طرح‌ها

ماشین‌حساب بردار: قائم یکه، برآیند و بردارهای عمود

این ماشین‌حساب بردار قائم یکهٔ N(t) یک منحنی در یک نقطه، بردار عمود بر دو بردار (ضرب خارجی)، برآیند دو بردار، و عمود بودن یا نبودن دو بردار را پیدا می‌کند. پاسخ‌ها دقیق‌اند، با رادیکال‌های ساده‌شده، و هر کدام بررسی می‌شود.

نتیجه

مقادیر را وارد کنید و برای دیدن نتیجه «محاسبه» را بزنید.

فرمول

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.

نحوه استفاده از این ماشین‌حساب

  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.

مثال‌های حل‌شده

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.

اشتباه‌های رایج

  • 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.

پرسش‌های متداول

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.