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Calculus III (Multivariable) Unit 1: Vectors & the geometry of space

Multivariable calculus starts in 3D space. Vectors describe direction and size, and dot and cross products measure angles, areas and perpendicular directions.

Covers: 3D coordinates, dot & cross product, lines & planes, quadric surfaces

Study guide

Multivariable calculus works in three dimensions. Points are (x, y, z), and the distance between two points is √(Δx² + Δy² + Δz²). Vectors ⟨a, b, c⟩ have magnitude √(a² + b² + c²) and can be added, scaled and normalized into unit vectors.

The dot product u·v = u₁v₁ + u₂v₂ + u₃v₃ = |u||v|cos θ is a number. It finds angles between vectors, and u·v = 0 means they're perpendicular. It also gives projections: the component of one vector along another.

The cross product u × v is a vector perpendicular to both u and v, found with a 3×3 determinant. Its direction follows the right-hand rule, and its magnitude |u × v| = |u||v|sin θ equals the area of the parallelogram the vectors span. It's not commutative: v × u = −(u × v).

A line in space through r₀ with direction v is r(t) = r₀ + tv, or in parametric form x = x₀ + at, etc. A plane through (x₀, y₀, z₀) with normal vector ⟨a, b, c⟩ has equation a(x − x₀) + b(y − y₀) + c(z − z₀) = 0. Cross two vectors in a plane to get its normal.

Quadric surfaces are 3D graphs of second-degree equations: spheres, ellipsoids, paraboloids, cones, cylinders and hyperboloids. Identify them by their traces, the curves formed by slicing with planes x = k, y = k or z = k.

The distance from a point P to the plane through Q with normal n is |PQ·n|/|n|. For the plane ax + by + cz = d and the point (x₁, y₁, z₁), that is |ax₁ + by₁ + cz₁ − d|/√(a² + b² + c²). To find where a line meets a plane, substitute the line's parametric equations into the plane's equation and solve for t. A line is parallel to a plane when its direction vector is perpendicular to the plane's normal (dot product 0), and perpendicular to the plane when the direction is a multiple of the normal.

A 2×2 determinant is ad − bc, and a 3×3 determinant expands along a row into three 2×2 determinants. A square matrix with determinant 0 is singular: its rows lie in a common plane, and Ax = 0 has nonzero solutions, such as the cross product of two rows of a 3×3 singular matrix. To multiply a matrix A by a vector x, take the dot product of each row of A with x. A is invertible exactly when det A ≠ 0, and then the system Ax = b has the single solution x = A⁻¹b.

The area of a triangle is half the magnitude of the cross product of two edge vectors. The absolute value |u·(v × w)| equals the volume of the parallelepiped the three vectors span, and equals the 3×3 determinant with rows u, v, w.

Key ideas

Formulas and rules

u·v = u₁v₁ + u₂v₂ + u₃v₃ = |u||v|cos θ
|u × v| = |u||v| sin θ
Line: r(t) = r₀ + t·v
Plane: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0

Common mistakes

Words to know

Orthogonal vectors
u · v = 0: the vectors are perpendicular.
Cross product
u × v: a vector perpendicular to both u and v.
Magnitude of the cross product
|u × v| = area of the parallelogram spanned by u and v.
Equation of a plane through (x₀, y₀, z₀) with normal ⟨a, b, c⟩
a(x − x₀) + b(y − y₀) + c(z − z₀) = 0
Vector equation of a line
r(t) = r₀ + t·v
Distance in 3D
√(Δx² + Δy² + Δz²)
Unit vector
A vector of length 1: v/|v|.
Dot product
u·v = u₁v₁ + u₂v₂ + u₃v₃ = |u||v|cos θ
Scalar projection
comp_u v = (u·v)/|u|
Right-hand rule
Gives the direction of u × v.
Normal vector (plane)
A vector perpendicular to the plane, ⟨a, b, c⟩ in ax + by + cz = d.
Quadric surface
The 3D graph of a second-degree equation, like an ellipsoid or paraboloid.
Trace
The curve where a surface meets a plane such as z = k.
3D coordinate system
Points (x, y, z) on three perpendicular axes, using the right-hand orientation.

Sample practice questions

41 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

⟨1, 2, 3⟩ · ⟨4, −1, 2⟩

  1. 6
  2. 10
  3. 8
  4. ⟨4, −2, 6⟩
Show the answer and steps

Answer: C) 8

  1. Dot product: multiply matching components and add.
  2. 1·4 + 2·(−1) + 3·2 = 4 − 2 + 6 = 8.
Question 2 (level 1)

Compute ⟨−6, −4, 0⟩ · ⟨−6, 3, 1⟩.

  1. −24
  2. ⟨36, −12, 0⟩
  3. 24
  4. 26
Show the answer and steps

Answer: C) 24

  1. Multiply matching components: 36, −12, 0.
  2. Add: 24.
Question 3 (level 3)

Distance between (1, 2, 3) and (4, 6, 3).

  1. 7
  2. √7
  3. 5
  4. 25
Show the answer and steps

Answer: C) 5

  1. Differences: 3, 4, 0.
  2. √(9 + 16 + 0) = 5.