9.3 Vectors
Understanding Vectors
A vector has both magnitude (size) and direction. Unlike numbers, vectors tell us how far and where things go. They're essential in physics, engineering, and navigation.
A vector has both magnitude (size) and direction. Unlike numbers, vectors tell us how far and where things go. They're essential in physics, engineering, and navigation.
⚡ Vector Notation:
Column Vector
$$\begin{pmatrix} 3 \\ 4 \end{pmatrix}$$
Most common notation
Bold Letter
a or v
Used in print
Underlined
a or v
Handwritten form
Arrow Notation
$$\vec{AB}$$
From A to B
What the numbers mean:
$$\begin{pmatrix} 3 \\ 4 \end{pmatrix}$$ means: Move 3 units right, 4 units up
$$\begin{pmatrix} -2 \\ 5 \end{pmatrix}$$ means: Move 2 units left, 5 units up
$$\begin{pmatrix} 4 \\ -3 \end{pmatrix}$$ means: Move 4 units right, 3 units down
$$\begin{pmatrix} 3 \\ 4 \end{pmatrix}$$ means: Move 3 units right, 4 units up
$$\begin{pmatrix} -2 \\ 5 \end{pmatrix}$$ means: Move 2 units left, 5 units up
$$\begin{pmatrix} 4 \\ -3 \end{pmatrix}$$ means: Move 4 units right, 3 units down
⚡ Vector Operations:
Addition
$$\begin{pmatrix} a_1 \\ a_2 \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \end{pmatrix}$$
Add corresponding components
Subtraction
$$\begin{pmatrix} a_1 \\ a_2 \end{pmatrix} - \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} a_1 - b_1 \\ a_2 - b_2 \end{pmatrix}$$
Subtract corresponding components
Scalar Multiplication
$$k \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} ka_1 \\ ka_2 \end{pmatrix}$$
Multiply each component by k
Example 1: Vector Addition
Given $\mathbf{a} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$, find $\mathbf{a} + \mathbf{b}$
Solution:
$$\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} + \begin{pmatrix} 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 3+1 \\ 2+4 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$$
Answer: $\begin{pmatrix} 4 \\ 6 \end{pmatrix}$
$$\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} + \begin{pmatrix} 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 3+1 \\ 2+4 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$$
Answer: $\begin{pmatrix} 4 \\ 6 \end{pmatrix}$
Example 2: Vector Subtraction
Given $\mathbf{p} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$, find $\mathbf{p} - \mathbf{q}$
Solution:
$$\mathbf{p} - \mathbf{q} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 5-2 \\ 3-1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$$
Answer: $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$
Solution:
$$\mathbf{p} - \mathbf{q} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 5-2 \\ 3-1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$$
Answer: $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$
Example 3: Vector Multiplication
Given $\mathbf{v} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$, find $3\mathbf{v}$
Solution:
$$3\mathbf{v} = 3 \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \times 4 \\ 3 \times (-2) \end{pmatrix} = \begin{pmatrix} 12 \\ -6 \end{pmatrix}$$
Note: The vector points in the same direction but is 3 times longer
Answer: $\begin{pmatrix} 12 \\ -6 \end{pmatrix}$
Solution:
$$3\mathbf{v} = 3 \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \times 4 \\ 3 \times (-2) \end{pmatrix} = \begin{pmatrix} 12 \\ -6 \end{pmatrix}$$
Note: The vector points in the same direction but is 3 times longer
Answer: $\begin{pmatrix} 12 \\ -6 \end{pmatrix}$
Example 4: Combined Operations
Given $\mathbf{a} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$, find $2\mathbf{a} + 3\mathbf{b}$
Step 1: Calculate $2\mathbf{a}$
$$2\mathbf{a} = 2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$$
Step 2: Calculate $3\mathbf{b}$
$$3\mathbf{b} = 3 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \end{pmatrix}$$
Step 3: Add them
$$2\mathbf{a} + 3\mathbf{b} = \begin{pmatrix} 4 \\ 6 \end{pmatrix} + \begin{pmatrix} 3 \\ -3 \end{pmatrix} = \begin{pmatrix} 7 \\ 3 \end{pmatrix}$$
Answer: $\begin{pmatrix} 7 \\ 3 \end{pmatrix}$
Step 1: Calculate $2\mathbf{a}$
$$2\mathbf{a} = 2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$$
Step 2: Calculate $3\mathbf{b}$
$$3\mathbf{b} = 3 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \end{pmatrix}$$
Step 3: Add them
$$2\mathbf{a} + 3\mathbf{b} = \begin{pmatrix} 4 \\ 6 \end{pmatrix} + \begin{pmatrix} 3 \\ -3 \end{pmatrix} = \begin{pmatrix} 7 \\ 3 \end{pmatrix}$$
Answer: $\begin{pmatrix} 7 \\ 3 \end{pmatrix}$
💡 Vector Arithmetic Tips:
• Addition: Head-to-tail method - draw vectors end-to-end
• Subtraction: $\mathbf{a} - \mathbf{b}$ means "from b to a"
• Scalar multiplication: Positive scalar = same direction, negative = opposite
• Zero vector: $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ has no direction
• Subtraction: $\mathbf{a} - \mathbf{b}$ means "from b to a"
• Scalar multiplication: Positive scalar = same direction, negative = opposite
• Zero vector: $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ has no direction
🎯 Vector Practice
Using Vectors to Prove Geometry
Vectors provide elegant ways to prove geometric properties like parallel lines, midpoints, and ratios.
Vectors provide elegant ways to prove geometric properties like parallel lines, midpoints, and ratios.
⚡ Key Proof Techniques:
Parallel Vectors:
Two vectors are parallel if one is a scalar multiple of the other
$$\mathbf{a} \parallel \mathbf{b} \iff \mathbf{a} = k\mathbf{b}$$ for some scalar $k$
Two vectors are parallel if one is a scalar multiple of the other
$$\mathbf{a} \parallel \mathbf{b} \iff \mathbf{a} = k\mathbf{b}$$ for some scalar $k$
Midpoint:
If M is the midpoint of AB, then:
$$\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB})$$
If M is the midpoint of AB, then:
$$\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB})$$
Position Vectors:
Any point P can be described by its position vector $\vec{OP}$ from origin O
$$\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$$
Any point P can be described by its position vector $\vec{OP}$ from origin O
$$\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$$
Example 1: Proving Lines are Parallel
Triangle ABC has points D and E on sides AB and AC such that AD:DB = 1:2 and AE:EC = 1:2. Prove that DE is parallel to BC.
Solution:
Let $\vec{AB} = \mathbf{b}$ and $\vec{AC} = \mathbf{c}$
Since AD:DB = 1:2, D divides AB in ratio 1:2
$$\vec{AD} = \frac{1}{3}\mathbf{b}$$
Since AE:EC = 1:2, E divides AC in ratio 1:2
$$\vec{AE} = \frac{1}{3}\mathbf{c}$$
Now find $\vec{DE}$:
$$\vec{DE} = \vec{AE} - \vec{AD} = \frac{1}{3}\mathbf{c} - \frac{1}{3}\mathbf{b} = \frac{1}{3}(\mathbf{c} - \mathbf{b})$$
And $\vec{BC}$:
$$\vec{BC} = \vec{AC} - \vec{AB} = \mathbf{c} - \mathbf{b}$$
We can see that:
$$\vec{DE} = \frac{1}{3}\vec{BC}$$
Since $\vec{DE}$ is a scalar multiple of $\vec{BC}$, DE is parallel to BC ✓
Let $\vec{AB} = \mathbf{b}$ and $\vec{AC} = \mathbf{c}$
Since AD:DB = 1:2, D divides AB in ratio 1:2
$$\vec{AD} = \frac{1}{3}\mathbf{b}$$
Since AE:EC = 1:2, E divides AC in ratio 1:2
$$\vec{AE} = \frac{1}{3}\mathbf{c}$$
Now find $\vec{DE}$:
$$\vec{DE} = \vec{AE} - \vec{AD} = \frac{1}{3}\mathbf{c} - \frac{1}{3}\mathbf{b} = \frac{1}{3}(\mathbf{c} - \mathbf{b})$$
And $\vec{BC}$:
$$\vec{BC} = \vec{AC} - \vec{AB} = \mathbf{c} - \mathbf{b}$$
We can see that:
$$\vec{DE} = \frac{1}{3}\vec{BC}$$
Since $\vec{DE}$ is a scalar multiple of $\vec{BC}$, DE is parallel to BC ✓
Example 2: Midpoint Proof
In triangle OAB, M is the midpoint of AB. Prove that $\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB})$
Solution:
Let $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$
To get from O to M, we can go via A:
$$\vec{OM} = \vec{OA} + \vec{AM}$$
Since M is the midpoint of AB:
$$\vec{AM} = \frac{1}{2}\vec{AB}$$
And $\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$
Therefore:
$$\vec{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a})$$
$$\vec{OM} = \mathbf{a} + \frac{1}{2}\mathbf{b} - \frac{1}{2}\mathbf{a}$$
$$\vec{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$$
$$\vec{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$$
Proved! ✓
Solution:
Let $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$
To get from O to M, we can go via A:
$$\vec{OM} = \vec{OA} + \vec{AM}$$
Since M is the midpoint of AB:
$$\vec{AM} = \frac{1}{2}\vec{AB}$$
And $\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$
Therefore:
$$\vec{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a})$$
$$\vec{OM} = \mathbf{a} + \frac{1}{2}\mathbf{b} - \frac{1}{2}\mathbf{a}$$
$$\vec{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$$
$$\vec{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$$
Proved! ✓
Example 3: Ratio Proof
Point P divides line segment AB in the ratio 2:3. If $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$, express $\vec{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
Solution:
P divides AB in ratio 2:3, so:
$$\vec{AP} = \frac{2}{5}\vec{AB}$$
We know that:
$$\vec{AB} = \mathbf{b} - \mathbf{a}$$
Therefore:
$$\vec{AP} = \frac{2}{5}(\mathbf{b} - \mathbf{a})$$
Using $\vec{OP} = \vec{OA} + \vec{AP}$:
$$\vec{OP} = \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a})$$
$$\vec{OP} = \mathbf{a} + \frac{2}{5}\mathbf{b} - \frac{2}{5}\mathbf{a}$$
$$\vec{OP} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}$$
Answer: $\vec{OP} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}$
Solution:
P divides AB in ratio 2:3, so:
$$\vec{AP} = \frac{2}{5}\vec{AB}$$
We know that:
$$\vec{AB} = \mathbf{b} - \mathbf{a}$$
Therefore:
$$\vec{AP} = \frac{2}{5}(\mathbf{b} - \mathbf{a})$$
Using $\vec{OP} = \vec{OA} + \vec{AP}$:
$$\vec{OP} = \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a})$$
$$\vec{OP} = \mathbf{a} + \frac{2}{5}\mathbf{b} - \frac{2}{5}\mathbf{a}$$
$$\vec{OP} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}$$
Answer: $\vec{OP} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}$
💡 Proof Tips:
• Define vectors clearly: Let $\vec{OA} = \mathbf{a}$, etc.
• Express everything from one point: Usually the origin O
• Use position vector formula: $\vec{AB} = \vec{OB} - \vec{OA}$
• Ratios: If P divides AB in ratio m:n, then $\vec{AP} = \frac{m}{m+n}\vec{AB}$
• Show your working: Every step should follow logically
• Express everything from one point: Usually the origin O
• Use position vector formula: $\vec{AB} = \vec{OB} - \vec{OA}$
• Ratios: If P divides AB in ratio m:n, then $\vec{AP} = \frac{m}{m+n}\vec{AB}$
• Show your working: Every step should follow logically
🎯 Vector Proofs
Finding Vector Lengths
The magnitude (or length) of a vector tells us how far the vector moves, ignoring direction.
The magnitude (or length) of a vector tells us how far the vector moves, ignoring direction.
⚡ Magnitude Formula:
2D Magnitude
$$\left|\begin{pmatrix} a \\ b \end{pmatrix}\right| = \sqrt{a^2 + b^2}$$
Uses Pythagoras' theorem
For 3D vectors:
$$\left|\begin{pmatrix} a \\ b \\ c \end{pmatrix}\right| = \sqrt{a^2 + b^2 + c^2}$$
$$\left|\begin{pmatrix} a \\ b \\ c \end{pmatrix}\right| = \sqrt{a^2 + b^2 + c^2}$$
Example 1: Finding Magnitude
Find the magnitude of $\mathbf{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$
Formula:
$$|\mathbf{v}| = \sqrt{a^2 + b^2}$$
Solution:
$$|\mathbf{v}| = \sqrt{3^2 + 4^2}$$
$$|\mathbf{v}| = \sqrt{9 + 16}$$
$$|\mathbf{v}| = \sqrt{25} = 5$$
Answer: The magnitude is 5 units
Formula:
$$|\mathbf{v}| = \sqrt{a^2 + b^2}$$
Solution:
$$|\mathbf{v}| = \sqrt{3^2 + 4^2}$$
$$|\mathbf{v}| = \sqrt{9 + 16}$$
$$|\mathbf{v}| = \sqrt{25} = 5$$
Answer: The magnitude is 5 units
Example 2: 3D Magnitude
Find the magnitude of $\mathbf{u} = \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix}$
Solution:
$$|\mathbf{u}| = \sqrt{2^2 + 3^2 + 6^2}$$
$$|\mathbf{u}| = \sqrt{4 + 9 + 36}$$
$$|\mathbf{u}| = \sqrt{49} = 7$$
Answer: The magnitude is 7 units
Solution:
$$|\mathbf{u}| = \sqrt{2^2 + 3^2 + 6^2}$$
$$|\mathbf{u}| = \sqrt{4 + 9 + 36}$$
$$|\mathbf{u}| = \sqrt{49} = 7$$
Answer: The magnitude is 7 units
Example 3: Unit Vector
Find the unit vector in the direction of $\mathbf{v} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$
Step 1: Find the magnitude
$$|\mathbf{v}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
Step 2: Divide by magnitude
$$\hat{\mathbf{v}} = \frac{1}{10}\begin{pmatrix} 6 \\ 8 \end{pmatrix} = \begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix}$$
Answer: $\begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix}$ (magnitude = 1)
Step 1: Find the magnitude
$$|\mathbf{v}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
Step 2: Divide by magnitude
$$\hat{\mathbf{v}} = \frac{1}{10}\begin{pmatrix} 6 \\ 8 \end{pmatrix} = \begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix}$$
Answer: $\begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix}$ (magnitude = 1)
💡 Magnitude Tips:
• Always square before adding: $\sqrt{a^2 + b^2}$ not $\sqrt{a + b}$
• Magnitude is always positive: It's a distance
• Unit vector: Divide by magnitude to get length 1
• Zero vector: $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ has magnitude 0
• Notation: $|\mathbf{v}|$ or $\|\mathbf{v}\|$ both mean magnitude
• Magnitude is always positive: It's a distance
• Unit vector: Divide by magnitude to get length 1
• Zero vector: $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ has magnitude 0
• Notation: $|\mathbf{v}|$ or $\|\mathbf{v}\|$ both mean magnitude
🎯 Vector Magnitudes Practice
Real Life Uses of Vectors:
• Physics: Velocity, force, acceleration all have magnitude and direction
• Navigation: GPS uses vectors to calculate routes and distances
• Engineering: Forces on structures analyzed using vectors
• Computer Graphics: 3D modeling and animations use vectors
• Aviation: Wind velocity, flight paths, landing approaches
• Gaming: Character movement, projectile paths, collision detection
• Physics: Velocity, force, acceleration all have magnitude and direction
• Navigation: GPS uses vectors to calculate routes and distances
• Engineering: Forces on structures analyzed using vectors
• Computer Graphics: 3D modeling and animations use vectors
• Aviation: Wind velocity, flight paths, landing approaches
• Gaming: Character movement, projectile paths, collision detection