Unit vectors — î, ĵ, k̂; constructing unit vector: Worked examples

Before we start, one reminder of the two symbols you must never confuse:

$A$ (arrow on top) = the whole vector — it has a size and a direction.
$∣ A ∣$ (vertical bars) = the magnitude — a single positive number, the vector's length.
$\hat{A}$ (hat) = the unit vector — same direction as $A$ , but length locked to $1$ .

The recipe glues them: $\hat{A} = \frac{A}{∣ A ∣}$ , and its reverse $A = ∣ A ∣ \hat{A}$ .

The scenario matrix

Every problem this topic can throw is one (or a blend) of these case classes:

#	Case class	What is tricky about it	Hit by
C1	All-positive 2D	The friendly baseline	Ex 1
C2	Mixed signs (a quadrant other than the first)	Signs must survive into $\hat{A}$	Ex 2
C3	3D with a negative component	Pythagoras twice; sign kept	Ex 3
C4	Already a unit vector	Dividing by $1$ changes nothing	Ex 4
C5	Zero vector (degenerate)	$\hat{A}$ is undefined — no direction	Ex 5
C6	Ugly / irrational magnitude	Answer must stay exact, not rounded away	Ex 6
C7	Real-world word problem (units!)	Strip units to get direction, restore later	Ex 7
C8	Exam twist: vector hidden in a difference	Build the vector first, then normalise	Ex 8
C9	Limiting behaviour: one component huge	$\hat{A}$ leans toward that axis	Ex 9

Each example below is tagged with the cell it fills.

Worked Examples

Worked example Example 2 — C2: mixed signs (third quadrant)

Find the unit vector of $A = - 3 \hat{i} - 4 \hat{j}$ .

Forecast: Both components negative $\Rightarrow$ this arrow points down-and-to-the-left (third quadrant). The magnitude cannot know that — it only cares about size. So I expect the signs to survive into $\hat{A}$ .

Step 1. $∣ A ∣ = (- 3)^{2} + (- 4)^{2} = 9 + 16 = 25 = 5$ . Why this step? Squaring kills the minus signs — magnitude is always positive. This is exactly why magnitude alone can never tell you the quadrant.

Step 2. $\hat{A} = \frac{- 3 i ^ - 4 j ^}{5} = - 0.6 \hat{i} - 0.8 \hat{j}$ . Why this step? We divide the whole vector, minus signs included. Dividing by a positive number preserves the sign of each component, so the pointer still aims into the third quadrant.

Verify: $(- 0.6)^{2} + (- 0.8)^{2} = 0.36 + 0.64 = 1. ✓$ Compare with Ex 1: same length- $1$ pointer, but flipped to the opposite direction — the signs did their job.

Worked example Example 3 — C3: 3D with a negative component

Find $\hat{B}$ for $B = 2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ .

Forecast: Three components, one negative. Magnitude uses all three squares, so I expect a whole number if I'm lucky ( $2, 3, 6$ smell like a Pythagorean triple). $\hat{B}$ should keep the minus on $\hat{j}$ .

Step 1. $∣ B ∣ = 2^{2} + (- 3)^{2} + 6^{2} = 4 + 9 + 36 = 49 = 7$ . Why this step? In 3D we apply Pythagoras twice — first across the base ( $x, y$ ), then up to $z$ — which collapses to one square root of the sum of all three squares. Signs vanish under squaring, so $- 3$ contributes $+ 9$ .

Step 2. $\hat{B} = \frac{2 i ^ - 3 j ^ + 6 k ^}{7} = \frac{2}{7} \hat{i} - \frac{3}{7} \hat{j} + \frac{6}{7} \hat{k}$ . Why this step? Same recipe as 2D; the third axis changes nothing about how we normalise.

Verify: $(\frac{2}{7})^{2} + (\frac{3}{7})^{2} + (\frac{6}{7})^{2} = \frac{4 + 9 + 36}{49} = \frac{49}{49} = 1. ✓$ The minus survived on $\hat{j}$ — direction intact.

Worked example Example 5 — C5: the zero vector (degenerate case)

Try to find the unit vector of $0 = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}$ .

Forecast: A vector of length $0$ has no direction — it's a single point, an arrow with no shaft. There is nothing to "point" at. So I predict the recipe breaks.

Step 1. $∣ 0 ∣ = 0^{2} + 0^{2} + 0^{2} = 0$ . Why this step? Measure the length as always — and it comes out zero.

Step 2. $\hat{0} = \frac{0}{0} = \frac{0 i ^ + 0 j ^ + 0 k ^}{0} \Rightarrow undefined$ . Why this step? Division by zero is undefined. Geometrically: to build a pointer you must shrink a real arrow, but a zero-length arrow has no way to say which way. There is no unit vector of the zero vector.

Verify: No arithmetic to plug back — the check is conceptual: any candidate $\hat{0}$ would need magnitude $1$ , but $\hat{0} = 0 /0$ produces $0/0$ , which is meaningless. $✓$ (This is the one case you must guard against in code and on exams.)

Worked example Example 7 — C7: real-world word problem (units matter)

A velocity is $v = 30 \hat{i} + 40 \hat{j}$ measured in $m/s$ . A drone is told to fly at $100 m/s$ in the same direction. Find the drone's velocity vector.

Forecast: "Same direction" screams unit vector — strip $v$ 's length to get the pure pointer, then re-stretch by $100$ . Since $30, 40$ scales to $3, 4, 5$ , expect a tidy answer.

Step 1. $∣ v ∣ = 3 0^{2} + 4 0^{2} = 900 + 1600 = 2500 = 50 m/s$ . Why this step? We only care about $v$ 's direction, so we must first know its size in order to divide it out. Units ( $m/s$ ) ride along on the magnitude.

Step 2. $\overset{v}{^} = \frac{30 i ^ + 40 j ^}{50} = 0.6 \hat{i} + 0.8 \hat{j}$ (dimensionless). Why this step? Dividing $m/s$ by $m/s$ cancels the units — a unit vector is dimensionless, pure direction. This matches the parent note's "a unit vector has no units" point.

Step 3. $v_{drone} = (100 m/s) \overset{v}{^} = 60 \hat{i} + 80 \hat{j} m/s$ . Why this step? Vector = magnitude × direction. The $100 m/s$ supplies the new size and re-attaches the units.

Verify: $∣ v_{drone} ∣ = 6 0^{2} + 8 0^{2} = 3600 + 6400 = 10000 = 100 m/s . ✓$ Right speed, and $\overset{v}{^}$ unchanged $\Rightarrow$ right direction.

Worked example Example 8 — C8: exam twist — vector hidden in a difference

Points $P = (1, 2, 2)$ and $Q = (4, 6, 2)$ . Find the unit vector pointing from $P$ to $Q$ .

Forecast: The vector isn't handed to me — I must build it as $Q - P$ first (this is the displacement). Only then do I normalise. If I forget the subtraction I'll normalise the wrong thing.

Step 1. $P Q = Q - P = (4 - 1) \hat{i} + (6 - 2) \hat{j} + (2 - 2) \hat{k} = 3 \hat{i} + 4 \hat{j} + 0 \hat{k}$ . Why this step? A direction "from $P$ to $Q$ " is the displacement $Q - P$ — head minus tail. See Position vector and displacement. Note the $\hat{k}$ -component is $0$ : the two points share the same height, so the displacement lies flat in a plane.

Step 2. $∣ P Q ∣ = 3^{2} + 4^{2} + 0^{2} = 9 + 16 + 0 = 25 = 5$ . Why this step? A zero component contributes nothing to the length — but you still include it so the recipe stays uniform.

Step 3. $P Q = \frac{3 i ^ + 4 j ^ + 0 k ^}{5} = 0.6 \hat{i} + 0.8 \hat{j}$ . Why this step? Same normalisation. The $0 \hat{k}$ simply drops out.

Verify: $0. 6^{2} + 0. 8^{2} + 0^{2} = 1 = 1. ✓$ Sanity check the direction: $Q$ is to the right and above $P$ , and indeed both non-zero components are positive.

Worked example Example 9 — C9: limiting behaviour — one component dominates

Compare the unit vectors of $E_{1} = 10 \hat{i} + 1 \hat{j}$ and $E_{2} = 1000 \hat{i} + 1 \hat{j}$ . What happens as the $x$ -component grows without bound?

Forecast: As $x$ dwarfs $y$ , the arrow lies almost flat along $+ x$ , so $\hat{E}$ should approach $\hat{i}$ itself. I expect the $\hat{i}$ -coefficient $\to 1$ and the $\hat{j}$ -coefficient $\to 0$ .

Step 1. For $E_{1}$ : $∣ E_{1} ∣ = 1 0^{2} + 1^{2} = 101 \approx 10.0499$ , so $\hat{E}_{1} = \frac{10 i ^ + 1 j ^}{101} \approx 0.9950 \hat{i} + 0.0995 \hat{j} .$ Why this step? Just the recipe; keep enough decimals to see the trend.

Step 2. For $E_{2}$ : $∣ E_{2} ∣ = 100 0^{2} + 1^{2} = 1000001 \approx 1000.0005$ , so $\hat{E}_{2} \approx 0.9999995 \hat{i} + 0.0009999995 \hat{j} .$ Why this step? The bigger $x$ gets, the more the $\hat{i}$ -coefficient hugs $1$ and the $\hat{j}$ -coefficient shrinks toward $0$ .

Step 3 (the limit). As the $x$ -component $\to \infty$ with $y$ fixed, $\frac{x}{x ^{2} + 1} \to 1$ and $\frac{1}{x ^{2} + 1} \to 0$ , hence $\hat{E} \to \hat{i}$ . Why this step? This shows $\hat{i}$ is the limiting pointer of any vector that leans ever more toward the $x$ -axis — the standard basis vectors are the extreme cases of normalisation.

Verify: $0.995 0^{2} + 0.099 5^{2} = 0.99005 + 0.00990 \approx 1.000$ (rounding), and exactly $\frac{1 0 ^{2} + 1 ^{2}}{101} = \frac{101}{101} = 1. ✓$

Recall Quick self-test across the matrix

Cover the answers and name the trap in each.

Which case gives an undefined unit vector, and why? ::: The zero vector (C5) — magnitude $0$ forces division by zero, and a zero-length arrow has no direction. Normalising an already-unit vector gives what? ::: The same vector back (C4) — dividing by $1$ is a no-op. When you're asked for the direction "from $P$ to $Q$ ", what must you build first? ::: The displacement $P Q = Q - P$ (C8), then normalise. As one component grows without bound, the unit vector approaches what? ::: The basis vector of that axis (C9), e.g. $\hat{i}$ . Do negative components survive into $\hat{A}$ ? ::: Yes — dividing by the positive magnitude keeps every sign (C2/C3).

Connections

Parent: Unit vectors
Magnitude and direction of a vector
Vectors — components and resolution
Pythagoras theorem
Position vector and displacement
Vector addition — triangle and parallelogram laws