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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
- Points in 3D are (x, y, z); distance generalizes the Pythagorean theorem.
- Dot product gives a number and measures how aligned two vectors are; zero means perpendicular.
- Cross product gives a vector perpendicular to both, with length equal to the parallelogram's area.
- A line needs a point and a direction vector; a plane needs a point and a normal vector.
Formulas and rules
Common mistakes
- Mixing up which product gives a scalar and which gives a vector.
- Using a direction vector as a plane's normal.
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.
⟨1, 2, 3⟩ · ⟨4, −1, 2⟩
- 6
- 10
- 8
- ⟨4, −2, 6⟩
Show the answer and steps
Answer: C) 8
- Dot product: multiply matching components and add.
- 1·4 + 2·(−1) + 3·2 = 4 − 2 + 6 = 8.
Compute ⟨−6, −4, 0⟩ · ⟨−6, 3, 1⟩.
- −24
- ⟨36, −12, 0⟩
- 24
- 26
Show the answer and steps
Answer: C) 24
- Multiply matching components: 36, −12, 0.
- Add: 24.
Distance between (1, 2, 3) and (4, 6, 3).
- 7
- √7
- 5
- 25
Show the answer and steps
Answer: C) 5
- Differences: 3, 4, 0.
- √(9 + 16 + 0) = 5.