} - 10 \hat{j}\) m.
Given:
\(A_x = -10\) m, \(A_y = -10\) m
Magnitude:
$$|\vec{A}| = \sqrt{(-10)^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200} = 14.14 \text{ m}$$
Direction:
$$\theta = \tan^{-1} (\frac{-10}{-10}) = \tan^{-1}(1) = 45^\circ$$
Since both components are negative, vector lies in third quadrant:
$$\theta = 180^\circ + 45^\circ = 225^\circ$$
Answer: Magnitude = 14.14 m, Direction = 225°
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 distinction between vectors and scalar quantities, emphasizing their roles in engineering mathematics. It then developed vector algebra, focusing on operations such as addition, subtraction, and multiplication, with detailed explanations of the dot product and cross product and their geometric and physical significance. The chapter further explored advanced vector calculus concepts including gradient, divergence, and curl, which describe how vector fields change in space. These operators are essential for analyzing fluid flow, electromagnetic fields, and structural forces. The mathematical definitions were followed by practical calculation methods to apply these concepts in engineering problems. Overall, the chapter provided a comprehensive foundation for understanding vector theory and its applications in solving real-world engineering challenges.
Given two vectors representing displacement in a survey site: \(\mathbf{A} = 3\mathbf{i} + 4\mathbf{j}\) metres and \(\mathbf{B} = -2\mathbf{i} + 6\mathbf{j}\) metres, calculate the scalar (dot) product \(\mathbf{A} \cdot \mathbf{B}\). (2 marks)
A surveyor records two vectors \(\mathbf{P} = 5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\) m and \(\mathbf{Q} = -1\mathbf{i} + 4\mathbf{j} + 0\mathbf{k}\) m. Find the vector (cross) product \(\mathbf{P} \times \mathbf{Q}\). (3 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.