Table of Contents
▼Quick Formula Reference
Core Concepts
Definition: A vector is a mathematical quantity that has both magnitude (size) and direction. Unlike scalars, which only have magnitude, vectors are used to represent physical quantities like displacement, velocity, force, and acceleration.
Vectors are typically represented by an arrow, where:
- The length of the arrow represents the magnitude
- The arrowhead indicates the direction
Examples of Vector Quantities:
- Displacement
- Velocity
- Acceleration
- Force
- Momentum
- Electric field
| Property | Scalar | Vector |
|---|---|---|
| Has magnitude | ✓ | ✓ |
| Has direction | ✗ | ✓ |
| Examples | Mass, Speed, Temperature, Time, Distance | Force, Velocity, Displacement, Acceleration |
| Mathematical operations | Simple arithmetic | Vector algebra |
Arrow Notation: \(\vec{a}\) or \(\overrightarrow{AB}\)
Component Form (2D):
\[\vec{a} = a_x\hat{i} + a_y\hat{j}\]
or
\[\vec{a} = (a_x, a_y)\]
Component Form (3D):
\[\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}\]
or
\[\vec{a} = (a_x, a_y, a_z)\]
Magnitude:
\[|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}\]
Standard Unit Vectors:
- \(\hat{i} = (1, 0, 0)\) - unit vector along x-axis
- \(\hat{j} = (0, 1, 0)\) - unit vector along y-axis
- \(\hat{k} = (0, 0, 1)\) - unit vector along z-axis
1. Zero Vector (\(\vec{0}\)): Magnitude = 0, no specific direction
2. Unit Vector (\(\hat{a}\)): Magnitude = 1
\[\hat{a} = \frac{\vec{a}}{|\vec{a}|}\]
3. Position Vector: Vector from origin to a point P(x, y, z)
\[\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\]
4. Equal Vectors: Same magnitude and direction
5. Negative Vector (\(-\vec{a}\)): Same magnitude, opposite direction
6. Collinear Vectors: Vectors along the same line (parallel or anti-parallel)
7. Coplanar Vectors: Vectors lying in the same plane
Triangle Law: Place vectors head-to-tail. The resultant is from tail of first to head of last.
Parallelogram Law: The resultant is the diagonal of the parallelogram formed by the two vectors.
Component Method:
\[\vec{R} = \vec{a} + \vec{b} = (a_x + b_x)\hat{i} + (a_y + b_y)\hat{j} + (a_z + b_z)\hat{k}\]
Properties:
- Commutative: \(\vec{a} + \vec{b} = \vec{b} + \vec{a}\)
- Associative: \((\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})\)
- Additive Identity: \(\vec{a} + \vec{0} = \vec{a}\)
Vector subtraction is defined as:
\[\vec{a} - \vec{b} = \vec{a} + (-\vec{b})\]
Component Method:
\[\vec{a} - \vec{b} = (a_x - b_x)\hat{i} + (a_y - b_y)\hat{j} + (a_z - b_z)\hat{k}\]
Multiplying a vector by a scalar \(k\):
\[k\vec{a} = ka_x\hat{i} + ka_y\hat{j} + ka_z\hat{k}\]
Properties:
- If \(k > 0\): Same direction, magnitude scaled by \(|k|\)
- If \(k < 0\): Opposite direction, magnitude scaled by \(|k|\)
- If \(k = 0\): Results in zero vector
Magnitude: \(|k\vec{a}| = |k||\vec{a}|\)
Definition:
\[\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\]
where \(\theta\) is the angle between the vectors.
Component Form:
\[\vec{a} \cdot \vec{b} = a_xb_x + a_yb_y + a_zb_z\]
Result: Scalar (number)
Key Properties:
- If \(\vec{a} \cdot \vec{b} = 0\), vectors are perpendicular
- \(\vec{a} \cdot \vec{a} = |\vec{a}|^2\)
- \(\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1\)
- \(\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0\)
Application: Work done = \(\vec{F} \cdot \vec{d}\)
Magnitude:
\[|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\]
Result: Vector perpendicular to both \(\vec{a}\) and \(\vec{b}\)
Direction: Right-hand rule
Determinant Method:
\[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}\]
Key Properties:
- \(\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}\) (anti-commutative)
- \(\vec{a} \times \vec{a} = \vec{0}\)
- \(\hat{i} \times \hat{j} = \hat{k}\), \(\hat{j} \times \hat{k} = \hat{i}\), \(\hat{k} \times \hat{i} = \hat{j}\)
Applications:
- Torque: \(\vec{\tau} = \vec{r} \times \vec{F}\)
- Area of parallelogram: \(|\vec{a} \times \vec{b}|\)
For a vector \(\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}\):
Direction Cosines:
- \(\cos\alpha = \frac{a_x}{|\vec{a}|}\) (angle with x-axis)
- \(\cos\beta = \frac{a_y}{|\vec{a}|}\) (angle with y-axis)
- \(\cos\gamma = \frac{a_z}{|\vec{a}|}\) (angle with z-axis)
Identity:
\[\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\]
Solved Examples
A vector has magnitude 10 units at 60° above horizontal. Find its components.
Given: Magnitude = 10 units, angle = 60°
Solution:
\[V_x = V\cos\theta = 10\cos60° = 10 \times \frac{1}{2} = 5\]
\[V_y = V\sin\theta = 10\sin60° = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66\]
Answer: \(V_x = 5\) units, \(V_y = 5\sqrt{3} \approx 8.66\) units
\(\vec{A} = \langle 4, -3 \rangle\), \(\vec{B} = \langle 1, 6 \rangle\). Find \(\vec{R} = \vec{A} + \vec{B}\) and its magnitude.
Solution:
\[\vec{R} = \vec{A} + \vec{B} = (4 + 1)\hat{i} + (-3 + 6)\hat{j} = 5\hat{i} + 3\hat{j}\]
\[\vec{R} = \langle 5, 3 \rangle\]
Magnitude:
\[|\vec{R}| = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83\]
Answer: \(\vec{R} = \langle 5, 3 \rangle\), \(|\vec{R}| = \sqrt{34} \approx 5.83\) units
Find the unit vector in the direction of \(\vec{a} = 3\hat{i} + 4\hat{j}\).
Solution:
\[|\vec{a}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]
\[\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{3\hat{i} + 4\hat{j}}{5} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}\]
Answer: \(\hat{a} = 0.6\hat{i} + 0.8\hat{j}\)
Find the angle between \(\vec{u} = \langle 1, 2, 0 \rangle\) and \(\vec{v} = \langle -2, 2, 2 \rangle\).
Solution:
\[\vec{u} \cdot \vec{v} = (1)(-2) + (2)(2) + (0)(2) = -2 + 4 + 0 = 2\]
\[|\vec{u}| = \sqrt{1^2 + 2^2 + 0^2} = \sqrt{5}\]
\[|\vec{v}| = \sqrt{(-2)^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}\]
\[\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} = \frac{2}{\sqrt{5} \times 2\sqrt{3}} = \frac{1}{\sqrt{15}}\]
\[\theta = \cos^{-1}\left(\frac{1}{\sqrt{15}}\right) \approx 75°\]
Answer: \(\theta \approx 75°\)
Find the projection of \(\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}\) on \(\vec{b} = \hat{i} + 2\hat{j} + \hat{k}\).
Solution:
\[\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10\]
\[|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}\]
Projection:
\[\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{10}{\sqrt{6}} = \frac{10\sqrt{6}}{6} = \frac{5\sqrt{6}}{3}\]
Answer: \(\frac{5\sqrt{6}}{3}\) units
Find the area of the parallelogram with sides \(\vec{a} = 2\hat{i} + 3\hat{j}\) and \(\vec{b} = \hat{i} - 2\hat{j}\).
Solution:
\[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 1 & -2 & 0 \end{vmatrix}\]
\[= \hat{i}(3 \times 0 - 0 \times (-2)) - \hat{j}(2 \times 0 - 0 \times 1) + \hat{k}(2 \times (-2) - 3 \times 1)\]
\[= \hat{k}(-4 - 3) = -7\hat{k}\]
\[|\vec{a} \times \vec{b}| = |-7\hat{k}| = 7\]
Area of parallelogram: \(|\vec{a} \times \vec{b}| = 7\) square units
Answer: 7 square units
Practice Questions
MCQs
Short Answer Questions
Numerical Problems
MCQs
Short Answer
Numericals
a) Draw a vector diagram [2 marks]
b) Calculate actual velocity [3 marks]
c) What direction should boat point to reach directly opposite? [3 marks]
a) Vector diagram shows boat velocity and river velocity
b) Actual velocity = \(\sqrt{4^2 + 3^2} = 5\) m/s at angle \(\theta = \tan^{-1}(3/4) = 36.9°\) downstream
c) Boat should point upstream at angle \(\sin^{-1}(3/4) = 48.6°\) to the perpendicular
Paper 1 Style
Paper 2 Style
a) Find vectors \(\vec{AB}\) and \(\vec{AC}\) [2 marks]
b) Calculate \(\vec{AB} \cdot \vec{AC}\) [2 marks]
c) Find the angle \(\angle BAC\) [3 marks]
a) \(\vec{AB} = \langle 2, -4, 2 \rangle\), \(\vec{AC} = \langle -1, -3, -3 \rangle\)
b) \(\vec{AB} \cdot \vec{AC} = 2(-1) + (-4)(-3) + 2(-3) = -2 + 12 - 6 = 4\)
c) \(|\vec{AB}| = \sqrt{24} = 2\sqrt{6}\), \(|\vec{AC}| = \sqrt{19}\), \(\cos\theta = \frac{4}{2\sqrt{6}\sqrt{19}}\), \(\theta \approx 78.5°\)
Quick Revision Tips
Key Points to Remember
- Vectors have both magnitude and direction; scalars have only magnitude
- Unit vector has magnitude 1: \(\hat{a} = \frac{\vec{a}}{|\vec{a}|}\)
- Dot product gives a scalar: \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\)
- Cross product gives a vector perpendicular to both: \(|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\)
- If \(\vec{a} \cdot \vec{b} = 0\), vectors are perpendicular
- Component method is often easiest for calculations
Common Mistakes to Avoid
- Confusing scalar and vector quantities
- Forgetting to take square root when finding magnitude
- Mixing up dot product (scalar) and cross product (vector)
- Not using correct signs in component calculations
- Forgetting units in final answers
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