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Vector Calculator: Unit Normal, Resultant at Orthogonal na Vector

Hinahanap ng calculator na ito ang unit normal vector N(t) ng isang curve sa isang punto, ang vector na normal sa dalawang vector (cross product), ang resultant ng dalawang vector, at kung orthogonal ang dalawang vector. Eksakto ang mga sagot, na may pinasimpleng surd, at sinusuri ang bawat isa.

Resulta

Ilagay ang mga halaga at pindutin ang “Kuwentahin” para makita ang resulta.

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.

Paano gamitin ang calculator na ito

  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.

Mga halimbawang may solusyon

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.

Mga karaniwang pagkakamali

  • 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.

Mga madalas itanong

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.