Unit vectors — î, ĵ, k̂; constructing unit vector: Question bank

This is a companion to the parent Unit Vectors note. No new formulas are introduced — only ways the old ones get misused.

True or false — justify

Every vector of length 1 is a unit vector.

True — that is the definition. "Unit" means magnitude exactly

1

; it says nothing about which direction it points.

\hat{i} + \hat{j}

is a unit vector.

False — its magnitude is

1^{2} + 1^{2} = 2 \approx 1.41

, not

1

. Adding two unit vectors does not give a unit vector unless they are the same vector.

A unit vector must point along one of the axes.

False —

0.6 \hat{i} + 0.8 \hat{j}

is a perfectly good unit vector pointing diagonally. Only

\hat{i}, \hat{j}, \hat{k}

happen to lie on axes; any direction has its own unit vector.

\hat{A}

is a unit vector, then

- \hat{A}

is also a unit vector.

True — negating flips the direction but multiplies length by

∣ - 1∣ = 1

, so the magnitude stays

1

. It points the opposite way with the same length.

A unit vector is dimensionless, so multiplying it by a length gives a length.

True — the unit vector carries only direction with no units, so the physical units come entirely from the scalar magnitude you multiply by (e.g. metres, newtons).

The zero vector

0

has a unit vector.

False —

0

has magnitude

0

, and

\hat{A} = 0 /0

is dividing by zero. A vector with no length has no direction to point, so no unit vector exists.

Two different vectors can share the same unit vector.

True —

3 \hat{i} + 4 \hat{j}

and

6 \hat{i} + 8 \hat{j}

both give

\hat{A} = 0.6 \hat{i} + 0.8 \hat{j}

. They point the same way and differ only in length, which the unit vector discards.

\hat{i} \cdot \hat{i} = 1

but

\hat{i} \cdot \hat{j} = 0

True — each basis vector dotted with itself gives

∣ \hat{i} ∣∣ \hat{i} ∣ cos 0^{\circ} = 1

, and dotted with a perpendicular one gives

cos 9 0^{\circ} = 0

. This is exactly the orthonormal property.

Spot the error

A = 3 \hat{i} + 4 \hat{j}

, so

∣ A ∣ = 3 + 4 = 7

The components are perpendicular, so you combine them with Pythagoras, not addition:

3^{2} + 4^{2} = 5

. Plain addition would only work for vectors along the same line.

"To shrink

A = 6 \hat{i} + 8 \hat{j}

to a unit vector, divide by

2

to get

3 \hat{i} + 4 \hat{j}

You must divide by the vector's own magnitude (

10

), not an arbitrary number. Dividing by

2

gives length

5

, not

1

; only

A /∣ A ∣

guarantees length exactly

1

\hat{B} = \frac{1}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k}

, but magnitude is positive so I'll write

+ \frac{2}{3} \hat{j}

The minus sign is not about magnitude — it encodes that the component points in

- y

. Dropping it changes the direction; only the overall magnitude is forced positive, never the individual signed components.

∣ A ∣ = A_{x}^{2} + A_{y}^{2} + A_{z}^{2}

, and

A_{y} = - 2

, so I get

A_{x}^{2} - 4 + A_{z}^{2}

Squaring a negative gives a positive:

(- 2)^{2} = + 4

. The square erases the sign, which is why magnitude is always non-negative regardless of component signs.

"A unit vector points, so it has no units — meaning it's not really a vector."

It is a genuine vector (it has direction and magnitude

1

); "no units" only means it is dimensionless. Dimensionless is not the same as "not a vector".

A = ∣ A ∣ \cdot \hat{A}

, so to rebuild I add the magnitude and the unit vector."

It is multiplication, not addition: magnitude (a scalar) times the direction (unit vector). Adding a scalar to a vector is not even a defined operation.

Why questions

Why do we divide by

∣ A ∣

specifically, and not some other quantity?

Because dividing by exactly the vector's own length rescales its magnitude to

∣ A ∣/∣ A ∣ = 1

while multiplying by the positive scalar

1/∣ A ∣

never rotates it — so direction is preserved and length becomes

1

Why does Pythagoras (squares), not simple addition, give the magnitude?

Because

\hat{i}, \hat{j}, \hat{k}

are mutually perpendicular, the components form the sides of a right-angled box; the vector is its diagonal, and diagonal lengths come from Pythagoras applied to perpendicular sides.

Why is the unit vector called a "pointer" or "compass needle"?

It strips off the "how much" (magnitude) and keeps only the "which way" (direction), like a compass needle that is always the same length and only ever tells you a bearing.

Why does multiplying

A

by the positive scalar

1/∣ A ∣

not change its direction?

A positive scalar only stretches or shrinks a vector along its own line; direction reverses only if the scalar is negative, and

1/∣ A ∣

is always positive since

∣ A ∣ > 0

Why can we write any vector as

A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}

Because the three basis vectors point along independent perpendicular axes, so scaling and adding them reaches every point in space — the hats supply direction, the numbers supply amount, together forming a sum.

Why does the sign of a component live in the number and not in the hat?

The hats are fixed pointers along

+ x, + y, + z

; to point the opposite way you attach a negative number, e.g.

- 2 \hat{j}

means "2 units in the

- y

direction". This keeps the basis constant.

Edge cases

What is the unit vector of a vector already having magnitude

1

The vector itself — dividing by

1

changes nothing, so

\hat{A} = A

. It is already a pure pointer.

What is the unit vector of

\hat{i}

alone?

Just

\hat{i}

, since

∣ \hat{i} ∣ = 1

. The standard basis vectors are unit vectors already; there is nothing to rescale.

Can a unit vector be exactly

\hat{i}

and

\hat{j}

at the same time?

No — a single unit vector has one direction.

\hat{i}

and

\hat{j}

point along different (perpendicular) axes, so they are two distinct unit vectors, not one.

What happens to the "construct a unit vector" recipe if one component is zero, e.g.

A = 0 \hat{i} + 5 \hat{j}

It still works:

∣ A ∣ = 0 + 25 = 5

, so

\hat{A} = \hat{j}

. A zero component simply means the pointer has no reach along that axis — nothing breaks.

Is there a unit vector pointing in "no direction"?

No — that would be the zero vector, which has magnitude

0

, and a unit vector by definition needs magnitude

1

. "No direction" and "unit length" are contradictory.

If two vectors have equal magnitude but opposite unit vectors, how are they related?

They are exact negatives of each other: same length, directly opposite directions, so

A = - B

(their displacements cancel).

Recall One-line takeaways to keep

Unit = magnitude 1, not "small", not "unitless-therefore-not-a-vector".
Magnitude always needs squares (Pythagoras), because components are perpendicular.
Signs live in the numbers; the hats are fixed positive-axis pointers.
The zero vector is the one vector with no unit vector.

Connections

Unit vectors — î, ĵ, k̂; constructing unit vector
Vectors — components and resolution
Magnitude and direction of a vector
Vector addition — triangle and parallelogram laws
Dot product
Position vector and displacement
Pythagoras theorem