UniMaths
Free chapterTry everything here: cards, proofs and practice. Sign up free to save your progress and unlock all 7 chapters.Sign up free
LA1 · 1

Vectors and geometry

13 concepts · 47 questions

Vectors in ℝⁿ, the dot product and length, Cauchy–Schwarz and the triangle inequality, angles and projections, the cross and triple products, lines and planes, distances and intersections.

Definition 1.1Vectors in ℝⁿ

Rn\mathbb{R}^n is the set of nn-tuples v=(v1,…,vn)\mathbf{v} = (v_1, \dots, v_n), written as columns. Vectors are added and scaled componentwise:

u+v=(u1+v1,…,un+vn),λv=(λv1,…,λvn).\mathbf{u} + \mathbf{v} = (u_1 + v_1, \dots, u_n + v_n),\qquad\lambda\mathbf{v} = (\lambda v_1, \dots, \lambda v_n).

Geometrically v\mathbf{v} is an arrow (displacement); u+v\mathbf{u} + \mathbf{v} is "u\mathbf{u} then v\mathbf{v}" (parallelogram rule). The standard basis is e1,…,en\mathbf{e}_1, \dots, \mathbf{e}_n, with v=∑viei\mathbf{v} = \sum v_i\mathbf{e}_i. A linear combination of v1,…,vk\mathbf{v}_1, \dots, \mathbf{v}_k is ∑λivi\sum\lambda_i\mathbf{v}_i.

Vectors add and scale componentwise. Picture: arrows placed head to tail.

Standard basis: (3,−1,2)=3e1−e2+2e3(3, -1, 2) = 3\mathbf{e}_1 - \mathbf{e}_2 + 2\mathbf{e}_3.

Linear combination: λ1v1+⋯+λkvk\lambda_1\mathbf{v}_1 + \cdots + \lambda_k\mathbf{v}_k — the central idea of the course.

Definition 1.2The dot product and length

u⋅v=∑i=1nuivi\mathbf{u}\cdot\mathbf{v} = \sum_{i=1}^nu_iv_i and ∥v∥=v⋅v\|\mathbf{v}\| = \sqrt{\mathbf{v}\cdot\mathbf{v}}. Properties: u⋅v=v⋅u\mathbf{u}\cdot\mathbf{v} = \mathbf{v}\cdot\mathbf{u}; linear in each argument; v⋅v≥0\mathbf{v}\cdot\mathbf{v}\ge0 with equality iff v=0\mathbf{v} = \mathbf{0}; ∥λv∥=∣λ∣∥v∥\|\lambda\mathbf{v}\| = |\lambda|\|\mathbf{v}\|. A unit vector has length 11; v∥v∥\frac{\mathbf{v}}{\|\mathbf{v}\|} is the unit vector in the direction of v≠0\mathbf{v}\ne\mathbf{0}. The distance between points is ∥u−v∥\|\mathbf{u} - \mathbf{v}\|.

u⋅v=∑uivi\mathbf{u}\cdot\mathbf{v} = \sum u_iv_i, a number. ∥v∥=v⋅v\|\mathbf{v}\| = \sqrt{\mathbf{v}\cdot\mathbf{v}} (Pythagoras).

Expand like algebra: ∥u+v∥2=∥u∥2+2u⋅v+∥v∥2\|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + 2\mathbf{u}\cdot\mathbf{v} + \|\mathbf{v}\|^2.

Normalise: v/∥v∥\mathbf{v}/\|\mathbf{v}\|.

Theorem 1.3The Cauchy–Schwarz inequality

For all u,v∈Rn\mathbf{u}, \mathbf{v}\in\mathbb{R}^n: ∣u⋅v∣≤∥u∥∥v∥|\mathbf{u}\cdot\mathbf{v}|\le\|\mathbf{u}\|\|\mathbf{v}\|, with equality iff u,v\mathbf{u}, \mathbf{v} are parallel (one is a multiple of the other). In coordinates: (∑uivi)2≤(∑ui2)(∑vi2)\big(\sum u_iv_i\big)^2\le\big(\sum u_i^2\big)\big(\sum v_i^2\big).

∣u⋅v∣≤∥u∥∥v∥|\mathbf{u}\cdot\mathbf{v}|\le\|\mathbf{u}\|\|\mathbf{v}\|, equality iff parallel.

Proof: ∥u−tv∥2≥0\|\mathbf{u} - t\mathbf{v}\|^2\ge0 for all tt; discriminant ≤0\le0.

Makes angles possible: u⋅v∥u∥∥v∥∈[−1,1]\frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}\in[-1, 1].

Proof

Corollary 1.4The triangle inequality and angles

∥u+v∥≤∥u∥+∥v∥\|\mathbf{u} + \mathbf{v}\|\le\|\mathbf{u}\| + \|\mathbf{v}\|. For non-zero u,v\mathbf{u}, \mathbf{v} the angle θ∈[0,π]\theta\in[0, \pi] between them is defined by cos⁡θ=u⋅v∥u∥∥v∥\cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}. They are orthogonal (u⊥v\mathbf{u}\perp\mathbf{v}) iff u⋅v=0\mathbf{u}\cdot\mathbf{v} = 0, and then ∥u+v∥2=∥u∥2+∥v∥2\|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 (Pythagoras).

cos⁡θ=u⋅v∥u∥∥v∥\cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}. Orthogonal iff u⋅v=0\mathbf{u}\cdot\mathbf{v} = 0.

Triangle inequality from Cauchy–Schwarz.

Pythagoras for orthogonal vectors.

Proof

Definition 1.5Projection onto a vector

For a≠0\mathbf{a}\ne\mathbf{0}, the projection of v\mathbf{v} onto a\mathbf{a} is proj⁡av=v⋅aa⋅aa\operatorname{proj}_{\mathbf{a}}\mathbf{v} = \frac{\mathbf{v}\cdot\mathbf{a}}{\mathbf{a}\cdot\mathbf{a}}\mathbf{a}, the "shadow" of v\mathbf{v} on the line through a\mathbf{a}.

proj⁡av=v⋅aa⋅aa\operatorname{proj}_{\mathbf{a}}\mathbf{v} = \frac{\mathbf{v}\cdot\mathbf{a}}{\mathbf{a}\cdot\mathbf{a}}\mathbf{a}: the shadow of v\mathbf{v} on the line of a\mathbf{a}. The remainder is orthogonal to a\mathbf{a}, and the projection is the closest point.

Preview of Gram–Schmidt and least squares.

Proposition 1.6Projection gives the closest point

v=proj⁡av+w\mathbf{v} = \operatorname{proj}_{\mathbf{a}}\mathbf{v} + \mathbf{w} with w⊥a\mathbf{w}\perp\mathbf{a}, and proj⁡av\operatorname{proj}_{\mathbf{a}}\mathbf{v} is the multiple of a\mathbf{a} closest to v\mathbf{v}.

Splits v\mathbf{v} into a part along a\mathbf{a} and a part perpendicular to it.

Closest point on the line: by Pythagoras.

Preview of Gram–Schmidt and least squares.

Proof

Definition 1.7The cross product in ℝ³

For u,v∈R3\mathbf{u}, \mathbf{v}\in\mathbb{R}^3, u×v=(u2v3−u3v2, u3v1−u1v3, u1v2−u2v1)\mathbf{u}\times\mathbf{v} = (u_2v_3 - u_3v_2,\ u_3v_1 - u_1v_3,\ u_1v_2 - u_2v_1). Properties: v×u=−u×v\mathbf{v}\times\mathbf{u} = -\mathbf{u}\times\mathbf{v}; bilinear; u×v\mathbf{u}\times\mathbf{v} is orthogonal to both u\mathbf{u} and v\mathbf{v}; ∥u×v∥=∥u∥∥v∥sin⁡θ\|\mathbf{u}\times\mathbf{v}\| = \|\mathbf{u}\|\|\mathbf{v}\|\sin\theta, the area of the parallelogram they span; u×v=0\mathbf{u}\times\mathbf{v} = \mathbf{0} iff parallel. Direction by the right-hand rule: e1×e2=e3\mathbf{e}_1\times\mathbf{e}_2 = \mathbf{e}_3. Not associative.

u×v\mathbf{u}\times\mathbf{v}: perpendicular to both, length = parallelogram area, right-hand rule.

Anticommutative: v×u=−u×v\mathbf{v}\times\mathbf{u} = -\mathbf{u}\times\mathbf{v}, so u×u=0\mathbf{u}\times\mathbf{u} = \mathbf{0}.

Mnemonic: a 3×33\times3 "determinant" with e1,e2,e3\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 in the top row.

Proposition 1.8The scalar triple product

[a,b,c]=a⋅(b×c)[\mathbf{a}, \mathbf{b}, \mathbf{c}] = \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) equals the determinant of the matrix with rows a,b,c\mathbf{a}, \mathbf{b}, \mathbf{c}. Its absolute value is the volume of the parallelepiped they span; it is 00 iff the three vectors are coplanar. It is unchanged by cyclic permutation and changes sign under a swap.

a⋅(b×c)=det⁡\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \det (rows a,b,c\mathbf{a}, \mathbf{b}, \mathbf{c}) =±= \pm volume.

Zero iff coplanar. Cyclic shifts don't change it.

Proof

Definition 1.9Lines

The line through a\mathbf{a} with direction d≠0\mathbf{d}\ne\mathbf{0} is {a+td:t∈R}\{\mathbf{a} + t\mathbf{d}: t\in\mathbb{R}\} (vector/parametric form). In R3\mathbb{R}^3 it can also be written x−a1d1=y−a2d2=z−a3d3\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3} (Cartesian form, when all di≠0d_i\ne0). The line through points p,q\mathbf{p}, \mathbf{q} has direction q−p\mathbf{q} - \mathbf{p}.

Line = point + multiples of a direction: r=a+td\mathbf{r} = \mathbf{a} + t\mathbf{d}.

Through two points: direction q−p\mathbf{q} - \mathbf{p}.

Same line, many equations: any point and any parallel direction.

Definition 1.10Planes in ℝ³

The plane through a\mathbf{a} with normal n≠0\mathbf{n}\ne\mathbf{0} is {x:n⋅(x−a)=0}\{\mathbf{x}: \mathbf{n}\cdot(\mathbf{x} - \mathbf{a}) = 0\}, i.e. n1x+n2y+n3z=dn_1x + n_2y + n_3z = d with d=n⋅ad = \mathbf{n}\cdot\mathbf{a}. Parametrically it is a+su+tv\mathbf{a} + s\mathbf{u} + t\mathbf{v} for non-parallel u,v\mathbf{u}, \mathbf{v} in the plane, and then n=u×v\mathbf{n} = \mathbf{u}\times\mathbf{v}.

Plane: n⋅x=d\mathbf{n}\cdot\mathbf{x} = d, where n\mathbf{n} is perpendicular to the plane.

Through three points: n=(q−p)×(r−p)\mathbf{n} = (\mathbf{q} - \mathbf{p})\times(\mathbf{r} - \mathbf{p}).

Read off the normal from the coefficients: 2x−y+3z=52x - y + 3z = 5 has n=(2,−1,3)\mathbf{n} = (2, -1, 3).

Proposition 1.11Distances to planes and lines

The distance from p\mathbf{p} to the plane n⋅x=d\mathbf{n}\cdot\mathbf{x} = d is ∣n⋅p−d∣∥n∥\frac{|\mathbf{n}\cdot\mathbf{p} - d|}{\|\mathbf{n}\|}. The distance from p\mathbf{p} to the line a+td\mathbf{a} + t\mathbf{d} in R3\mathbb{R}^3 is ∥(p−a)×d∥∥d∥\frac{\|(\mathbf{p} - \mathbf{a})\times\mathbf{d}\|}{\|\mathbf{d}\|}.

Point to plane: ∣n⋅p−d∣∥n∥\frac{|\mathbf{n}\cdot\mathbf{p} - d|}{\|\mathbf{n}\|} — plug the point into the equation and divide by ∥n∥\|\mathbf{n}\|.

Point to line: ∥(p−a)×d∥∥d∥\frac{\|(\mathbf{p} - \mathbf{a})\times\mathbf{d}\|}{\|\mathbf{d}\|}.

Proof

Example 1.12Intersections of lines and planes

  • Line and plane: substitute a+td\mathbf{a} + t\mathbf{d} into n⋅x=d\mathbf{n}\cdot\mathbf{x} = d and solve for tt (no solution if n⋅d=0\mathbf{n}\cdot\mathbf{d} = 0 and the line is off the plane).
  • Two planes: if the normals aren't parallel they meet in a line with direction n1×n2\mathbf{n}_1\times\mathbf{n}_2.
  • Two lines in R3\mathbb{R}^3: equate a+td=b+se\mathbf{a} + t\mathbf{d} = \mathbf{b} + s\mathbf{e}; they may meet, be parallel, or be skew (neither).

Line ∩ plane: substitute, solve for tt.

Plane ∩ plane: a line in direction n1×n2\mathbf{n}_1\times\mathbf{n}_2; find one common point.

Lines in 3D can be skew: not parallel, never meeting.

Check 1.13Check: vector geometry problems

  • Perpendicular? dot product 00. Parallel? cross product 0\mathbf{0} (in R3\mathbb{R}^3) or one a multiple of the other.
  • Need a normal? cross two directions in the plane. Check it by dotting with both.
  • Distances: project onto a normal (plane) or use the cross product (line).
  • Sanity checks: substitute points back into equations; distances are non-negative; angles in [0,π][0, \pi].

Perpendicular: dot = 0. Parallel: cross = 0. Normal = cross of two in-plane directions. Check by dotting.

Always substitute points back into your equations.