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When you need a vector that is perpendicular to a given one, the dot‑product and cross‑product techniques provide clear, reliable methods. A zero dot product signals orthogonality, while the cross product of two non‑parallel vectors yields a vector that is perpendicular to both.

Two Dimensions — Dot Product

Step 1

Assume an unknown vector V = (v₁, v₂). This vector will be perpendicular to the known vector U = (u₁, u₂).

Step 2

Compute the dot product: V · U = u₁v₁ + u₂v₂. For example, if U = (–3, 10), then V · U = –3v₁ + 10v₂.

Step 3

Set the dot product to zero and solve for one component: –3v₁ + 10v₂ = 0 ⇒ v₂ = (3/10)v₁.

Step 4

Select any value for v₁; for instance, let v₁ = 1.

Step 5

Compute v₂ = 0.3. Thus V = (1, 0.3) is perpendicular to U = (–3, 10). Choosing v₁ = –1 gives V′ = (–1, –0.3), the opposite direction. Any scalar multiple of either vector remains perpendicular, and normalising to unit length yields W = V / √(1² + 0.3²) = (1/√10, 0.3/√10).

Three Dimensions — Dot Product

Step 1

Define an unknown vector V = (v₁, v₂, v₃).

Step 2

Compute the dot product with a known vector U = (10, 4, –1): V · U = 10v₁ + 4v₂ – v₃.

Step 3

Set the dot product to zero, yielding the plane equation 10v₁ + 4v₂ – v₃ = 0. Any vector satisfying this relation is perpendicular to U.

Step 4

Choose convenient values, e.g., v₁ = 1 and v₂ = 1, then solve for v₃ = 10 + 4 = 14. This gives V = (1, 1, 14).

Step 5

Verify orthogonality: V · U = 10(1) + 4(1) – 14 = 0. Thus V is indeed perpendicular to U.

Three Dimensions — Cross Product

Step 1

Select any vector not parallel to U. A convenient choice is a basis vector, such as X = (1, 0, 0).

Step 2

Compute the cross product: W = X × U = (0, 1, 4) when U = (10, 4, –1).

Step 3

Confirm perpendicularity: W · U = 0·10 + 1·4 + 4·(–1) = 0. Using different non‑parallel vectors like (0, 1, 0) or (0, 0, 1) will produce other perpendicular vectors, all lying in the plane defined by 10v₁ + 4v₂ – v₃ = 0.