Motion in a Plane · Lesson 3 of 13

Unit Vectors and Components

“Resolve vectors along chosen directions and connect signed components with magnitude and direction.”

Learning Objectives

• Express vectors using unit vectors and signed components. • Resolve a vector using sine and cosine. • Recover magnitude and direction with the correct quadrant. • Extend components to three dimensions and to rotated axes.

A direction multiplied by a number

A unit vector has magnitude one and specifies direction. It has no physical dimension or unit. The symbols i-hat, j-hat and k-hat denote unit directions along the positive x, y and z axes of a fixed right-handed Cartesian system. A physical vector is built by multiplying those unit directions by components carrying the vector’s units.

Unit vector along a nonzero vectorLaTeX

\overset{n}{^} = \frac{A}{∣ A ∣}, A = ∣ A ∣ \overset{n}{^}

This is defined only if |A| ≠ 0. A zero vector has no unique unit direction.

Cartesian componentsLaTeX

A = A_{x} \hat{i} + A_{y} \hat{j}, A_{x} = A cos θ, A_{y} = A sin θ

θ is measured from the positive x direction. Components are signed numbers with units; A_x i-hat and A_y j-hat are component vectors.

Draw perpendiculars from the arrowhead to the axes. They form a right triangle: the x-component is the signed horizontal projection and the y-component the signed vertical projection. If the angle is given from the y-axis instead, the adjacent side is the y-component. Decide which side is adjacent before choosing sine or cosine; do not attach sine permanently to a particular compass direction.

A vector and its perpendicular components— The orange and green component vectors add head to tail to give the blue vector. A_x and A_y are signed scalar components; the arrows are component vectors.

Recover magnitude and direction

Magnitude and direction in a planeLaTeX

A = A_{x}^{2} + A_{y}^{2}, tan θ = \frac{A _{y}}{A _{x}}

The tangent ratio alone does not select a quadrant. Use component signs, or an atan2(A_y,A_x) calculation. If A_x = 0, read the vertical direction directly.

Quadrant	A_x	A_y	Direction from +x
I	Positive	Positive	Between 0° and 90°
II	Negative	Positive	Between 90° and 180°
III	Negative	Negative	Between 180° and 270°
IV	Positive	Negative	Between 270° and 360°, or a negative angle

A vector in the second quadrant

Problem
A displacement has components −3 m eastward and +4 m northward. Find magnitude and direction, taking east and north as positive axes.

1.The vector is −3 i-hat + 4 j-hat metres. Its magnitude is √(9 + 16) = 5 m.
2.The acute reference angle is arctan(4/3) = 53.13°. The signs place the vector northwest, in quadrant II.
3.The angle from positive x is 180° − 53.13° = 126.87°, equivalently 53.13° north of west. A calculator’s bare arctan(−4/3) would give the wrong quadrant.

Components change when axes rotate

The vector itself does not change when you rotate the coordinate axes. Its numerical components generally do, because the projections are now measured in different directions. The magnitude remains the same. A component is often called a scalar component because it is a signed number multiplying a basis vector; it is not a coordinate-independent scalar physical quantity. This distinction explains why a component can change with axis orientation while mass or temperature does not.

Projection along a chosen unit directionLaTeX

A_{\overset{n}{^}} = A cos ϕ

φ is the angle between A and the unit direction n-hat. The signed projection has the units of A. The component vector is A_n n-hat.

Resolving along diagonal directions

Problem
For A = 2 i-hat + 3 j-hat, find components along i-hat + j-hat and i-hat − j-hat.

1.Each specified direction vector has magnitude √2. Their unit vectors are n-plus = (i-hat + j-hat)/√2 and n-minus = (i-hat − j-hat)/√2. They are perpendicular.
2.The signed scalar projections are (2 + 3)/√2 = 5/√2 and (2 − 3)/√2 = −1/√2.
3.The component vectors are (5/2)(i-hat + j-hat) and (−1/2)(i-hat − j-hat). Their sum is 2 i-hat + 3 j-hat.
4.If writing A = λ(i-hat + j-hat) + μ(i-hat − j-hat), the coefficients are λ = 5/2 and μ = −1/2. These coefficients differ from projections because the direction vectors are not unit vectors.

Two nonparallel directions can form a basis

A vector in a plane can be written as λa + μb for any two nonzero, nonparallel vectors a and b in that plane. The head-to-tail construction identifies the required multiples. Perpendicular directions are convenient because projection and Pythagoras work directly; for an oblique basis, the coefficients generally are not the simple perpendicular projections. Parallel basis vectors cannot represent every direction in a plane.

The extension to three dimensions

Three componentsLaTeX

A = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}, A = A_{x}^{2} + A_{y}^{2} + A_{z}^{2}

The same component approach works in space. A position vector is r = x i-hat + y j-hat + z k-hat.

Direction cosinesLaTeX

A_{x} = A cos α, A_{y} = A cos β, A_{z} = A cos γ, cos^{2} α + cos^{2} β + cos^{2} γ = 1

α, β and γ are angles with the positive x, y and z axes. They are not three ordinary angles in one plane.

The direction-cosine identity follows by substituting the three component expressions into A² = A_x² + A_y² + A_z² and dividing by A². It is valid for a nonzero vector. It also checks whether three proposed direction angles can describe one direction.

Quick check

A vector has A_x < 0 and A_y > 0. Where does it point?

Key Takeaways

• Unit vectors specify direction and carry no physical units. • Signed components multiply unit vectors to build a physical vector. • Use component signs to choose the correct direction quadrant. • Normalise a direction vector before interpreting a coefficient as a scalar projection. • A nonparallel pair spans a plane; three perpendicular components describe space.

Previous · Lesson 2

Vector Addition, Subtraction and Scaling

Next · Lesson 4

Adding Vectors with Components