VECTOR MATH
Self-Study Notebook — Addition · Subtraction · Dot Product · Magnitude · Direction
📌 Read the examples first, then try the quiz! ✏️
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① Vector Addition & Subtraction
ADD = tip to tail / SUBTRACT = flip then add
Component-wise: (a₁, a₂) ± (b₁, b₂) = (a₁±b₁, a₂±b₂)
⚠️ Order matters in subtraction! ← tricky!
📖 EXAMPLE

If \(\vec{a} = (3, -1)\) and \(\vec{b} = (2, 4)\), then
\(\vec{a} + \vec{b} = (3+2,\; -1+4) = (5, 3)\)
\(\vec{a} - \vec{b} = (3-2,\; -1-4) = (1, -5)\)

1
Given \(\vec{u} = (4, -2)\) and \(\vec{v} = (-1, 5)\),
what is \(\vec{u} + \vec{v}\)?
Easy
Answer: B — (3, 3)
Add component-wise: \((4 + (-1),\; -2 + 5) = (3, 3)\)
✏️ Don't mix up the signs on the second component!
2
\(\vec{a} = (6, 2)\), \(\vec{b} = (3, 8)\). Find \(\vec{a} - \vec{b}\). ⚡ Common Mistake
Easy
Answer: A — (3, −6)
\((6-3,\; 2-8) = (3, -6)\)
⚠️ Option B is \(\vec{b} - \vec{a}\)! Order matters!
3
Which expression equals \(\vec{a} - \vec{b} + \vec{c}\)?
Medium
Answer: C
Subtracting \(\vec{b}\) is the same as adding \(-\vec{b}\).
B is a trap: \(\vec{a} - (\vec{b}+\vec{c}) = \vec{a} - \vec{b} - \vec{c}\) — signs change inside parentheses!
4
3D vectors: \(\vec{p} = (1, 0, -3)\), \(\vec{q} = (2, -1, 4)\).
Find \(\vec{p} + \vec{q}\).
Easy
Answer: B — (3, −1, 1)
\((1+2,\; 0+(-1),\; -3+4) = (3, -1, 1)\)
✏️ \(-3 + 4 = 1\), not \(-1\)! Careful with negative numbers.
② Magnitude (Length)
|v| = √(x² + y²) — Pythagorean Theorem in disguise!
3D: \(|\vec{v}| = \sqrt{x^2 + y^2 + z^2}\)
⚠️ Magnitude is always ≥ 0! Square first, then add, then root.
📖 EXAMPLE

\(\vec{v} = (3, 4)\)
\(|\vec{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
Classic 3-4-5 right triangle!

5
Find \(|\vec{v}|\) where \(\vec{v} = (5, 12)\).
Easy
Answer: C — 13
\(\sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13\)
✏️ 5-12-13 is a Pythagorean triple — memorise it!
6
\(\vec{w} = (-3, 4)\). What is \(|\vec{w}|\)? ⚡ Trap!
Easy
Answer: B — 5
\(\sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5\)
⚠️ \((-3)^2 = 9\), NOT \(-9\)! Squaring removes the negative sign.
7
3D vector \(\vec{r} = (2, 2, 1)\). Find \(|\vec{r}|\).
Easy
Answer: A — 3
\(\sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3\)
③ Direction — Unit Vector
Unit = direction only, no size → divide by magnitude
\(\hat{v} = \dfrac{\vec{v}}{|\vec{v}|}\)    Always length = 1
⚠️ Check: \(|\hat{v}|\) must equal 1 — use this to verify!
📖 EXAMPLE

\(\vec{v} = (3, 4)\), \(|\vec{v}| = 5\)
\(\hat{v} = \left(\dfrac{3}{5}, \dfrac{4}{5}\right) = (0.6,\; 0.8)\)
Check: \(\sqrt{0.6^2 + 0.8^2} = \sqrt{0.36 + 0.64} = \sqrt{1} = 1\) ✓

8
Find the unit vector of \(\vec{v} = (0, -7)\).
Easy
Answer: C — (0, −1)
\(|\vec{v}| = 7\), so \(\hat{v} = \left(\frac{0}{7}, \frac{-7}{7}\right) = (0, -1)\)
⚠️ Preserve the direction (negative sign stays!).
9
\(\vec{v} = (1, 1)\). Which is the correct unit vector? ⚡ Fraction alert
Medium
Answer: B
\(|\vec{v}| = \sqrt{1^2+1^2} = \sqrt{2}\), so \(\hat{v} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
⚠️ Option C: \(|\frac{1}{2}, \frac{1}{2}| = \frac{\sqrt{2}}{2} \ne 1\) — not a unit vector!
10
Which statement about a unit vector is FALSE?
Medium
Answer: D — FALSE statement
Unit vectors exist in any dimension (2D, 3D, even higher).
A, B, C are all TRUE facts about unit vectors.
④ Dot Product
DOT = multiply matching components, then ADD
\(\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2\) → result is a NUMBER (scalar!)
\(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\)
⚠️ Dot product ≠ vector — it's a scalar! No direction.
PERPENDICULAR → DOT = 0  |  PARALLEL → |DOT| = |a||b|
\(\theta = 90°\) → \(\cos 90° = 0\) → dot product = 0    ← KEY FACT
\(\theta = 0°\) → \(\cos 0° = 1\) → dot product is maximised
📖 EXAMPLE

\(\vec{a} = (2, 3)\), \(\vec{b} = (4, -1)\)
\(\vec{a} \cdot \vec{b} = (2)(4) + (3)(-1) = 8 - 3 = 5\)
Since \(5 \ne 0\), the vectors are not perpendicular.

11
Compute \(\vec{a} \cdot \vec{b}\) for \(\vec{a} = (3, -2)\) and \(\vec{b} = (1, 4)\).
Easy
Answer: C — −5
\((3)(1) + (-2)(4) = 3 - 8 = -5\)
⚠️ Option B is wrong — dot product is a scalar, never a vector!
12
Which pair of vectors is perpendicular? ⚡ Key concept
Medium
Answer: B
\((3)(-4) + (4)(3) = -12 + 12 = 0\) ✓ Perpendicular!
A: same direction (parallel). C: also parallel. D: \(1+2=3 \ne 0\).
✏️ Quick trick: swap components and negate one to get a perpendicular vector!
13
If \(\vec{a} \cdot \vec{b} > 0\), what can we conclude about angle \(\theta\) between them?
Medium
Answer: A — acute angle
\(\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta\). Since magnitudes > 0, dot > 0 means \(\cos\theta > 0\), so \(\theta < 90°\).
Memory: positive dot → acute (pointing same-ish way)
14
3D: \(\vec{a} = (1, -1, 2)\), \(\vec{b} = (3, 1, 0)\). Find \(\vec{a} \cdot \vec{b}\).
Easy
Answer: A — 2
\((1)(3) + (-1)(1) + (2)(0) = 3 - 1 + 0 = 2\)
✏️ Don't forget the z-component! \((2)(0) = 0\), adds nothing.
15
\(\vec{a} = (k, 3)\) and \(\vec{b} = (2, -1)\) are perpendicular. Find \(k\). ⚡ Algebra + Vectors
Tricky
Answer: B — k = 3/2
Perpendicular → dot = 0:
\(2k + (3)(-1) = 0 \Rightarrow 2k - 3 = 0 \Rightarrow k = \frac{3}{2}\)
⑤ Mixed Challenges — Don't Get Tricked!
BIG PICTURE — 4 KEY FACTS
1. ADD/SUBTRACT: component-wise → vector result
2. MAGNITUDE: √(sum of squares) → scalar ≥ 0
3. UNIT VECTOR: divide by magnitude → length = 1
4. DOT PRODUCT: multiply + sum → scalar; = 0 means ⊥
16
\(\vec{v} = (6, 8)\). Find the unit vector, then compute its magnitude. ⚡ Verify yourself
Medium
Answer: D — 1 (always!)
\(|\vec{v}| = 10\), \(\hat{v} = (0.6, 0.8)\)
\(|\hat{v}| = \sqrt{0.36+0.64} = \sqrt{1} = 1\)
The whole point of a unit vector — magnitude is always 1.
17
If \(|\vec{a}| = 5\) and \(|\vec{b}| = 4\), what is the maximum possible value of \(\vec{a} \cdot \vec{b}\)?
Tricky
Answer: B — 20
\(\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta\). Max when \(\cos\theta = 1\) (i.e., \(\theta = 0°\), same direction).
Max = \(5 \times 4 \times 1 = 20\)
18
\(\vec{a} = (2, 1)\), \(\vec{b} = (1, 3)\). Find \(|\vec{a} + \vec{b}|\). ⚡ Two-step!
Medium
Answer: C — 5
Step 1: \(\vec{a}+\vec{b} = (3, 4)\)
Step 2: \(|(3,4)| = \sqrt{9+16} = 5\)
⚠️ A is wrong! \(|\vec{a}+\vec{b}| \ne |\vec{a}|+|\vec{b}|\) in general — Triangle Inequality!
19
Which vector is parallel to \(\vec{v} = (2, -6)\)? ⚡ Ratio test
Tricky
Answer: C — (−1, 3)
Parallel means one is a scalar multiple of the other.
\((-1, 3) = -\frac{1}{2}(2, -6)\) ✓
A is perpendicular (dot = 12+12 = 24... no wait: \(2 \times 6 + (-6)(-2) = 12+12=24\ne0\)). Check ratios: \(\frac{-1}{2} = \frac{3}{-6} = -\frac{1}{2}\) ✓
20
\(\vec{a} \cdot \vec{a}\) equals what? ⚡ Mind-bender! ← Think carefully!
Tricky
Answer: B — |a|²
\(\vec{a}\cdot\vec{a} = |\vec{a}||\vec{a}|\cos 0° = |\vec{a}|^2 \times 1 = |\vec{a}|^2\)
Or component-wise: \((a_1^2 + a_2^2) = |\vec{a}|^2\)
✏️ This is a useful shortcut: dot with itself = square of magnitude!

🎉 All Done!