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Orthogonal Vector Calculator

Solve orthogonal vector problems with clear steps, notation, and a final check.

Calculate without using AI.

Evaluate the governing formula locally in your browser. Define each known quantity once, then change values to test another case instantly.

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Vectors

Orthogonal Vector Calculator explained

The short version

  • Two vectors are orthogonal when their dot product is exactly zero, which means they meet at 90°.
  • The dot product multiplies matching components and adds the results, giving a single number.
  • The sign of that number tells you the angle at a glance: positive is sharp, zero is square, negative is wide.

The formula this page uses

u · v = u₁v₁ + u₂v₂ + u₃v₃ cos θ = (u · v) / (|u| |v|) orthogonal ⟺ u · v = 0

What each part means

SymbolWhat it means
u, v — The two vectorsThey must have the same number of components to be dotted.
u · v — Dot productA single number, not a vector. Its unit is the product of the two vectors' units.
θ — Angle between themBetween 0° and 180°, found from the cosine formula.
|u|, |v| — The two magnitudesNeeded only for the angle, not for the orthogonality test.

Show your work: a full example

  1. Two vectorsu = (2, −3, 4) and v = (5, 6, 2)
  2. Multiply matching components(2)(5) = 10, (−3)(6) = −18, (4)(2) = 8
  3. Add them10 − 18 + 8 = 0
  4. Read the resultthe dot product is zero, so u and v are orthogonal
  5. Confirm with the angle formula|u| = √29 ≈ 5.385 and |v| = √65 ≈ 8.062
  6. Cosine of the angle0 ÷ (5.385 × 8.062) = 0
  7. The anglecos⁻¹(0) = 90°, exactly as the zero predicted

A second, different case

  1. A different case: not orthogonala = (3, 4) and b = (5, 12)
  2. Dot product(3)(5) + (4)(12) = 15 + 48 = 63
  3. Magnitudes|a| = √25 = 5 and |b| = √169 = 13
  4. Cosine of the angle63 ÷ (5 × 13) = 63/65 = 0.9692
  5. The anglecos⁻¹(0.9692) = 14.25°, a narrow angle, matching the positive dot product
  6. Build a vector perpendicular to aswap the components and flip one sign: (−4, 3)
  7. Check it(3)(−4) + (4)(3) = −12 + 12 = 0, so (−4, 3) is orthogonal to (3, 4)
Copy-ready example

u = (2, −3, 4) and v = (5, 6, 2)

Two vectors

What the dot product's value tells you about the angle

uvu · vAngleRelationship
(3, 4)(6, 8)50parallel, same direction
(3, 4)(5, 12)6314.25°narrow angle
(1, 0)(1, 1)145°sharp angle
(2, −3, 4)(5, 6, 2)090°orthogonal
(3, 4)(−4, 3)090°orthogonal
(1, 0)(−1, 1)−1135°wide angle
(1, 0)(−1, 0)−1180°parallel, opposite directions

Three mistakes to check for

What students writeWhy it's wrongDo this instead
u · v = (10, −18, 8)The three products were left as a list. A dot product finishes by adding them into one number.10 − 18 + 8 = 0. It is the cross product that returns a vector.
u · v = 0, so one of the vectors must be zeroTwo perfectly ordinary non-zero vectors give zero whenever they point at right angles.Both (2, −3, 4) and (5, 6, 2) have positive length; the zero reports the angle, not a missing vector.
A negative dot product means a sign slip somewhereNegative simply means the angle is wider than 90°.(1, 0) · (−1, 1) = −1 correctly reports 135°.

Questions about the Orthogonal Vector Calculator

Are orthogonal and perpendicular different words?

For two vectors they mean the same thing. Orthogonal is preferred in higher dimensions and for functions, where drawing a right angle stops making sense but the dot product still works.

How do I build a vector orthogonal to a given one?

In 2D, swap the components and negate one: (3, 4) becomes (−4, 3). In 3D there is a whole plane of choices, so take the cross product with any vector that is not parallel to yours.

Is the zero vector orthogonal to everything?

Its dot product with anything is 0, so by the algebraic definition yes. Geometrically it is meaningless, because the zero vector has no direction and there is no angle to measure.

What does the dot product mean physically?

Work. Push with force u along a displacement v and the energy transferred is u · v. Push at right angles and the dot product is zero, which is why carrying a bag horizontally does no work against gravity.

Where to go next

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