18.3 Vector Representations, Operations, Magnitude, Direction & Dot Product

Key Takeaways

  • A vector in $\mathbb{R}^2$ is expressed in component form as $\mathbf{v} = \langle v_1, v_2 \rangle = v_1 \mathbf{i} + v_2 \mathbf{j}$, with magnitude $\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2}$ and direction angle $\theta = \text{atan2}(v_2, v_1)$.
  • The unit vector in the direction of any non-zero vector $\mathbf{v}$ is $\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|}$, which has length 1 and shares the identical direction angle.
  • Vector addition $\mathbf{u} + \mathbf{v} = \langle u_1 + v_1, u_2 + v_2 \rangle$ corresponds geometrically to the tip-to-tail method and the parallelogram law, while scalar multiplication scales magnitude by $|c|$ and reverses direction if $c < 0$.
  • The dot product evaluates algebraically as $\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2$ and geometrically as $\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos \theta$; two non-zero vectors are orthogonal if and only if $\mathbf{u} \cdot \mathbf{v} = 0$.
  • In navigation, resultant velocity is the vector sum of airspeed and wind velocity ($\mathbf{v}_{\text{ground}} = \mathbf{v}_{\text{air}} + \mathbf{v}_{\text{wind}}$), with bearings measured clockwise from true North.
Last updated: September 2026

18.3 Vector Representations, Operations, Magnitude, Direction & Dot Product

Vector Fundamentals in the Euclidean Plane ($\mathbb{R}^2$)

A vector in the two-dimensional Euclidean plane is a geometric entity possessing both magnitude (length) and direction. Geometrically, a vector is depicted as a directed line segment with an initial point $P(x_1, y_1)$ and a terminal point $Q(x_2, y_2)$. The vector $\vec{PQ}$ is invariant under translation; placing its initial point at the origin $(0, 0)$ establishes its standard position as a position vector.

The vector is uniquely characterized by its component form $\mathbf{v} = \langle v_1, v_2 \rangle$, where: v1=x2x1,v2=y2y1v_1 = x_2 - x_1, \quad v_2 = y_2 - y_1

The standard unit basis vectors in $\mathbb{R}^2$ are defined along the coordinate axes as $\mathbf{i} = \langle 1, 0 \rangle$ and $\mathbf{j} = \langle 0, 1 \rangle$. Any vector $\mathbf{v} = \langle v_1, v_2 \rangle$ can be written uniquely as a linear combination of these basis vectors: v=v1i+v2j\mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j}


Magnitude, Direction Angle & Normalization

The magnitude (or Euclidean norm) of vector $\mathbf{v} = \langle v_1, v_2 \rangle$, denoted $|\mathbf{v}|$, is calculated via the Pythagorean distance formula: v=v12+v22\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2}

The direction angle $\theta$ is the counterclockwise angle measured from the positive $x$-axis to the vector in standard position ($0^\circ \le \theta < 360^\circ$ or $0 \le \theta < 2\pi$). The components satisfy: v1=vcosθ,v2=vsinθv_1 = \|\mathbf{v}\| \cos \theta, \quad v_2 = \|\mathbf{v}\| \sin \theta

To find $\theta$, compute the reference angle $\alpha = \arctan\left|\frac{v_2}{v_1}\right|$ for $v_1 \neq 0$ and adjust for the quadrant containing $(v_1, v_2)$:

  • Quadrant I ($v_1 > 0, v_2 > 0$): $\theta = \alpha = \arctan(v_2 / v_1)$
  • Quadrant II ($v_1 < 0, v_2 > 0$): $\theta = 180^\circ - \alpha = 180^\circ + \arctan(v_2 / v_1)$
  • Quadrant III ($v_1 < 0, v_2 < 0$): $\theta = 180^\circ + \alpha = 180^\circ + \arctan(v_2 / v_1)$
  • Quadrant IV ($v_1 > 0, v_2 < 0$): $\theta = 360^\circ - \alpha = 360^\circ + \arctan(v_2 / v_1)$

A unit vector is any vector with magnitude 1. The process of dividing a non-zero vector by its magnitude to obtain a unit vector in the identical direction is normalization: u=vv=v1v12+v22,v2v12+v22\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \left\langle \frac{v_1}{\sqrt{v_1^2 + v_2^2}}, \frac{v_2}{\sqrt{v_1^2 + v_2^2}} \right\rangle


Vector Operations: Addition, Subtraction & Scalar Multiplication

Vector algebra in $\mathbb{R}^2$ operates componentwise:

  1. Vector Addition: For $\mathbf{u} = \langle u_1, u_2 \rangle$ and $\mathbf{v} = \langle v_1, v_2 \rangle$: u+v=u1+v1,u2+v2\mathbf{u} + \mathbf{v} = \langle u_1 + v_1, u_2 + v_2 \rangle Geometrically, vector addition corresponds to the tip-to-tail method (translating the tail of $\mathbf{v}$ to the tip of $\mathbf{u}$) and the parallelogram law (the resultant is the directed diagonal of the parallelogram formed by $\mathbf{u}$ and $\mathbf{v}$ from a common origin).
  2. Vector Subtraction: uv=u+(v)=u1v1,u2v2\mathbf{u} - \mathbf{v} = \mathbf{u} + (-\mathbf{v}) = \langle u_1 - v_1, u_2 - v_2 \rangle Geometrically, $\mathbf{u} - \mathbf{v}$ represents the directed segment pointing from the terminal point of $\mathbf{v}$ to the terminal point of $\mathbf{u}$.
  3. Scalar Multiplication: For scalar $c \in \mathbb{R}$: cv=cv1,cv2c\mathbf{v} = \langle c v_1, c v_2 \rangle The magnitude scales to $|c\mathbf{v}| = |c| \cdot |\mathbf{v}|$. If $c > 0$, direction is preserved; if $c < 0$, direction reverses by $180^\circ$; if $c = 0$, the product is the zero vector $\mathbf{0} = \langle 0, 0 \rangle$.

The Dot Product & Geometric Angle Determination

The dot product (or Euclidean inner product) is an algebraic operation between two vectors that returns a scalar: uv=u1v1+u2v2\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2

The algebraic dot product connects to geometry through the Law of Cosines: uv=uvcosθ\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos \theta

where $\theta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$ ($0^\circ \le \theta \le 180^\circ$). Solving for $\cos \theta$ yields: cosθ=uvuv=u1v1+u2v2u12+u22v12+v22\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} = \frac{u_1 v_1 + u_2 v_2}{\sqrt{u_1^2 + u_2^2} \sqrt{v_1^2 + v_2^2}}

The sign of the dot product categorizes the angle $\theta$:

  • $\mathbf{u} \cdot \mathbf{v} > 0 \iff 0^\circ \le \theta < 90^\circ$ (acute angle)
  • $\mathbf{u} \cdot \mathbf{v} = 0 \iff \theta = 90^\circ$ (orthogonal / perpendicular vectors)
  • $\mathbf{u} \cdot \mathbf{v} < 0 \iff 90^\circ < \theta \le 180^\circ$ (obtuse angle)

Thus, two non-zero vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if and only if $\mathbf{u} \cdot \mathbf{v} = 0$.


Real-World Applications: Static Equilibrium & Navigation

  1. Resultant Force and Static Equilibrium: Concurrent forces acting on a body sum vectorially to form the resultant force $\mathbf{F}{\text{net}} = \sum \mathbf{F}i$. A system is in static equilibrium if and only if $\mathbf{F}{\text{net}} = \mathbf{0}$. The equilibrant force required to balance a system is $\mathbf{F}{\text{equilibrant}} = -\mathbf{F}_{\text{net}}$.
  2. Aviation & Marine Navigation: Navigation defines direction using bearings measured clockwise from true North ($000^\circ$). A bearing of $\phi$ converts to standard Cartesian angle $\theta$ via $\theta = 90^\circ - \phi$ (mod $360^\circ$), with components $\mathbf{v} = \langle |\mathbf{v}| \sin \phi, |\mathbf{v}| \cos \phi \rangle$. The actual ground path (track and ground speed) is the resultant vector sum of the vehicle's airspeed vector and the atmospheric wind vector: vground=vair+vwind\mathbf{v}_{\text{ground}} = \mathbf{v}_{\text{air}} + \mathbf{v}_{\text{wind}}

2D Vector Operations Summary

OperationAlgebraic FormulaGeometric MeaningScalar or Vector
Component Form$\mathbf{v} = \langle v_1, v_2 \rangle = \langle x_2 - x_1, y_2 - y_1 \rangle$Displacement from initial to terminal pointVector
Magnitude$|\mathbf{v}| = \sqrt{v_1^2 + v_2^2}$Euclidean length of the directed segmentNon-negative Scalar
Unit Vector$\mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|}$Pure direction vector with magnitude 1Vector
Vector Addition$\mathbf{u} + \mathbf{v} = \langle u_1 + v_1, u_2 + v_2 \rangle$Parallelogram diagonal / Tip-to-tail resultantVector
Scalar Product$c\mathbf{v} = \langle c v_1, c v_2 \rangle$Stretches length by $c
Dot Product$\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2$$|\mathbf{u}| |\mathbf{v}| \cos \theta$; zero if orthogonalScalar
Angle Between$\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}$Smallest non-negative angle between vectorsScalar Angle

Worked Exemplar: Resultant Velocity & Course in Navigation

Problem: An airplane flies on a compass heading (bearing) of $060^\circ$ with an airspeed of $450\text{ mph}$. A crosswind blows toward bearing $045^\circ$ (northeast) at $60\text{ mph}$.

  1. Express the airspeed and wind velocity vectors in component form.
  2. Determine the airplane's resultant ground velocity vector $\mathbf{v}_g$.
  3. Calculate the aircraft's actual ground speed and ground track heading.

Step 1: Convert Navigation Bearings to Cartesian Components. For bearing $\phi$, components are $v_x = |\mathbf{v}| \sin \phi$ and $v_y = |\mathbf{v}| \cos \phi$ (or using Cartesian angle $\theta = 90^\circ - \phi$):

  • Airspeed Vector ($\mathbf{v}_a$): Bearing $\phi = 60^\circ \implies \theta_a = 90^\circ - 60^\circ = 30^\circ$. va=450cos30,450sin30=450(32),450(12)=2253,225389.71,225.00\mathbf{v}_a = \langle 450 \cos 30^\circ, 450 \sin 30^\circ \rangle = \left\langle 450 \left(\frac{\sqrt{3}}{2}\right), 450 \left(\frac{1}{2}\right) \right\rangle = \langle 225\sqrt{3}, 225 \rangle \approx \langle 389.71, 225.00 \rangle
  • Wind Vector ($\mathbf{v}_w$): Bearing $\phi = 45^\circ \implies \theta_w = 90^\circ - 45^\circ = 45^\circ$. vw=60cos45,60sin45=60(22),60(22)=302,30242.43,42.43\mathbf{v}_w = \langle 60 \cos 45^\circ, 60 \sin 45^\circ \rangle = \left\langle 60 \left(\frac{\sqrt{2}}{2}\right), 60 \left(\frac{\sqrt{2}}{2}\right) \right\rangle = \langle 30\sqrt{2}, 30\sqrt{2} \rangle \approx \langle 42.43, 42.43 \rangle

Step 2: Calculate Resultant Ground Velocity Vector ($\mathbf{v}_g$). Add components algebraically: vg=va+vw=2253+302,225+302389.71+42.43,225.00+42.43=432.14,267.43\mathbf{v}_g = \mathbf{v}_a + \mathbf{v}_w = \langle 225\sqrt{3} + 30\sqrt{2}, 225 + 30\sqrt{2} \rangle \approx \langle 389.71 + 42.43, 225.00 + 42.43 \rangle = \langle 432.14, 267.43 \rangle

Step 3: Calculate Ground Speed & Direction. Compute the magnitude of $\mathbf{v}_g$: vg=(432.14)2+(267.43)2=186,745.0+71,519.2=258,264.2508.20 mph\|\mathbf{v}_g\| = \sqrt{(432.14)^2 + (267.43)^2} = \sqrt{186,745.0 + 71,519.2} = \sqrt{258,264.2} \approx 508.20\text{ mph}

Compute the standard Cartesian direction angle $\theta_g$: θg=arctan(267.43432.14)arctan(0.61885)31.75\theta_g = \arctan\left(\frac{267.43}{432.14}\right) \approx \arctan(0.61885) \approx 31.75^\circ

Convert back to navigational bearing $\phi_g$: ϕg=9031.75=58.25058\phi_g = 90^\circ - 31.75^\circ = 58.25^\circ \approx 058^\circ

The airplane travels at an actual ground speed of approximately $508.2\text{ mph}$ along a true course heading of $058.3^\circ$.

Test Your Knowledge

Find the measure of the angle theta between the vectors u = <3, -4> and v = <5, 12> in the 2D plane.

A
B
C
D
Test Your Knowledge

For what real value of the scalar k are the vectors a = <2k - 1, 4> and b = <3, k + 5> orthogonal?

A
B
C
D
Test Your Knowledge

Which of the following represents the unit vector u that has the same direction as the vector v = <-5, 12>?

A
B
C
D
Test Your Knowledge

Two forces F_1 = <40, -15> N and F_2 = <-16, 85> N act concurrently on a particle at the origin. What is the magnitude of the resultant force acting on the particle, and what equilibrant force F_3 is required to maintain static equilibrium?

A
B
C
D