Orthogonal Vectors and Subspaces

April 13, 2026

Problem

Verify that u = [1, 2, −1] and v = [3, 0, 3] are orthogonal by computing u · v. Show the right angle geometrically.

Explanation

Orthogonality from the dot product

Two vectors u,v∈Rn\mathbf{u}, \mathbf{v} \in \mathbb{R}^n are orthogonal (written u⊥v\mathbf{u} \perp \mathbf{v}) if their dot product is zero: u⋅v=u1v1+u2v2+⋯+unvn=0\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + \cdots + u_n v_n = 0

This is equivalent to the geometric statement "the angle between them is 90°90°" — because u⋅v=∥u∥ ∥v∥ cos⁡θ\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \, \|\mathbf{v}\| \, \cos \theta

and cos⁡90°=0\cos 90° = 0.

Step-by-step

u=(1,2,−1),v=(3,0,3)\mathbf{u} = (1, 2, -1), \quad \mathbf{v} = (3, 0, 3).

Step 1 — Compute the dot product. u⋅v=(1)(3)+(2)(0)+(−1)(3)\mathbf{u} \cdot \mathbf{v} = (1)(3) + (2)(0) + (-1)(3)

Step 2 — Evaluate. u⋅v=3+0−3=0\mathbf{u} \cdot \mathbf{v} = 3 + 0 - 3 = \boxed{0}

So u⊥v\mathbf{u} \perp \mathbf{v} ✓

Angle: cos⁡θ=0/(∥u∥∥v∥)=0\cos \theta = 0 / (\|\mathbf{u}\| \|\mathbf{v}\|) = 0, so θ=90°\theta = 90°.

Norms and normalization

The norm (length) of a vector is ∥u∥=u⋅u\|\mathbf{u}\| = \sqrt{\mathbf{u} \cdot \mathbf{u}}.

For our u\mathbf{u}: ∥u∥=1+4+1=6\|\mathbf{u}\| = \sqrt{1 + 4 + 1} = \sqrt{6}. For v\mathbf{v}: ∥v∥=9+0+9=18=32\|\mathbf{v}\| = \sqrt{9 + 0 + 9} = \sqrt{18} = 3\sqrt{2}.

A unit vector u^=u/∥u∥\hat{\mathbf{u}} = \mathbf{u}/\|\mathbf{u}\| has norm 1. Two unit vectors are orthonormal iff they are orthogonal.

Properties of the dot product

  • Commutative: u⋅v=v⋅u\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}.
  • Distributive: u⋅(v+w)=u⋅v+u⋅w\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w}.
  • Scalar: (cu)⋅v=c(u⋅v)(c \mathbf{u}) \cdot \mathbf{v} = c (\mathbf{u} \cdot \mathbf{v}).
  • Positive-definite: u⋅u≥0\mathbf{u} \cdot \mathbf{u} \ge 0, with equality iff u=0\mathbf{u} = \mathbf{0}.

Orthogonal subspaces

Two subspaces U,WU, W are orthogonal if every vector in UU is orthogonal to every vector in WW. They need not span the whole space; they just can't share any direction.

The orthogonal complement of UU, written U⊥U^\perp, is the set of all vectors orthogonal to every vector in UU. It satisfies: dim⁡U+dim⁡U⊥=dim⁡(ambient space)\dim U + \dim U^\perp = \dim (\text{ambient space})

Important identities for any m×nm \times n matrix AA:

  • Row⁡(A)⊥=Null⁡(A)\operatorname{Row}(A)^\perp = \operatorname{Null}(A)
  • Col⁡(A)⊥=Null⁡(AT)\operatorname{Col}(A)^\perp = \operatorname{Null}(A^T)

These are the four fundamental subspaces.

Why orthogonality matters

  • Pythagorean theorem: u⊥v  ⟹  ∥u+v∥2=∥u∥2+∥v∥2\mathbf{u} \perp \mathbf{v} \implies \|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2.
  • Projection: projecting onto a subspace uses orthogonal decomposition.
  • Gram-Schmidt: builds an orthogonal basis from any set of linearly independent vectors.
  • Fourier analysis: the sines and cosines {1,cos⁡nx,sin⁡nx}\{1, \cos nx, \sin nx\} are mutually orthogonal under the integral inner product.

Common mistakes

  • Calling vectors orthogonal when one is zero. By convention, 0\mathbf{0} is orthogonal to every vector (dot product is zero).
  • Confusing orthogonality with parallelism. Orthogonal: dot product 0. Parallel: cross product 0 (in R3\mathbb{R}^3) or vectors are scalar multiples.
  • Forgetting to square-root the norm. ∥u∥2=u⋅u\|\mathbf{u}\|^2 = \mathbf{u} \cdot \mathbf{u}; the length itself takes a square root.

Try it in the visualization

Drag two vectors in 3D. The angle display updates; the two vectors are colored to show orthogonality (green) vs. acute/obtuse angles (yellow/red). A small right-angle square appears at the origin when the dot product is zero.

Interactive Visualization

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Orthogonal Vectors and Subspaces | MathSpin