Vectors in Precalculus
Component form, magnitude, direction, vector operations, dot product, and real-world applications. Master vectors from first principles through applied problems.
Quick Reference — Key Formulas
Vector Basics
Component Form
A 2D vector from point A to point B is written as:
From A(1, 3) to B(4, 7): v = ⟨3, 4⟩
Standard Unit Vectors
Any vector can be written using i and j:
⟨3, −2⟩ = 3i − 2j
Vector Properties
Vector Operations
| Operation | Formula | Example |
|---|---|---|
| Addition | ⟨a, b⟩ + ⟨c, d⟩ = ⟨a+c, b+d⟩ | ⟨2, 3⟩ + ⟨1, −5⟩ = ⟨3, −2⟩ |
| Subtraction | ⟨a, b⟩ − ⟨c, d⟩ = ⟨a−c, b−d⟩ | ⟨5, 2⟩ − ⟨3, 4⟩ = ⟨2, −2⟩ |
| Scalar multiplication | k⟨a, b⟩ = ⟨ka, kb⟩ | 3⟨2, −1⟩ = ⟨6, −3⟩ |
| Magnitude | |⟨a, b⟩| = √(a² + b²) | |⟨3, 4⟩| = √(9 + 16) = 5 |
| Unit vector | û = v / |v| = ⟨a/|v|, b/|v|⟩ | û for ⟨3, 4⟩: ⟨3/5, 4/5⟩ |
| Dot product | ⟨a, b⟩ · ⟨c, d⟩ = ac + bd | ⟨2, 3⟩ · ⟨4, −1⟩ = 8 + (−3) = 5 |
Dot Product — Two Forms
Component Form
Multiply corresponding components, then add. This form is easier for computation.
Geometric Form
Reveals the angle between vectors. Set the two forms equal to find θ.
Perpendicular Test
Vectors are perpendicular (orthogonal) when their dot product = 0.
u = ⟨2, 3⟩ and v = ⟨−3, 2⟩: u · v = (2)(−3) + (3)(2) = −6 + 6 = 0 → perpendicular ✓
Real-World Applications
Resultant Force
Two forces act on an object: F₁ = 40 N at 0°, F₂ = 30 N at 90°. Find the resultant.
- 1.Convert to components: F₁ = ⟨40, 0⟩, F₂ = ⟨0, 30⟩
- 2.Add: R = ⟨40 + 0, 0 + 30⟩ = ⟨40, 30⟩
- 3.Magnitude: |R| = √(1600 + 900) = √2500 = 50 N
- 4.Direction: θ = arctan(30/40) = arctan(0.75) ≈ 36.9°
Navigation / Bearing
A boat travels 12 mph at N30°E. What are the north and east components of velocity?
- 1.N30°E = 30° from North toward East
- 2.East component: 12 sin(30°) = 12(0.5) = 6 mph East
- 3.North component: 12 cos(30°) = 12(√3/2) ≈ 10.4 mph North
- 4.Velocity vector: ⟨6, 10.4⟩ (East, North)
Angle Between Vectors
Find the angle between u = ⟨1, 2⟩ and v = ⟨3, −1⟩.
- 1.Dot product: u · v = (1)(3) + (2)(−1) = 3 − 2 = 1
- 2.|u| = √(1 + 4) = √5, |v| = √(9 + 1) = √10
- 3.cos(θ) = 1 / (√5 · √10) = 1/√50 = 1/(5√2)
- 4.θ = arccos(1/(5√2)) ≈ 81.9°
Frequently Asked Questions
What is a vector in precalculus?
A vector is a quantity with both magnitude (size) and direction, unlike a scalar which only has magnitude. In precalculus, vectors are typically written in component form ⟨a, b⟩ where a is the horizontal component and b is the vertical component. For example, ⟨3, 4⟩ represents a vector that moves 3 units right and 4 units up. Vectors are used to model forces, velocity, displacement, and navigation problems.
How do you find the magnitude and direction of a vector?
For vector v = ⟨a, b⟩: Magnitude = |v| = √(a² + b²). Direction angle θ = arctan(b/a), measured from the positive x-axis (adjust for quadrant). Example: v = ⟨3, 4⟩ has magnitude = √(9 + 16) = √25 = 5, and direction = arctan(4/3) ≈ 53.1°.
What does the dot product tell you?
The dot product of two vectors gives a scalar (number), not a vector. For u = ⟨a, b⟩ and v = ⟨c, d⟩: u · v = ac + bd. The dot product equals |u||v|cos(θ) where θ is the angle between the vectors. This means: if u · v = 0, the vectors are perpendicular. If u · v > 0, the angle is acute. If u · v < 0, the angle is obtuse.
Practice vector problems
Work through magnitude, dot products, angles, and applications with step-by-step solutions. Free to start — no account required.
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