Vector Calculator: Unit Normal, Resultant and Orthogonal Vectors

This vector calculator finds the unit normal vector N(t) of a curve at a point, the normal vector to two vectors (the cross product), the resultant of two vectors, and whether two vectors are orthogonal. Answers are exact, with simplified square roots, and each one is checked.

Result

Enter values and press Calculate to see the result.

Formula

N(t)=T′(t)∥T′(t)∥\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\lVert \mathbf{T}'(t) \rVert}

The unit normal vector of a curve. At a point it points the same way as the part of r″ perpendicular to r′, which is how the calculator finds it exactly.

n=u×v\mathbf{n} = \mathbf{u} \times \mathbf{v}

A normal vector to the plane containing u and v.

u⋅v=0  ⟺  u⊥v\mathbf{u} \cdot \mathbf{v} = 0 \iff \mathbf{u} \perp \mathbf{v}

Two non-zero vectors are orthogonal exactly when their dot product is 0.

How to use this calculator

  1. Choose what you want: a unit normal vector of a curve, the normal to two vectors, a resultant, or an orthogonality check.
  2. For a curve, type each component as a polynomial in t and the value of t. For two vectors, type their components separated by commas.
  3. Press Calculate. Magnitudes are found with Pythagoras — the magnitude of a vector ⟨a, b, c⟩ is √(a² + b² + c²) — and every step is shown.

Worked examples

Unit normal vector of a parabola

r′(t) = ⟨1, 2t⟩ and r″(t) = ⟨0, 2⟩, so at t = 1: r′ = ⟨1, 2⟩, r″ = ⟨0, 2⟩.

The part of r″ perpendicular to r′: ⟨0, 2⟩ − (4/5)⟨1, 2⟩ = ⟨−4/5, 2/5⟩.

Divide by its length 2√5/5: N(1) = ⟨−2√5/5, √5/5⟩ ≈ ⟨−0.894, 0.447⟩, pointing into the curve.

Unit normal vector of a 3D curve

At t = 0: r′ = ⟨1, 0, 0⟩ and r″ = ⟨0, 2, 0⟩, already perpendicular.

So N(0) = ⟨0, 1, 0⟩ and T(0) = ⟨1, 0, 0⟩.

Normal vector to a plane (cross product)

u × v = ⟨2·6 − 3·5, 3·4 − 1·6, 1·5 − 2·4⟩ = ⟨−3, 6, −3⟩.

Its length is √54 = 3√6, so the unit normal is ⟨−√6/6, √6/3, −√6/6⟩.

Resultant of two forces

Add component by component: ⟨3 + 1, 4 − 2⟩ = ⟨4, 2⟩.

Magnitude √(16 + 4) = √20 = 2√5 ≈ 4.47; direction about 26.57° above the x-axis.

Are two vectors orthogonal?

u · v = 2 − 1 − 1 = 0, so they are orthogonal (at 90°).

u × v = ⟨0, −3, −3⟩ is orthogonal to both.

Common mistakes

  • Using r″ itself as the normal direction: only the part of r″ perpendicular to r′ points along N.
  • Getting the middle sign of the cross product wrong — the j-component is a₃b₁ − a₁b₃.
  • Adding magnitudes instead of vectors: |u + v| is not |u| + |v| unless they point the same way.
  • Calling the zero vector orthogonal to everything without noticing it has no direction.

Frequently asked questions

What is the unit normal vector of a curve?

N(t) = T′(t)/|T′(t)|: a vector of length 1, perpendicular to the direction of motion, pointing towards the inside of the bend.

How do I find a normal vector to a plane?

Take two vectors that lie in the plane and find their cross product u × v. Divide by its length for a unit normal vector.

How do I calculate a resultant vector?

Add the vectors component by component. Its magnitude comes from Pythagoras and, in 2D, its direction from the angle with the x-axis.

How can I tell if two vectors are orthogonal?

Their dot product is 0. For ⟨2, 1, −1⟩ and ⟨1, −1, 1⟩: 2 − 1 − 1 = 0, so they are orthogonal.

How do I find the magnitude of a vector?

Square each component, add, and take the square root: |⟨3, 4⟩| = √(9 + 16) = 5.