By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems involving forces and directions effectively.
Vectors are fundamental in chemical engineering for describing quantities that have both magnitude and direction, such as velocity fields in fluid flow, forces on equipment, or gradients in concentration and temperature. This chapter focuses on vector theory in two and three dimensions, providing the mathematical tools needed to analyze these quantities precisely. Mastery of vector operations, including addition, subtraction, dot and cross products, and resolution into components, is essential for solving practical chemical engineering problems like fluid dynamics, process control, and mechanical stresses.
Vectors have both magnitude and direction, while scalars have only magnitude. In chemical engineering, vectors represent quantities such as velocity, force, and displacement, which vary directionally in 2D or 3D space. Scalars include temperature, pressure, and mass. A vector in two dimensions is commonly written as \(\mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j}\), and in three dimensions as \(\mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j} + A_z \mathbf{k}\), where \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) are unit vectors along the x, y, and z axes respectively.
$$ \mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j} + A_z \mathbf{k} $$
Example 1: A fluid velocity vector in a pipeline is given by components \(4\, \text{m/s}\) along x and \(3\, \text{m/s}\) along y. Find the magnitude of the velocity vector.
Given: \(A_x = 4\, \text{m/s}\), \(A_y = 3\, \text{m/s}\)
$$ |\mathbf{A}| = \sqrt{A_x^2 + A_y^2} $$
$$ = \sqrt{4^2 + 3^2} $$
$$ = \sqrt{16 + 9} $$
$$ = \sqrt{25} = 5\, \text{m/s} $$
Answer: 5 m/s
Example 2: A force vector acting on a tank is \(\mathbf{F} = 5 \mathbf{i} + 7 \mathbf{j} + 2 \mathbf{k}\) kN. Calculate the magnitude of the force.
Given: \(F_x = 5\, \text{kN}\), \(F_y = 7\, \text{kN}\), \(F_z = 2\, \text{kN}\)
$$ |\mathbf{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2} $$
$$ = \sqrt{5^2 + 7^2 + 2^2} $$
$$ = \sqrt{25 + 49 + 4} $$
$$ = \sqrt{78} = 8.83\, \text{kN} $$
Answer: 8.83 kN
Example 3: A displacement vector for a chemical plant component is given as \(\mathbf{D} = -3 \mathbf{i} + 4 \mathbf{j}\) m. Find its magnitude and direction angle \(\theta\) from the positive x-axis.
Given: \(D_x = -3\, \text{m}\), \(D_y = 4\, \text{m}\)
Magnitude:
$$ |\mathbf{D}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, \text{m} $$
Direction angle:
$$ \theta = \tan^{-1} (\frac{D_y}{D_x}) = \tan^{-1} (\frac{4}{-3}) $$
$$ = \tan^{-1}(-1.333) = -53.13^\circ $$
Since \(D_x\) is negative and \(D_y\) positive, vector lies in the second quadrant:
$$ \theta = 180^\circ - 53.13^\circ = 126.87^\circ $$
Answer: Magnitude = 5 m, Direction = 126.87° from positive x-axis
Example 4: At Bamburi Cement Ltd, a conveyor belt moves with a velocity vector given by components \(5\, \text{m/s}\) along x and \(12\, \text{m/s}\) along y. Find the magnitude of the velocity vector.
Given: \(A_x = 5\, \text{m/s}\), \(A_y = 12\, \text{m/s}\)
$$ |\mathbf{A}| = \sqrt{A_x^2 + A_y^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\, \text{m/s} $$
Answer: 13 m/s
Vector addition combines two or more vectors to find a resultant vector, important in chemical engineering for net forces or velocities. Vector subtraction finds the difference between vectors, useful in relative velocity or force calculations.
Vector addition formula for two vectors \(\mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j} + A_z \mathbf{k}\) and \(\mathbf{B} = B_x \mathbf{i} + B_y \mathbf{j} + B_z \mathbf{k}\):
$$ \mathbf{R} = \mathbf{A} + \mathbf{B} = (A_x + B_x) \mathbf{i} + (A_y + B_y) \mathbf{j} + (A_z + B_z) \mathbf{k} $$
Vector subtraction:
$$ \mathbf{D} = \mathbf{A} - \mathbf{B} = (A_x - B_x) \mathbf{i} + (A_y - B_y) \mathbf{j} + (A_z - B_z) \mathbf{k} $$
Example 1: Two velocity vectors in a mixing tank are \(\mathbf{V}_1 = 3 \mathbf{i} + 2 \mathbf{j}\) m/s and \(\mathbf{V}_2 = -1 \mathbf{i} + 4 \mathbf{j}\) m/s. Find the resultant velocity vector.
Given: \(\mathbf{V}_1\), \(\mathbf{V}_2\)
$$ \mathbf{R} = \mathbf{V}_1 + \mathbf{V}_2 = (3 + (-1)) \mathbf{i} + (2 + 4) \mathbf{j} $$
$$ = 2 \mathbf{i} + 6 \mathbf{j} $$
Magnitude:
$$ |\mathbf{R}| = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 6.32\, \text{m/s} $$
Answer: \(\mathbf{R} = 2 \mathbf{i} + 6 \mathbf{j}\) m/s, Magnitude = 6.32 m/s
Example 2: A force \(\mathbf{F}_1 = 7 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k}\) kN acts on a vessel, opposed by \(\mathbf{F}_2 = 2 \mathbf{i} + 6 \mathbf{j} + 1 \mathbf{k}\) kN. Find the net force vector.
Given: \(\mathbf{F}_1\), \(\mathbf{F}_2\)
$$ \mathbf{F}_{net} = \mathbf{F}_1 - \mathbf{F}_2 = (7 - 2) \mathbf{i} + (3 - 6) \mathbf{j} + (4 - 1) \mathbf{k} $$
$$ = 5 \mathbf{i} - 3 \mathbf{j} + 3 \mathbf{k} $$
Magnitude:
$$ |\mathbf{F}_{net}| = \sqrt{5^2 + (-3)^2 + 3^2} = \sqrt{25 + 9 + 9} = \sqrt{43} = 6.56\, \text{kN} $$
Answer: \(\mathbf{F}_{net} = 5 \mathbf{i} - 3 \mathbf{j} + 3 \mathbf{k}\) kN, Magnitude = 6.56 kN
Example 3: Two displacement vectors are \(\mathbf{D}_1 = 4 \mathbf{i} - 2 \mathbf{j} + 1 \mathbf{k}\) m and \(\mathbf{D}_2 = -3 \mathbf{i} + 5 \mathbf{j} + 2 \mathbf{k}\) m. Find \(\mathbf{D}_1 + \mathbf{D}_2\) and \(\mathbf{D}_1 - \mathbf{D}_2\).
Given: \(\mathbf{D}_1\), \(\mathbf{D}_2\)
Addition:
$$ \mathbf{D}_{sum} = (4 + (-3)) \mathbf{i} + (-2 + 5) \mathbf{j} + (1 + 2) \mathbf{k} = 1 \mathbf{i} + 3 \mathbf{j} + 3 \mathbf{k} $$
Subtraction:
$$ \mathbf{D}_{diff} = (4 - (-3)) \mathbf{i} + (-2 - 5) \mathbf{j} + (1 - 2) \mathbf{k} = 7 \mathbf{i} - 7 \mathbf{j} - 1 \mathbf{k} $$
Magnitudes:
$$ |\mathbf{D}_{sum}| = \sqrt{1^2 + 3^2 + 3^2} = \sqrt{1 + 9 + 9} = \sqrt{19} = 4.36\, \text{m} $$
$$ |\mathbf{D}_{diff}| = \sqrt{7^2 + (-7)^2 + (-1)^2} = \sqrt{49 + 49 + 1} = \sqrt{99} = 9.95\, \text{m} $$
Answer:
\(\mathbf{D}_{sum} = 1 \mathbf{i} + 3 \mathbf{j} + 3 \mathbf{k}\) m, Magnitude = 4.36 m
\(\mathbf{D}_{diff} = 7 \mathbf{i} - 7 \mathbf{j} - 1 \mathbf{k}\) m, Magnitude = 9.95 m
Vector multiplication includes the dot product, producing a scalar, and the cross product, producing a vector. The dot product is useful for work done by a force or projection of vectors, while the cross product is used to find moments or perpendicular vectors.
For vectors \(\mathbf{A}\) and \(\mathbf{B}\):
$$ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta = A_x B_x + A_y B_y + A_z B_z $$
where \(\theta\) is the angle between \(\mathbf{A}\) and \(\mathbf{B}\).
$$ \mathbf{A} \times \mathbf{B} = (A_y B_z - A_z B_y) \mathbf{i} - (A_x B_z - A_z B_x) \mathbf{j} + (A_x B_y - A_y B_x) \mathbf{k} $$
The magnitude is
$$ |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin \theta $$
Example 1 (Dot product): A force \(\mathbf{F} = 6 \mathbf{i} + 2 \mathbf{j}\) kN acts on a pipe segment. The displacement vector is \(\mathbf{d} = 3 \mathbf{i} + 4 \mathbf{j}\) m. Find the work done by the force.
Given: \(\mathbf{F}\), \(\mathbf{d}\)
$$ W = \mathbf{F} \cdot \mathbf{d} = F_x d_x + F_y d_y = 6 \times 3 + 2 \times 4 $$
$$ = 18 + 8 = 26\, \text{kN·m} $$
Answer: 26 kN·m
Example 2 (Dot product): Two velocity vectors in a reactor are \(\mathbf{V}_1 = 5 \mathbf{i} + 3 \mathbf{j} + 2 \mathbf{k}\) m/s and \(\mathbf{V}_2 = 1 \mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k}\) m/s. Find the dot product and the angle between them.
Given: \(\mathbf{V}_1\), \(\mathbf{V}_2\)
Dot product:
$$ \mathbf{V}_1 \cdot \mathbf{V}_2 = 5 \times 1 + 3 \times 4 + 2 \times 3 = 5 + 12 + 6 = 23 $$
Magnitudes:
$$ |\mathbf{V}_1| = \sqrt{5^2 + 3^2 + 2^2} = \sqrt{25 + 9 + 4} = \sqrt{38} = 6.16 $$
$$ |\mathbf{V}_2| = \sqrt{1^2 + 4^2 + 3^2} = \sqrt{1 + 16 + 9} = \sqrt{26} = 5.10 $$
Angle:
$$ \cos \theta = \frac{\mathbf{V}_1 \cdot \mathbf{V}_2}{|\mathbf{V}_1| |\mathbf{V}_2|} = \frac{23}{6.16 \times 5.10} = \frac{23}{31.42} = 0.732 $$
$$ \theta = \cos^{-1} (0.732) = 43.0^\circ $$
Answer: Dot product = 23, Angle = 43.0°
Example 3 (Cross product): Find the moment vector \(\mathbf{M} = \mathbf{r} \times \mathbf{F}\) where \(\mathbf{r} = 2 \mathbf{i} + 3 \mathbf{j} + \mathbf{k}\) m is the position vector from the pivot and \(\mathbf{F} = 4 \mathbf{i} + 5 \mathbf{j} + 6 \mathbf{k}\) N is the applied force.
Given: \(\mathbf{r}\), \(\mathbf{F}\)
Compute components:
$$ M_x = (3)(6) - (1)(5) = 18 - 5 = 13 $$
$$ M_y = - [(2)(6) - (1)(4)] = - (12 - 4) = -8 $$
$$ M_z = (2)(5) - (3)(4) = 10 - 12 = -2 $$
So,
$$ \mathbf{M} = 13 \mathbf{i} - 8 \mathbf{j} - 2 \mathbf{k} \, \text{N·m} $$
Magnitude:
$$ |\mathbf{M}| = \sqrt{13^2 + (-8)^2 + (-2)^2} = \sqrt{169 + 64 + 4} = \sqrt{237} = 15.4\, \text{N·m} $$
Answer: \(\mathbf{M} = 13 \mathbf{i} - 8 \mathbf{j} - 2 \mathbf{k}\) N·m, Magnitude = 15.4 N·m
Example 4 (Cross product): Two velocity vectors in 3D are \(\mathbf{V}_1 = 3 \mathbf{i} + 2 \mathbf{j} + \mathbf{k}\) m/s and \(\mathbf{V}_2 = \mathbf{i} + 4 \mathbf{j} + 5 \mathbf{k}\) m/s. Find \(\mathbf{V}_1 \times \mathbf{V}_2\).
Given: \(\mathbf{V}_1\), \(\mathbf{V}_2\)
Compute components:
$$ C_x = (2)(5) - (1)(4) = 10 - 4 = 6 $$
$$ C_y = -[(3)(5) - (1)(1)] = - (15 - 1) = -14 $$
$$ C_z = (3)(4) - (2)(1) = 12 - 2 = 10 $$
So,
$$ \mathbf{V}_1 \times \mathbf{V}_2 = 6 \mathbf{i} - 14 \mathbf{j} + 10 \mathbf{k} $$
Magnitude:
$$ |\mathbf{V}_1 \times \mathbf{V}_2| = \sqrt{6^2 + (-14)^2 + 10^2} = \sqrt{36 + 196 + 100} = \sqrt{332} = 18.22\, \text{m}^2/\text{s}^2 $$
Answer: \(\mathbf{V}_1 \times \mathbf{V}_2 = 6 \mathbf{i} - 14 \mathbf{j} + 10 \mathbf{k}\), Magnitude = 18.22
Position vectors describe the location of a point relative to an origin, essential in chemical engineering for locating equipment or fluid particles in space. A position vector \(\mathbf{r}\) from origin O to point P is defined by its coordinates in 2D or 3D.
$$ \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} $$
The position vector can be used to calculate distance between points or relative positions.
Example 1: Find the position vector of point P with coordinates (3, 4) m from origin O in 2D.
Given: \(x = 3\, \text{m}\), \(y = 4\, \text{m}\)
$$ \mathbf{r} = 3 \mathbf{i} + 4 \mathbf{j} $$
Magnitude (distance OP):
$$ |\mathbf{r}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, \text{m} $$
Answer: \(\mathbf{r} = 3 \mathbf{i} + 4 \mathbf{j}\), Distance = 5 m
Example 2: In a chemical plant, a valve is located at coordinates (5, -2, 3) m. Find the position vector of the valve from the origin.
Given: \(x=5\, \text{m}\), \(y=-2\, \text{m}\), \(z=3\, \text{m}\)
$$ \mathbf{r} = 5 \mathbf{i} - 2 \mathbf{j} + 3 \mathbf{k} $$
Distance:
$$ |\mathbf{r}| = \sqrt{5^2 + (-2)^2 + 3^2} = \sqrt{25 + 4 + 9} = \sqrt{38} = 6.16\, \text{m} $$
Answer: \(\mathbf{r} = 5 \mathbf{i} - 2 \mathbf{j} + 3 \mathbf{k}\), Distance = 6.16 m
Example 3: Find the vector from point A(2, 3, 1) m to point B(5, 7, 4) m.
Given: Coordinates of A and B
$$ \mathbf{r}_{AB} = \mathbf{r}_B - \mathbf{r}_A = (5 - 2) \mathbf{i} + (7 - 3) \mathbf{j} + (4 - 1) \mathbf{k} = 3 \mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k} $$
Magnitude:
$$ |\mathbf{r}_{AB}| = \sqrt{3^2 + 4^2 + 3^2} = \sqrt{9 + 16 + 9} = \sqrt{34} = 5.83\, \text{m} $$
Answer: \(\mathbf{r}_{AB} = 3 \mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k}\), Distance AB = 5.83 m
Example 4: At Kenya Breweries Ltd, a pump is located at coordinates (6, 2, -4) m from the origin. Find the position vector and distance from the origin.
Given: \(x=6\, \text{m}\), \(y=2\, \text{m}\), \(z=-4\, \text{m}\)
$$ \mathbf{r} = 6 \mathbf{i} + 2 \mathbf{j} - 4 \mathbf{k} $$
Distance:
$$ |\mathbf{r}| = \sqrt{6^2 + 2^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 7.48\, \text{m} $$
Answer: \(\mathbf{r} = 6 \mathbf{i} + 2 \mathbf{j} - 4 \mathbf{k}\), Distance = 7.48 m
Resolution of a vector means expressing it as components along mutually perpendicular axes, typically x, y, and z. This is crucial in chemical engineering to analyze forces, velocities, or fluxes along different directions.
If a vector \(\mathbf{A}\) has magnitude \(A\) and makes angles \(\alpha, \beta, \gamma\) with x, y, z axes respectively:
$$ A_x = A \cos \alpha, \quad A_y = A \cos \beta, \quad A_z = A \cos \gamma $$
Example 1: Resolve a force of magnitude 10 kN acting at 30° to the horizontal (x-axis) in 2D into components along x and y.
Given: \(A = 10\, \text{kN}\), \(\alpha = 30^\circ\)
Components:
$$ A_x = 10 \cos 30^\circ = 10 \times 0.866 = 8.66\, \text{kN} $$
$$ A_y = 10 \sin 30^\circ = 10 \times 0.5 = 5\, \text{kN} $$
Answer: \(A_x = 8.66\, \text{kN}\), \(A_y = 5\, \text{kN}\)
Example 2: A velocity vector of magnitude 15 m/s makes angles 45°, 60°, and 90° with x, y, and z axes respectively. Find its components.
Given: \(A = 15\, \text{m/s}\), \(\alpha = 45^\circ\), \(\beta = 60^\circ\), \(\gamma = 90^\circ\)
Components:
$$ A_x = 15 \cos 45^\circ = 15 \times 0.707 = 10.61\, \text{m/s} $$
$$ A_y = 15 \cos 60^\circ = 15 \times 0.5 = 7.5\, \text{m/s} $$
$$ A_z = 15 \cos 90^\circ = 15 \times 0 = 0\, \text{m/s} $$
Answer: \(A_x = 10.61\, \text{m/s}\), \(A_y = 7.5\, \text{m/s}\), \(A_z = 0\)
Example 3: Resolve a displacement vector of magnitude 20 m that makes equal angles with the x, y, and z axes into its components.
Given: \(A = 20\, \text{m}\), \(\alpha = \beta = \gamma\)
Since the vector makes equal angles with all axes, \(\cos \alpha = \cos \beta = \cos \gamma\).
Using the property for direction cosines:
$$ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 $$
$$ 3 \cos^2 \alpha = 1 \Rightarrow \cos \alpha = \frac{1}{\sqrt{3}} = 0.577 $$
Components:
$$ A_x = 20 \times 0.577 = 11.54\, \text{m} $$
$$ A_y = 11.54\, \text{m} $$
$$ A_z = 11.54\, \text{m} $$
Answer: \(A_x = A_y = A_z = 11.54\, \text{m}\)
Example 4: At Kenya Electricity Generating Company (KenGen), a force of magnitude 18 kN acts at 40° to the horizontal (x-axis). Resolve the force into x and y components.
Given: \(A = 18\, \text{kN}\), \(\alpha = 40^\circ\)
Components:
$$ A_x = 18 \cos 40^\circ = 18 \times 0.766 = 13.79\, \text{kN} $$
$$ A_y = 18 \sin 40^\circ = 18 \times 0.643 = 11.57\, \text{kN} $$
Answer: \(A_x = 13.79\, \text{kN}\), \(A_y = 11.57\, \text{kN}\)
A chemical fluid flows in a pipe with velocity components \(6 \mathbf{i} + 8 \mathbf{j}\) m/s. Calculate the magnitude and direction of velocity relative to the x-axis. (5 marks)
Two forces act on a reactor: \(\mathbf{F}_1 = 8 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k}\) kN and \(\mathbf{F}_2 = 5 \mathbf{i} + 7 \mathbf{j} + 6 \mathbf{k}\) kN. Find the resultant force vector and its magnitude. (7 marks)
Calculate the work done by a force \(\mathbf{F} = 10 \mathbf{i} + 5 \mathbf{j}\) kN when moving a point through displacement \(\mathbf{d} = 2 \mathbf{i} + 3 \mathbf{j}\) m. (5 marks)
A position vector locates a pump at (4, -3, 5) m. Find the distance from origin and write the position vector. (5 marks)
Resolve a force of magnitude 25 kN acting at angles 60°, 45°, and 30° with the x, y, and z axes respectively into its components. (8 marks)
This chapter explored the fundamental concepts of vectors and scalars in both two and three dimensions, highlighting their distinct characteristics. It detailed the various operations on vectors including addition, subtraction, and different types of multiplication, providing the basis for manipulating vector quantities. The concept of position vectors was introduced to describe the location of points in space relative to an origin. The chapter then examined the dot product, explaining how it produces a scalar and its applications in determining angles between vectors. Following this, the cross product was discussed, showing how it results in a vector perpendicular to the plane containing the original vectors. Finally, the resolution of vectors into components was covered, enabling the analysis of vector quantities along specified directions. Together, these topics form the foundation for applying vector theory in engineering mathematics.
A chemical engineer represents the velocity of fluid flow in a pipe as vector \(\vec{A} = 3\hat{i} + 4\hat{j}\) m/s and the velocity of a second fluid stream as \(\vec{B} = 1\hat{i} + 2\hat{j}\) m/s. Calculate the resultant velocity vector \(\vec{R} = \vec{A} + \vec{B}\).
Given two vectors \(\vec{P} = 5\hat{i} - 2\hat{j} + 3\hat{k}\) and \(\vec{Q} = -3\hat{i} + 4\hat{j} + \hat{k}\), find the vector \(\vec{S} = \vec{P} - \vec{Q}\).
A position vector \(\vec{r}\) points from the origin to a mixing tank located at coordinates (6 m, 8 m, 2 m). Calculate the magnitude of \(\vec{r}\).
Calculate the dot product of vectors \(\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k}\) and \(\vec{B} = -1\hat{i} + 4\hat{j} + 2\hat{k}\).
Find the cross product \(\vec{C} = \vec{A} \times \vec{B}\) where \(\vec{A} = \hat{i} + 2\hat{j} + 3\hat{k}\) and \(\vec{B} = 4\hat{i} - \hat{j} + 2\hat{k}\).
A chemical pipeline runs from point \(P(2, -1, 3)\) to point \(Q(5, 3, 7)\). Find the position vector from \(P\) to \(Q\).
Resolve the vector \(\vec{V} = 10\hat{i} + 15\hat{j}\) into components along \(\hat{i}\) and \(\hat{j}\) unit vectors, and calculate the magnitude of \(\vec{V}\).
Two forces act on a chemical reactor: \(\vec{F_1} = 8\hat{i} + 6\hat{j} + 2\hat{k}\) N and \(\vec{F_2} = -3\hat{i} + 4\hat{j} + 5\hat{k}\) N. Calculate the angle between the two forces.
A fluid velocity vector is \(\vec{u} = 3\hat{i} + 6\hat{j} + 2\hat{k}\) m/s. Find the component of \(\vec{u}\) along the vector \(\vec{v} = \hat{i} + 2\hat{j} + 2\hat{k}\).
A chemical plant has a vector force \(\vec{F} = 12\hat{i} - 5\hat{j} + 9\hat{k}\) N acting on a valve. Calculate the moment (torque) about the origin if the position vector of the valve is \(\vec{r} = 4\hat{i} + 3\hat{j} - 2\hat{k}\) m.
1. Resultant velocity vector \(\vec{R} = \vec{A} + \vec{B}\):
Given:
\(\vec{A} = 3\hat{i} + 4\hat{j}\) m/s
\(\vec{B} = 1\hat{i} + 2\hat{j}\) m/s
$$\vec{R} = \vec{A} + \vec{B} = (3 + 1)\hat{i} + (4 + 2)\hat{j} = 4\hat{i} + 6\hat{j} \text{ m/s}$$
Answer: \(\boxed{4\hat{i} + 6\hat{j}\ \text{m/s}}\)
2. Vector subtraction \(\vec{S} = \vec{P} - \vec{Q}\):
Given:
\(\vec{P} = 5\hat{i} - 2\hat{j} + 3\hat{k}\)
\(\vec{Q} = -3\hat{i} + 4\hat{j} + \hat{k}\)
$$\vec{S} = (5 - (-3))\hat{i} + (-2 - 4)\hat{j} + (3 - 1)\hat{k} = (8)\hat{i} + (-6)\hat{j} + (2)\hat{k}$$
Answer: \(\boxed{8\hat{i} - 6\hat{j} + 2\hat{k}}\)
3. Magnitude of position vector \(\vec{r}\):
Given coordinates: (6 m, 8 m, 2 m)
$$|\vec{r}| = \sqrt{6^2 + 8^2 + 2^2} = \sqrt{36 + 64 + 4} = \sqrt{104}$$
$$|\vec{r}| = 10.198 \text{ m}$$
Answer: \(\boxed{10.20\ \text{m}}\)
4. Dot product \(\vec{A} \cdot \vec{B}\):
Given:
\(\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k}\)
\(\vec{B} = -1\hat{i} + 4\hat{j} + 2\hat{k}\)
$$\vec{A} \cdot \vec{B} = (2)(-1) + (3)(4) + (1)(2) = -2 + 12 + 2 = 12$$
Answer: \(\boxed{12}\)
5. Cross product \(\vec{C} = \vec{A} \times \vec{B}\):
Given:
\(\vec{A} = \hat{i} + 2\hat{j} + 3\hat{k}\)
\(\vec{B} = 4\hat{i} - \hat{j} + 2\hat{k}\)
Calculate determinant:
$$\vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 4 & -1 & 2 \end{vmatrix} = \hat{i}(2 \times 2 - 3 \times (-1)) - \hat{j}(1 \times 2 - 3 \times 4) + \hat{k}(1 \times (-1) - 2 \times 4)$$
$$\vec{C} = \hat{i}(4 + 3) - \hat{j}(2 - 12) + \hat{k}(-1 - 8) = 7\hat{i} - (-10)\hat{j} - 9\hat{k} = 7\hat{i} + 10\hat{j} - 9\hat{k}$$
Answer: \(\boxed{7\hat{i} + 10\hat{j} - 9\hat{k}}\)
6. Position vector from \(P\) to \(Q\):
Given:
\(P(2, -1, 3)\), \(Q(5, 3, 7)\)
$$\vec{PQ} = (5 - 2)\hat{i} + (3 - (-1))\hat{j} + (7 - 3)\hat{k} = 3\hat{i} + 4\hat{j} + 4\hat{k}$$
Answer: \(\boxed{3\hat{i} + 4\hat{j} + 4\hat{k}}\)
7. Resolution and magnitude of \(\vec{V} = 10\hat{i} + 15\hat{j}\):
Components are already along \(\hat{i}\) and \(\hat{j}\):
Magnitude:
$$|\vec{V}| = \sqrt{10^2 + 15^2} = \sqrt{100 + 225} = \sqrt{325} = 18.03$$
Answer: Components are \(10\hat{i}\) and \(15\hat{j}\), magnitude \(\boxed{18.03}\)
8. Angle between forces \(\vec{F_1}\) and \(\vec{F_2}\):
Given:
\(\vec{F_1} = 8\hat{i} + 6\hat{j} + 2\hat{k}\), \(\vec{F_2} = -3\hat{i} + 4\hat{j} + 5\hat{k}\)
Calculate dot product:
$$\vec{F_1} \cdot \vec{F_2} = (8)(-3) + (6)(4) + (2)(5) = -24 + 24 + 10 = 10$$
Magnitudes:
$$|\vec{F_1}| = \sqrt{8^2 + 6^2 + 2^2} = \sqrt{64 + 36 + 4} = \sqrt{104} = 10.198$$
$$|\vec{F_2}| = \sqrt{(-3)^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 7.071$$
Calculate angle \(\theta\):
$$\cos\theta = \frac{\vec{F_1} \cdot \vec{F_2}}{|\vec{F_1}||\vec{F_2}|} = \frac{10}{10.198 \times 7.071} = \frac{10}{72.10} = 0.1387$$
$$\theta = \cos^{-1}(0.1387) = 82.04^\circ$$
Answer: \(\boxed{82.0^\circ}\)
9. Component of \(\vec{u}\) along \(\vec{v}\):
Given:
\(\vec{u} = 3\hat{i} + 6\hat{j} + 2\hat{k}\)
\(\vec{v} = \hat{i} + 2\hat{j} + 2\hat{k}\)
Calculate dot product:
$$\vec{u} \cdot \vec{v} = 3 \times 1 + 6 \times 2 + 2 \times 2 = 3 + 12 + 4 = 19$$
Magnitude of \(\vec{v}\):
$$|\vec{v}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$
Component magnitude:
$$\text{Comp}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|} = \frac{19}{3} = 6.333 \text{ m/s}$$
Answer: \(\boxed{6.33\ \text{m/s}}\)
10. Moment (torque) \(\vec{M} = \vec{r} \times \vec{F}\):
Given:
\(\vec{r} = 4\hat{i} + 3\hat{j} - 2\hat{k}\) m
\(\vec{F} = 12\hat{i} - 5\hat{j} + 9\hat{k}\) N
Calculate:
$$\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & -2 \\ 12 & -5 & 9 \end{vmatrix} = \hat{i}(3 \times 9 - (-2) \times (-5)) - \hat{j}(4 \times 9 - (-2) \times 12) + \hat{k}(4 \times (-5) - 3 \times 12)$$
$$= \hat{i}(27 - 10) - \hat{j}(36 - (-24)) + \hat{k}(-20 - 36)$$
$$= 17\hat{i} - (36 + 24)\hat{j} - 56\hat{k} = 17\hat{i} - 60\hat{j} - 56\hat{k}$$
Answer: \(\boxed{17\hat{i} - 60\hat{j} - 56\hat{k}\ \text{Nm}}\)
A chemical engineer at a Nairobi refinery represents the velocity of two fluid streams as vectors \(\mathbf{A} = 3\mathbf{i} + 2\mathbf{j}\) m/s and \(\mathbf{B} = -\mathbf{i} + 4\mathbf{j}\) m/s. Find the resultant velocity vector \(\mathbf{R} = \mathbf{A} + \mathbf{B}\). (4 marks)
Given two position vectors \(\mathbf{P} = 5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\) m and \(\mathbf{Q} = 2\mathbf{i} + 7\mathbf{j} + \mathbf{k}\) m, find the vector \(\mathbf{PQ}\) from point \(P\) to \(Q\). (4 marks)
Calculate the dot product of vectors \(\mathbf{A} = 4\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) and \(\mathbf{B} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}\), and state its significance in the context of fluid flow direction in a chemical plant. (4 marks)
For vectors \(\mathbf{A} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\) and \(\mathbf{B} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}\), calculate the cross product \(\mathbf{A} \times \mathbf{B}\). (4 marks)
A pipeline runs from point \(A(1,2,3)\) to point \(B(4,6,8)\) within a chemical processing plant. Find the unit vector in the direction from \(A\) to \(B\). (4 marks)
Two forces act on a valve in a chemical plant: \(\mathbf{F}_1 = 6\mathbf{i} + 3\mathbf{j}\) N and \(\mathbf{F}_2 = -2\mathbf{i} + 5\mathbf{j}\) N. Find the magnitude of the resultant force. (4 marks)
Resolve the vector \(\mathbf{V} = 5\mathbf{i} + 12\mathbf{j}\) N into components parallel and perpendicular to the vector \(\mathbf{U} = 3\mathbf{i} + 4\mathbf{j}\) N. (4 marks)
Find the angle between vectors \(\mathbf{A} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}\) and \(\mathbf{B} = \mathbf{i} - \mathbf{j} + 4\mathbf{k}\). (4 marks)
Calculate the scalar projection of vector \(\mathbf{A} = 7\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}\) on vector \(\mathbf{B} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k}\). (4 marks)
A chemical engineer needs to find the moment of force about a point due to force \(\mathbf{F} = 10\mathbf{i} + 15\mathbf{j}\) N applied at position vector \(\mathbf{r} = 2\mathbf{i} + 3\mathbf{j}\) m. Calculate the moment vector \(\mathbf{M} = \mathbf{r} \times \mathbf{F}\). (4 marks)
\(\mathbf{R} = (3 - 1)\mathbf{i} + (2 + 4)\mathbf{j} = 2\mathbf{i} + 6\mathbf{j}\) m/s
\(\mathbf{PQ} = \mathbf{Q} - \mathbf{P} = (2 - 5)\mathbf{i} + (7 - 3)\mathbf{j} + (1 - 2)\mathbf{k} = -3\mathbf{i} + 4\mathbf{j} - \mathbf{k}\) m
Dot product \(= 4 \times 2 + (-1) \times 1 + 3 \times (-1) = 8 - 1 - 3 = 4\). Significance: Positive value indicates vectors have a component in the same direction, relevant for flow alignment.
\(\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 4 & -1 & 1 \end{vmatrix} = (2 \times 1 - 3 \times (-1))\mathbf{i} - (1 \times 1 - 3 \times 4)\mathbf{j} + (1 \times (-1) - 2 \times 4)\mathbf{k} = (2 + 3)\mathbf{i} - (1 - 12)\mathbf{j} + (-1 - 8)\mathbf{k} = 5\mathbf{i} + 11\mathbf{j} - 9\mathbf{k}\)
\(\mathbf{AB} = (4-1)\mathbf{i} + (6-2)\mathbf{j} + (8-3)\mathbf{k} = 3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\)
Magnitude \(= \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 7.071\)
Unit vector \(= \frac{1}{7.071}(3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) = 0.424\mathbf{i} + 0.566\mathbf{j} + 0.707\mathbf{k}\)
Resultant force \(\mathbf{F} = (6 - 2)\mathbf{i} + (3 + 5)\mathbf{j} = 4\mathbf{i} + 8\mathbf{j}\) N
Magnitude \(= \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 8.944\) N
\(\mathbf{U}\) magnitude \(= \sqrt{3^2 + 4^2} = 5\)
Scalar projection of \(\mathbf{V}\) on \(\mathbf{U}\) \(= \frac{\mathbf{V} \cdot \mathbf{U}}{|\mathbf{U}|} = \frac{5 \times 3 + 12 \times 4}{5} = \frac{15 + 48}{5} = \frac{63}{5} = 12.6\)
Parallel component \(= \frac{\mathbf{V} \cdot \mathbf{U}}{|\mathbf{U}|^2} \mathbf{U} = \frac{63}{25}(3\mathbf{i} + 4\mathbf{j}) = 7.56\mathbf{i} + 10.08\mathbf{j}\)
Perpendicular component \(= \mathbf{V} -\) parallel component \(= (5 - 7.56)\mathbf{i} + (12 - 10.08)\mathbf{j} = -2.56\mathbf{i} + 1.92\mathbf{j}\)
Dot product \(= 2 \times 1 + 3 \times (-1) + 1 \times 4 = 2 - 3 + 4 = 3\)
Magnitudes: \(|\mathbf{A}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} = 3.742\)
\(|\mathbf{B}| = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 4.243\)
Angle \(\theta = \cos^{-1} (\frac{3}{3.742 \times 4.243}) = \cos^{-1}(0.189) = 79.1^\circ\)
Scalar projection \(= \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|}\)
\(\mathbf{A} \cdot \mathbf{B} = 7 \times 3 + 2 \times 1 + 6 \times 2 = 21 + 2 + 12 = 35\)
\(|\mathbf{B}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} = 3.742\)
Scalar projection \(= \frac{35}{3.742} = 9.354\)
Moment \(\mathbf{M} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 0 \\ 10 & 15 & 0 \end{vmatrix} = (3 \times 0 - 0 \times 15)\mathbf{i} - (2 \times 0 - 0 \times 10)\mathbf{j} + (2 \times 15 - 3 \times 10)\mathbf{k} = 0\mathbf{i} - 0\mathbf{j} + (30 - 30)\mathbf{k} = \mathbf{0}\)
Question 11 (Compulsory - 20 marks)
A chemical engineer at Kenya Pipeline Company is analyzing forces acting on a storage tank support system. The position vectors of two points on the support beams are \(\mathbf{A} = 4\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) m and \(\mathbf{B} = 7\mathbf{i} + 5\mathbf{j} + 4\mathbf{k}\) m. Forces at these points are \(\mathbf{F}_A = 10\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\) N and \(\mathbf{F}_B = -3\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}\) N respectively.
a) Calculate the resultant force vector acting on the system. (10 marks)
b) Determine the total moment about the origin due to the forces at points \(A\) and \(B\). (10 marks)
Question 12 (20 marks)
In a chemical reactor, the velocity vectors of two reactant streams entering at a junction are \(\mathbf{V}_1 = 6\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\) m/s and \(\mathbf{V}_2 = -2\mathbf{i} + 5\mathbf{j} + \mathbf{k}\) m/s. Determine:
a) The resultant velocity vector of the mixed stream.
b) The angle between the two velocity vectors.
c) The magnitude of the cross product of \(\mathbf{V}_1\) and \(\mathbf{V}_2\).
Question 13 (20 marks)
A chemical engineer measures the force \(\mathbf{F} = 8\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}\) N applied at position vector \(\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) m on a pump flange.
a) Calculate the moment vector about the origin.
b) Find the component of the force parallel to the position vector \(\mathbf{r}\).
c) Find the component of the force perpendicular to \(\mathbf{r}\).
Question 14 (20 marks)
At a chemical processing plant, a vector \(\mathbf{A} = 5\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\) represents the direction of heat flow, and vector \(\mathbf{B} = 1\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\) represents the orientation of a thermocouple sensor.
a) Determine the dot product and interpret its meaning.
b) Calculate the unit vector perpendicular to both \(\mathbf{A}\) and \(\mathbf{B}\).
c) Find the angle between \(\mathbf{A}\) and \(\mathbf{B}\).
Question 11
a) Resultant force:
\(\mathbf{F}_R = \mathbf{F}_A + \mathbf{F}_B = (10 - 3)\mathbf{i} + (5 + 6)\mathbf{j} + (2 + 4)\mathbf{k} = 7\mathbf{i} + 11\mathbf{j} + 6\mathbf{k}\) N
b) Moments:
\(\mathbf{M}_A = \mathbf{r}_A \times \mathbf{F}_A = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 2 & 1 \\ 10 & 5 & 2 \end{vmatrix} = (2 \times 2 - 1 \times 5)\mathbf{i} - (4 \times 2 - 1 \times 10)\mathbf{j} + (4 \times 5 - 2 \times 10)\mathbf{k} = (-1)\mathbf{i} - (-2)\mathbf{j} + (0)\mathbf{k} = -\mathbf{i} + 2\mathbf{j} + 0\mathbf{k}\) Nm
\(\mathbf{M}_B = \mathbf{r}_B \times \mathbf{F}_B = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 7 & 5 & 4 \\ -3 & 6 & 4 \end{vmatrix} = (5 \times 4 - 4 \times 6)\mathbf{i} - (7 \times 4 - 4 \times (-3))\mathbf{j} + (7 \times 6 - 5 \times (-3))\mathbf{k} = (20 - 24)\mathbf{i} - (28 + 12)\mathbf{j} + (42 + 15)\mathbf{k} = -4\mathbf{i} - 40\mathbf{j} + 57\mathbf{k}\) Nm
Total moment:
\(\mathbf{M} = \mathbf{M}_A + \mathbf{M}_B = (-1 - 4)\mathbf{i} + (2 - 40)\mathbf{j} + (0 + 57)\mathbf{k} = -5\mathbf{i} - 38\mathbf{j} + 57\mathbf{k}\) Nm
Question 12
a) Resultant velocity:
\(\mathbf{V}_R = \mathbf{V}_1 + \mathbf{V}_2 = (6 - 2)\mathbf{i} + (2 + 5)\mathbf{j} + (3 + 1)\mathbf{k} = 4\mathbf{i} + 7\mathbf{j} + 4\mathbf{k}\) m/s
b) Dot product:
\(\mathbf{V}_1 \cdot \mathbf{V}_2 = 6 \times (-2) + 2 \times 5 + 3 \times 1 = -12 + 10 + 3 = 1\)
Magnitudes:
\(|\mathbf{V}_1| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7\)
\(|\mathbf{V}_2| = \sqrt{(-2)^2 + 5^2 + 1^2} = \sqrt{4 + 25 + 1} = \sqrt{30} = 5.477\)
Angle:
\(\theta = \cos^{-1}(\frac{1}{7 \times 5.477}) = \cos^{-1}(0.0261) = 88.5^\circ\)
c) Cross product magnitude:
\(|\mathbf{V}_1 \times \mathbf{V}_2| = |\mathbf{V}_1||\mathbf{V}_2| \sin \theta = 7 \times 5.477 \times \sin 88.5^\circ = 38.339 \times 0.9996 = 38.324\) m²/s²
Question 13
a) Moment:
\(\mathbf{M} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & 1 \\ 8 & 6 & 4 \end{vmatrix} = (2 \times 4 - 1 \times 6)\mathbf{i} - (3 \times 4 - 1 \times 8)\mathbf{j} + (3 \times 6 - 2 \times 8)\mathbf{k} = (8 - 6)\mathbf{i} - (12 - 8)\mathbf{j} + (18 - 16)\mathbf{k} = 2\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}\) Nm
b) Parallel component:
\(\text{Scalar projection} = \frac{\mathbf{F} \cdot \mathbf{r}}{|\mathbf{r}|} = \frac{8 \times 3 + 6 \times 2 + 4 \times 1}{\sqrt{3^2 + 2^2 + 1^2}} = \frac{24 + 12 + 4}{\sqrt{14}} = \frac{40}{3.742} = 10.69\) N
Unit vector \(\hat{r} = \frac{1}{3.742}(3\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = 0.802\mathbf{i} + 0.535\mathbf{j} + 0.267\mathbf{k}\)
Parallel vector \(= 10.69 \times \hat{r} = 8.58\mathbf{i} + 5.72\mathbf{j} + 2.85\mathbf{k}\) N
c) Perpendicular component:
\(\mathbf{F}_\perp = \mathbf{F} - \mathbf{F}_\parallel = (8 - 8.58)\mathbf{i} + (6 - 5.72)\mathbf{j} + (4 - 2.85)\mathbf{k} = -0.58\mathbf{i} + 0.28\mathbf{j} + 1.15\mathbf{k}\) N
Question 14
a) Dot product:
\(\mathbf{A} \cdot \mathbf{B} = 5 \times 1 + 2 \times 4 + 3 \times 2 = 5 + 8 + 6 = 19\)
Interpretation: Positive dot product indicates heat flow direction and sensor orientation have a component in the same direction, useful for sensor placement.
b) Cross product:
\(\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 2 & 3 \\ 1 & 4 & 2 \end{vmatrix} = (2 \times 2 - 3 \times 4)\mathbf{i} - (5 \times 2 - 3 \times 1)\mathbf{j} + (5 \times 4 - 2 \times 1)\mathbf{k} = (4 - 12)\mathbf{i} - (10 - 3)\mathbf{j} + (20 - 2)\mathbf{k} = -8\mathbf{i} - 7\mathbf{j} + 18\mathbf{k}\)
Magnitude \(= \sqrt{(-8)^2 + (-7)^2 + 18^2} = \sqrt{64 + 49 + 324} = \sqrt{437} = 20.9\)
Unit vector \(= \frac{1}{20.9}(-8\mathbf{i} -7\mathbf{j} + 18\mathbf{k}) = -0.383\mathbf{i} - 0.335\mathbf{j} + 0.861\mathbf{k}\)
c) Angle:
\(|\mathbf{A}| = \sqrt{5^2 + 2^2 + 3^2} = \sqrt{25 + 4 + 9} = \sqrt{38} = 6.164\)
\(|\mathbf{B}| = \sqrt{1^2 + 4^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21} = 4.583\)
\(\theta = \cos^{-1} (\frac{19}{6.164 \times 4.583}) = \cos^{-1}(0.674) = 47.6^\circ\)
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