Chapter 13: Vectors in Two Dimensions
13.1 Introduction to Vectors
Definition:
A vector is a quantity that has both magnitude (size) and direction.
Examples: displacement, velocity, force, acceleration.
A scalar is a quantity that has magnitude only.
Examples: speed, mass, temperature, time.
Vector notation:
A vector can be written in several forms:
- Column vector: \(\begin{pmatrix} a \\ b \end{pmatrix}\) or \((a, b)\)
- Unit vector notation: \(a\mathbf{i} + b\mathbf{j}\)
- Position vector: \(\overrightarrow{OP} = (a, b)\) where O is the origin
- Vector between two points: \(\overrightarrow{AB}\)
Key terms:
- Position vector: A vector that starts at the origin.
- Unit vector: A vector with magnitude 1. \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) are unit vectors in the x and y directions.
- Magnitude (modulus): For vector \(\mathbf{v} = (a, b)\): \(|\mathbf{v}| = \sqrt{a^2 + b^2}\)
13.2 Vector Operations
Addition:
To add two vectors, add the corresponding components:
\[
\mathbf{a} + \mathbf{b} = (a_1 + b_1, a_2 + b_2)
\]
Subtraction:
\[
\mathbf{a} - \mathbf{b} = (a_1 - b_1, a_2 - b_2)
\]
Scalar multiplication:
To multiply a vector by a scalar \(k\), multiply each component:
\[
k\mathbf{a} = (ka_1, ka_2)
\]
Geometric interpretation:
- Addition: Triangle law or parallelogram law
- Subtraction: \(\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b})\)
- Scalar multiplication: Stretches or shrinks the vector; reverses direction if \(k\) is negative
Worked Example 13.1 (Vector operations – non-calculator)
Given \(\mathbf{a} = (3, 4)\) and \(\mathbf{b} = (1, -2)\), find:
- \(\mathbf{a} + \mathbf{b}\)
- \(\mathbf{a} - \mathbf{b}\)
- \(3\mathbf{a}\)
Solution:
(a) \(\mathbf{a} + \mathbf{b} = (3+1, 4+(-2)) = (4, 2)\)
(b) \(\mathbf{a} - \mathbf{b} = (3-1, 4-(-2)) = (2, 6)\)
(c) \(3\mathbf{a} = 3(3, 4) = (9, 12)\)
Worked Example 13.2 (Mauritian context – boat velocity)
A boat is sailing with velocity \((4, 3)\) m/s relative to the water. The current is \((1, -2)\) m/s.
- Find the resultant velocity of the boat.
- Find the speed of the boat relative to the ground.
- Find the direction of travel relative to the positive x-axis.
Solution:
(a) Resultant velocity = \((4+1, 3+(-2)) = (5, 1)\) m/s
(b) Speed = magnitude = \(\sqrt{5^2 + 1^2} = \sqrt{26}\) m/s
(c) Direction: \(\tan \theta = \frac{1}{5} \Rightarrow \theta = \tan^{-1}\left(\frac{1}{5}\right) \approx 11.3^\circ\)
Answers: (a) \((5, 1)\) m/s, (b) \(\sqrt{26}\) m/s, (c) \(11.3^\circ\) above the x-axis.
13.3 Parallel Vectors
Two vectors are parallel if one is a scalar multiple of the other:
\[
\mathbf{a} \parallel \mathbf{b} \iff \mathbf{a} = k\mathbf{b} \text{ for some scalar } k
\]
Proof: If \(\mathbf{a} = k\mathbf{b}\), then the vectors have the same or opposite direction. The scalar \(k\) determines the magnitude and direction. If \(k > 0\), they point in the same direction; if \(k < 0\), they point in opposite directions.
Worked Example 13.3 (Parallel vectors – non-calculator)
Given \(\mathbf{a} = (2, 4)\) and \(\mathbf{b} = (3, 6)\), show that \(\mathbf{a}\) and \(\mathbf{b}\) are parallel.
Solution:
\(\mathbf{b} = \frac{3}{2}(2, 4) = \frac{3}{2}\mathbf{a}\)
Since \(\mathbf{b} = \frac{3}{2}\mathbf{a}\), the vectors are parallel.
Worked Example 13.4 (Finding scalar for parallel vectors)
Given \(\mathbf{a} = (3, -2)\) and \(\mathbf{b} = (1, 4)\), find the value of \(k\) such that \(\mathbf{a} + k\mathbf{b}\) is parallel to the vector \((5, 2)\).
Solution:
\(\mathbf{a} + k\mathbf{b} = (3, -2) + k(1, 4) = (3 + k, -2 + 4k)\)
For this to be parallel to \((5, 2)\), there must be a scalar \(t\) such that:
\[
(3 + k, -2 + 4k) = t(5, 2)
\]
From the first component: \(3 + k = 5t\)
From the second component: \(-2 + 4k = 2t\)
Substitute \(t = \frac{3 + k}{5}\) into the second equation:
\[
-2 + 4k = 2\left(\frac{3 + k}{5}\right)
\]
\[
-2 + 4k = \frac{6 + 2k}{5}
\]
\[
-10 + 20k = 6 + 2k
\]
\[
18k = 16 \Rightarrow k = \frac{8}{9}
\]
Answer: \(k = \frac{8}{9}\)
13.4 Magnitude and Unit Vectors
Magnitude of \(\mathbf{v} = (a, b)\):
\[
|\mathbf{v}| = \sqrt{a^2 + b^2}
\]
Unit vector in the direction of \(\mathbf{v}\):
\[
\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}
\]
\(|\mathbf{v}| = \sqrt{a^2 + b^2}\)
\(\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}\)
Worked Example 13.5 (Magnitude and unit vector – non-calculator)
Points A and B have coordinates \((2, 5)\) and \((8, 13)\) respectively. Find:
- \(\overrightarrow{AB}\)
- \(|\overrightarrow{AB}|\)
- The unit vector in the direction of \(\overrightarrow{AB}\)
Solution:
(a) \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (8-2, 13-5) = (6, 8)\)
(b) \(|\overrightarrow{AB}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)
(c) Unit vector = \(\frac{(6, 8)}{10} = \left(\frac{3}{5}, \frac{4}{5}\right)\)
13.5 Position Vectors
Position vector of point \(P\) is \(\overrightarrow{OP}\), where \(O\) is the origin.
Vector between two points:
\[
\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}
\]
Midpoint:
\[
\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB})
\]
Worked Example 13.6 (Position vectors – non-calculator)
The position vectors of points A and B are \(\mathbf{a} = (2, 3)\) and \(\mathbf{b} = (8, 11)\). Point C lies on AB such that AC:CB = 3:1. Find the position vector of C.
Solution:
The point C divides AB in the ratio 3:1, so:
\[
\mathbf{c} = \frac{1( \mathbf{a}) + 3( \mathbf{b})}{3+1} = \frac{\mathbf{a} + 3\mathbf{b}}{4}
\]
\[
\mathbf{c} = \frac{(2, 3) + 3(8, 11)}{4} = \frac{(2, 3) + (24, 33)}{4} = \frac{(26, 36)}{4} = \left(\frac{13}{2}, 9\right)
\]
Answer: \(\left(\frac{13}{2}, 9\right)\)
13.6 Resolving Velocities
A velocity vector can be resolved into horizontal and vertical components:
\[
v_x = v \cos \theta
\]
\[
v_y = v \sin \theta
\]
where \(\theta\) is the angle the vector makes with the horizontal.
\(v_x = v \cos \theta\)
\(v_y = v \sin \theta\)
Worked Example 13.7 (Resolving velocities – non-calculator)
A particle moves with velocity \(10\) m/s at an angle of \(30^\circ\) above the horizontal. Find the horizontal and vertical components of the velocity.
Solution:
\(v_x = 10 \cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\) m/s
\(v_y = 10 \sin 30^\circ = 10 \times \frac{1}{2} = 5\) m/s
Answer: \(v_x = 5\sqrt{3}\) m/s, \(v_y = 5\) m/s.
Chapter 13 Summary
| Concept |
Key points |
| Vector notation |
Column, i-j form, position vectors |
| Magnitude |
\(|\mathbf{v}| = \sqrt{a^2 + b^2}\) |
| Unit vector |
\(\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}\) |
| Addition & subtraction |
Add/subtract corresponding components |
| Scalar multiplication |
\(k(a, b) = (ka, kb)\) |
| Parallel vectors |
One is a scalar multiple of the other |
| Position vector |
\(\overrightarrow{OP} = (a, b)\) |
| Vector between two points |
\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\) |
| Resolving velocities |
\(v_x = v \cos \theta\), \(v_y = v \sin \theta\) |
Exercises – Chapter 13
Easy (Drill – Non-Calculator)
- Write the following vectors in the form \(a\mathbf{i} + b\mathbf{j}\):
(a) \((3, 4)\)
(b) \((-2, 5)\)
(c) \((0, -7)\)
- Find the magnitude of:
(a) \((3, 4)\)
(b) \((-6, 8)\)
(c) \((1, 1)\)
- Given \(\mathbf{a} = (2, -3)\) and \(\mathbf{b} = (5, 1)\), find:
(a) \(\mathbf{a} + \mathbf{b}\)
(b) \(\mathbf{a} - \mathbf{b}\)
(c) \(3\mathbf{a}\)
Medium (Examination Style)
- Points A and B have coordinates \((2, 5)\) and \((8, 13)\) respectively. Find:
(a) \(\overrightarrow{AB}\)
(b) \(|\overrightarrow{AB}|\)
(c) The unit vector in the direction of \(\overrightarrow{AB}\)
- Given \(\mathbf{a} = (3, -2)\) and \(\mathbf{b} = (1, 4)\), find the value of \(k\) such that \(\mathbf{a} + k\mathbf{b}\) is parallel to the vector \((5, 2)\).
- Mauritian context: A boat is sailing with velocity \((4, 3)\) m/s relative to the water. The current is \((1, -2)\) m/s.
(a) Find the resultant velocity of the boat.
(b) Find the speed of the boat relative to the ground.
(c) Find the direction of travel relative to the positive x-axis.
- The points P, Q and R have position vectors \((2, 1)\), \((6, 3)\) and \((4, 7)\) respectively. Show that PQR is an isosceles triangle.
Hard (Challenge for A*)
- The position vectors of points A and B are \(\mathbf{a} = (2, 3)\) and \(\mathbf{b} = (8, 11)\). Point C lies on AB such that AC:CB = 3:1. Find the position vector of C.
- Mauritian context: A plane flies at 200 km/h in still air. The pilot wants to fly due east, but a wind is blowing from the north at 50 km/h.
(a) In what direction should the pilot head?
(b) What is the ground speed of the plane?
- Prove that the midpoints of the sides of any quadrilateral form a parallelogram. (Use position vectors.)