Magnitude and Direction of Vectors
CSEC Mathematics: The Size and Angle of Movement
Essential Understanding: Every vector has two fundamental properties: its magnitude (how long it is) and its direction (where it points). These properties turn abstract numbers into meaningful physical quantities like velocity, force, and displacement.
1. The Vector as a Triangle
The most important shift for a student is to stop seeing a vector \(\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}\) as a single line and start seeing it as a right-angled triangle where:
Horizontal Component (\(x\))
The base of the triangle. This is how far the vector moves horizontally.
Positive: Rightward movement
Negative: Leftward movement
Vertical Component (\(y\))
The height of the triangle. This is how far the vector moves vertically.
Positive: Upward movement
Negative: Downward movement
The Vector Itself
The hypotenuse of the triangle. This is the actual path of the vector.
Magnitude: Length of the hypotenuse
Direction: Angle of the hypotenuse
Visual Representation
Every vector \(\begin{pmatrix} x \\ y \end{pmatrix}\) can be drawn as a right-angled triangle:
2. Calculating Magnitude (Size)
The magnitude of a vector is its length. In CSEC notation, the magnitude of vector \(\mathbf{u}\) is written as \(|\mathbf{u}|\).
The Pythagorean Formula
Derived directly from Pythagoras' Theorem (\(a^2 + b^2 = c^2\)):
Key Tip: Even if the components are negative (e.g., \(-3\)), the magnitude is always positive because squaring a negative number results in a positive value (\((-3)^2 = 9\)).
3. Calculating Direction (Angle)
The direction of a vector is the angle \(\theta\) it makes with the positive \(x\)-axis.
The Tangent Formula
We use the Tangent ratio because we already know the opposite (\(y\)) and adjacent (\(x\)) sides:
Quadrant Awareness
Students must be taught to check their graph. The calculator's \(\tan^{-1}\) function only gives answers between \(-90^\circ\) and \(90^\circ\). You must adjust based on the quadrant:
Quadrant I
\(x > 0\), \(y > 0\)
\(\theta = \tan^{-1}(y/x)\)
\(0^\circ < \theta < 90^\circ\)
Quadrant II
\(x < 0\), \(y > 0\)
\(\theta = 180^\circ + \tan^{-1}(y/x)\)
\(90^\circ < \theta < 180^\circ\)
Quadrant III
\(x < 0\), \(y < 0\)
\(\theta = 180^\circ + \tan^{-1}(y/x)\)
\(180^\circ < \theta < 270^\circ\)
Quadrant IV
\(x > 0\), \(y < 0\)
\(\theta = 360^\circ + \tan^{-1}(y/x)\)
\(270^\circ < \theta < 360^\circ\)
Memory Trick: "All Students Take Calculus" - starting from Quadrant I and moving counterclockwise:
All (All trig functions positive in Quadrant I)
Students (Sine positive in Quadrant II)
Take (Tangent positive in Quadrant III)
Calculus (Cosine positive in Quadrant IV)
4. Interactive "Magnitude & Angle" Lab
Explore Vector Components
Objective: Adjust the \(x\) and \(y\) components using the sliders. Watch how the vector's magnitude and direction change in real-time.
Vector Components
\(\begin{pmatrix} 5.0 \\ 3.0 \end{pmatrix}\)
Magnitude (\(|\vec{v}|\))
5.83 units
\(\sqrt{5.0^2 + 3.0^2} = \sqrt{34.0} = 5.83\)
Direction (\(\theta\))
30.96°
\(\tan^{-1}(3.0/5.0) = 30.96^\circ\)
Current Quadrant
Quadrant I
Both x and y are positive. Angle = calculator result.
5. Unit Vectors and Standard Directions
Briefly introduce the concept of "Unit Direction" - vectors with magnitude 1 that point in standard directions.
Standard Unit Vectors
\(\mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) - One unit in the \(x\) direction
\(\mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) - One unit in the \(y\) direction
Notation Conversion
\(\begin{pmatrix} 3 \\ 4 \end{pmatrix} = 3\begin{pmatrix} 1 \\ 0 \end{pmatrix} + 4\begin{pmatrix} 0 \\ 1 \end{pmatrix} = 3\mathbf{i} + 4\mathbf{j}\)
This notation is common in physics and advanced mathematics.
Unit Vector Calculation
A unit vector in the direction of \(\vec{v}\) is:
Example: For \(\vec{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\):
\(|\vec{v}| = \sqrt{3^2 + 4^2} = 5\)
\(\hat{v} = \frac{1}{5}\begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix}\)
Applications in Physics
Unit vectors are used to separate forces, velocities, and accelerations into components.
Force Example: A 10N force at 30° above horizontal:
\(F_x = 10\cos 30^\circ \approx 8.66N\)
\(F_y = 10\sin 30^\circ = 5N\)
\(\vec{F} = 8.66\mathbf{i} + 5\mathbf{j}\)
6. CSEC Exam Mastery Tips
Avoid These Common Mistakes
Calculator Mode
- Ensure your calculator is in DEG (Degrees) mode.
- CSEC rarely uses radians for vector direction.
- Check: \(\tan^{-1}(1)\) should give \(45^\circ\), not \(0.785\) radians.
Exact Values
- If a question asks for the "exact magnitude," leave your answer as a surd.
- Example: \(\sqrt{34}\) not \(5.83095...\)
- Only round when instructed to do so.
Zero Vector
- A vector with zero magnitude \(\begin{pmatrix} 0 \\ 0 \end{pmatrix}\) has no specific direction.
- Don't try to calculate \(\tan^{-1}(0/0)\) - it's undefined.
- The zero vector is the only vector without a direction.
7. Worked Example: Finding \(|\vec{v}|\) and \(\theta\)
Problem: Find the magnitude and direction of the vector \(\vec{v} = \begin{pmatrix} -4 \\ 3 \end{pmatrix}\).
Magnitude:
\[ |\vec{v}| = \sqrt{(-4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = \mathbf{5 \text{ units}} \]
Initial Angle Calculation:
\[ \tan \theta = \frac{3}{-4} = -0.75 \]
\[ \theta = \tan^{-1}(-0.75) \approx -36.87^\circ \]
Quadrant Analysis:
\(x = -4\) (negative), \(y = 3\) (positive) → Quadrant II
The calculator gave \(-36.87^\circ\), which is in Quadrant IV.
Angle Adjustment:
For Quadrant II: \(\theta = 180^\circ + \text{(calculator result)}\)
\[ \theta = 180^\circ + (-36.87^\circ) = 180^\circ - 36.87^\circ = \mathbf{143.13^\circ} \]
8. Practice Mission: "The Flight Path"
Speed (Magnitude):
\[ |\mathbf{v}| = \sqrt{120^2 + 50^2} = \sqrt{14400 + 2500} = \sqrt{16900} = 130 \text{ km/h} \]
Angle (Direction):
\[ \tan \theta = \frac{50}{120} = \frac{5}{12} \approx 0.4167 \]
\[ \theta = \tan^{-1}(0.4167) \approx 22.6^\circ \text{ above horizontal} \]
Since both components are positive, the vector is in Quadrant I, so no adjustment is needed.
Magnitude:
\[ |\mathbf{u}| = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]
Initial Calculation:
\[ \tan \theta = \frac{-8}{-6} = \frac{4}{3} \approx 1.333 \]
\[ \theta = \tan^{-1}(1.333) \approx 53.1^\circ \]
Quadrant Analysis:
\(x = -6\) (negative), \(y = -8\) (negative) → Quadrant III
For Quadrant III: \(\theta = 180^\circ + 53.1^\circ = 233.1^\circ\)
Final Answer: Magnitude = 10, Direction = 233.1°
Calculate magnitudes:
1. \(\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)
2. \(\sqrt{8^2 + 9^2} = \sqrt{64 + 81} = \sqrt{145} \approx 12.04\)
3. \(\sqrt{(-7)^2 + (-10)^2} = \sqrt{49 + 100} = \sqrt{149} \approx 12.21\)
4. \(\sqrt{0^2 + 13^2} = \sqrt{169} = 13\)
Vectors 1 and 4 both have magnitude 13, which is the greatest among the options.
Note: Both \(\begin{pmatrix} 5 \\ 12 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 13 \end{pmatrix}\) have magnitude 13.
Formula Recap
Magnitude
\[ |\vec{v}| = \sqrt{x^2 + y^2} \]
Always positive
Direction
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]
Adjust for quadrant
Quadrant Rules
QII: \(+180^\circ\)
QIII: \(+180^\circ\)
QIV: \(+360^\circ\)
Unit Vector
\[ \hat{v} = \frac{\vec{v}}{|\vec{v}|} \]
Magnitude = 1