This material sits right at the start of multivariable calculus (Calc III). The course extends everything you already know about vectors in two dimensions into three-dimensional space, then builds the tools you will use for the rest of the quarter: curves, surfaces, partial derivatives, multiple integrals. If you are comfortable with vector addition and scalar multiplication in R2, you have the right foundation. If those feel shaky, revisit your Calc I vector notes before continuing.
The concepts here (magnitude, unit vectors, distance, sphere equations) are not difficult on their own, but they reappear constantly in later topics. Getting the notation and the formulas into muscle memory now saves a lot of scrambling later.
Vectors in R3 work the same way as in R2, just with a third component. The magnitude formula gains a third squared term under the radical. The 3D distance formula and the equation of a sphere both follow directly from that same idea, and unit vectors are found by dividing any vector by its magnitude.
Vector (in R3)
An ordered triple of real numbers, written in component form as ⟨v₁, v₂, v₃⟩ or in standard basis form as v₁i + v₂j + v₃k.
Think of it as an arrow in 3D space that has both a direction and a length.
Magnitude (norm, length) of a vector
For v = ⟨v₁, v₂, v₃⟩, the magnitude is ||v|| = √(v₁² + v₂² + v₃²).
In simple terms, this is how long the arrow is. It is always non-negative.
Component form
The representation of a vector as an ordered list of its coordinates: ⟨v₁, v₂, v₃⟩.
In simple terms, this means writing the vector as a tuple of numbers rather than using i, j, k notation.
Standard basis vectors (i, j, k)
The three unit-length vectors along the x, y, and z axes: i = ⟨1,0,0⟩, j = ⟨0,1,0⟩, k = ⟨0,0,1⟩.
Think of them as the building blocks: any vector in R3 can be written as a combination of these three.
Unit vector
A vector with magnitude 1. The unit vector in the direction of v is v / ||v||.
In simple terms, this is the "direction only" version of a vector, with all the length information stripped away.
Distance formula (3D)
The distance between points p = (x₁, y₁, z₁) and q = (x₂, y₂, z₂) is d = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²).
Think of it as the 3D version of the Pythagorean theorem, applied twice.
Equation of a sphere
A sphere centred at (a, b, c) with radius r satisfies r² = (x − a)² + (y − b)² + (z − c)².
In simple terms, it is the set of all points that are exactly distance r from the centre point.
In R2: ||v|| = √(v₁² + v₂²). This is the 2D Pythagorean theorem applied directly.
In R3: ||v|| = √(v₁² + v₂² + v₃²). Same idea, one more squared term under the radical.
Example: for v = ⟨3, −2, 5⟩, the magnitude is √(9 + 4 + 25) = √38.
A unit vector has magnitude 1.
To find the unit vector in the direction of v, divide v by its magnitude: u = v / ||v||.
Example: the unit vector in the direction of ⟨3, −2, 5⟩ is (1/√38)⟨3, −2, 5⟩ = ⟨3/√38, −2/√38, 5/√38⟩.
Component form writes the vector as an ordered tuple: ⟨3, −2, 5⟩.
Standard basis form writes it as a linear combination: 3i − 2j + 5k.
These are two notations for the same object. Your professor may prefer one over the other on exams, so be comfortable switching between them.
Addition and scalar multiplication work component by component, exactly as in R2.
Example: if v = ⟨3, −2, 5⟩ and w = ⟨−2, 4, 1⟩, then v − 2w = ⟨3 − (−4), −2 − 8, 5 − 2⟩ = ⟨7, −10, 3⟩.
For p = (x₁, y₁, z₁) and q = (x₂, y₂, z₂): d = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²).
This is equivalent to the magnitude of the vector from p to q.
A sphere centred at (a, b, c) with radius r: (x − a)² + (y − b)² + (z − c)² = r².
To write the equation when given a centre and a point on the surface, first compute the radius using the distance formula.
Example: centre at (1, −3, 5), passing through (2, 4, 6). Radius = √((2−1)² + (4−(−3))² + (6−5)²) = √(1 + 49 + 1) = √51. Equation: (x − 1)² + (y + 3)² + (z − 5)² = 51.
\|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}r^2 = (x - a)^2 + (y - b)^2 + (z - c)^2\hat{u} = \frac{\vec{v}}{\|\vec{v}\|}The distance formula in 3D is used everywhere spatial measurements matter: GPS navigation computes distances between latitude/longitude/altitude coordinates, and game engines use the same formula to detect collisions between objects. The sphere equation describes everything from weather radar coverage volumes to the signal range of a Wi-Fi router in a building.
Students often forget to square the radius when writing the sphere equation. The standard form has r² on the left side, not r.
When computing the unit vector, students sometimes multiply by the magnitude instead of dividing. The unit vector is v / ||v||, not v × ||v||.
Students sometimes confuse the distance formula (which gives a scalar) with the vector from one point to another (which gives a vector). The distance formula is the magnitude of that vector, but the two are different objects.
When converting between component form and standard basis form, watch the signs. The vector ⟨3, −2, 5⟩ becomes 3i − 2j + 5k, not 3i + 2j + 5k.
⚠️ Magnitude and unit vector calculations appear on nearly every Calc III exam, often as the first step in a larger problem (e.g. finding a directional derivative later in the course).
⚠️ The sphere equation is a common "write the equation" exam question. You will be given a centre and either a radius or a point on the surface.
⚠️ Vector arithmetic (addition, scalar multiplication) is assumed fluency from this point forward. Errors here cascade through every later problem.
⚠️ Be comfortable moving between component form and standard basis form. Some problems state vectors one way and ask for the answer in the other.
True or false: the magnitude of the zero vector ⟨0, 0, 0⟩ is 0. (True.)
Fill in the blank: the unit vector in the direction of v is v divided by ______. (||v||, the magnitude of v.)
True or false: the equation (x − 2)² + (y + 1)² + (z − 3)² = 9 describes a sphere with radius 9. (False. The radius is 3, because 9 = r².)
Fill in the blank: the vector ⟨0, 5, −1⟩ written in standard basis form is ______. (5j − k.)
True or false: ||v|| can be negative. (False. The magnitude is always non-negative.)
Q: Find the magnitude of the vector v = ⟨4, −3, 12⟩.
A: ||v|| = √(16 + 9 + 144) = √169 = 13.
Q: Write the unit vector in the direction of v = ⟨4, −3, 12⟩ in component form.
A: u = (1/13)⟨4, −3, 12⟩ = ⟨4/13, −3/13, 12/13⟩.
Q: Find the distance between the points (3, 1, −2) and (−1, 4, 2).
A: d = √((−1 − 3)² + (4 − 1)² + (2 − (−2))²) = √(16 + 9 + 16) = √41.
Q: Write the equation of the sphere with centre (−2, 0, 3) and radius 5.
A: (x + 2)² + y² + (z − 3)² = 25.
Q: If v = ⟨1, −1, 2⟩ and w = ⟨3, 0, −4⟩, compute 2v + w.
A: 2v = ⟨2, −2, 4⟩, so 2v + w = ⟨2 + 3, −2 + 0, 4 + (−4)⟩ = ⟨5, −2, 0⟩.
The magnitude formula reappears when you compute dot products and cross products (Sections 12.3 and 12.4), because both formulas involve vector lengths. The distance formula is the foundation of every "closest point" and optimisation problem in multivariable calculus. Spheres come back when you set up triple integrals in spherical coordinates later in the course.
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