Learning Objectives
Distinguish a point from a displacement vector.
Compute vector sums, scalar multiples, lengths, and unit vectors.
Use coordinates to find distance and a sphere equation.
Prerequisites: coordinate geometry, the Pythagorean theorem, and algebra with square roots. These geometric tools support later work with vector-valued functions and multivariable derivatives.
1. Points Locate; Vectors Describe Displacement
A point P = (x, y, z) gives a location relative to three perpendicular coordinate axes. A vector v = ⟨a, b, c⟩ records a displacement: a units in the x direction, b in the y direction, and c in the z direction. The same vector can start at different points while retaining its components.
For points A = (x₁, y₁, z₁) and B = (x₂, y₂, z₂), the vector from A to B is AB = ⟨x₂ − x₁, y₂ − y₁, z₂ − z₁⟩. Subtract starting coordinates from ending coordinates. Reversing the direction negates every component.
Worked Example: Displacement and Length
Let A = (1, −2, 3) and B = (4, 2, 3). Then AB = ⟨4 − 1, 2 − (−2), 3 − 3⟩ = ⟨3, 4, 0⟩.
Its magnitude is |AB| = √(3² + 4² + 0²) = 5. Thus the distance between the points is 5 coordinate units.
The reverse displacement BA = ⟨−3, −4, 0⟩ has the same length but opposite direction.
2. Calculate Component by Component
Operation | Rule |
|---|---|
Addition | ⟨a,b,c⟩ + ⟨d,e,f⟩ = ⟨a+d,b+e,c+f⟩ |
Scalar multiplication | k⟨a,b,c⟩ = ⟨ka,kb,kc⟩ |
Magnitude | |⟨a,b,c⟩| = √(a²+b²+c²) |
Adding vectors models successive displacements. Multiplying by a positive scalar stretches or shrinks a vector without changing its direction. A negative scalar also reverses its direction. Multiplication by zero produces the zero vector, which has length zero and no defined direction.
Worked Example: Combining Displacements
Let u = ⟨2, −1, 3⟩ and v = ⟨−4, 2, 1⟩. Compute 2u + v.
First, 2u = ⟨4, −2, 6⟩. Then add matching components: 2u + v = ⟨0, 0, 7⟩.
The result points in the positive z direction and has length 7. The x and y displacements cancel.
3. Separate Length from Direction
For a nonzero vector v, the unit vector in its direction is v/|v|. A unit vector has length 1. To construct a vector of specified length L in that direction, multiply the unit vector by L, where L ≥ 0.
Example: A Unit Direction
For v = ⟨3, 4, 0⟩, divide by its magnitude 5 to obtain u = ⟨3/5, 4/5, 0⟩.
Check: |u| = √(9/25 + 16/25) = 1. A vector of length 10 in that direction is 10u = ⟨6, 8, 0⟩.
4. A Sphere Is a Distance Condition
All points at distance R from a center (a, b, c) form a sphere when R > 0. Squaring the distance formula gives (x − a)² + (y − b)² + (z − c)² = R². For example, a sphere centered at (1, −2, 0) with radius 3 has equation (x − 1)² + (y + 2)² + z² = 9.
Common Mistakes
Use end minus start consistently when finding displacement.
Do not divide the zero vector by its magnitude.
The constant in a sphere equation is the radius squared, not the radius.
5. Practice
1. Find PQ and the distance between P = (−1, 0, 2) and Q = (1, 3, 8).
2. Find a unit vector in the direction ⟨0, −5, 12⟩.
3. Write the equation of the sphere centered at (−2, 1, 4) with radius 5.
Worked Solutions
1. PQ = ⟨2, 3, 6⟩ and its length is √(4 + 9 + 36) = 7.
2. The magnitude is √(25 + 144) = 13, so the unit vector is ⟨0, −5/13, 12/13⟩.
3. The equation is (x + 2)² + (y − 1)² + (z − 4)² = 25.
6. Summary
- A vector records displacement rather than a fixed starting location.
- Vector arithmetic uses matching coordinates; magnitude uses the three-dimensional distance formula.
- Normalize a nonzero vector by dividing by its magnitude.
- A sphere equation expresses a fixed distance from its center.