Mathematics · Analytic Geometry / Vectors

Magnitude of a Vector (Length)

The magnitude of a vector is its length: the square root of the sum of the squared coordinates, a twice-applied Pythagoras in space.

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Formula

LaTeX: |\vec{v}| = \sqrt{v_{1}^{2} + v_{2}^{2} + v_{3}^{2}}
|v⃗| in length units (LU) · coordinates in LU

Variables & units – Magnitude of a Vector (Length)

SymbolMeaningUnit
v⃗Vector in R² or R³length units
v₁, v₂, v₃Coordinates of the vectorlength units
|v⃗|Magnitude (length) of the vectorlength units

Derivation & background – Magnitude of a Vector (Length)

In the plane the magnitude is the Pythagorean theorem in the slope triangle; in space you apply it twice (space diagonal of a cuboid). The distance of two points A and B is the magnitude of the connection vector |AB⃗|. Division by the magnitude normalizes: v⃗⁰ = v⃗/|v⃗| is the unit vector of length 1. Important for vector algebra: |v⃗|² = v⃗·v⃗ links magnitude and dot product.

Exam blueprint

Validity range

Holds for vectors in R² (without third coordinate) and R³ with Cartesian coordinates; the magnitude is never negative and 0 exactly when v⃗ is the zero vector.

Derivation steps

Pythagoras twice: first in the base plane, then with the height.

  1. 1In the xy-plane (v₁|v₂) has length √(v₁² + v₂²) (Pythagoras in the slope triangle).
  2. 2With the third coordinate as height |v⃗| = √((√(v₁²+v₂²))² + v₃²) = √(v₁² + v₂² + v₃²) follows.

Rearrangements

Unit vector

Length 1, same direction; basis for HNF and direction angles.

Distance of two points

Form the connection vector, then take the magnitude.

Magnitude via dot product

Useful in proofs and for computing without the root.

Task variant

Compute the distance of the points A(1|2|3) and B(4|6|3).

AB⃗ = (3|4|0), |AB⃗| = √(9 + 16 + 0) = √25 = 5 LU.

Determine the unit vector of v⃗ = (2|−1|2).

|v⃗| = √(4 + 1 + 4) = 3, so v⃗⁰ = (2/3|−1/3|2/3). Check: (2/3)² + (1/3)² + (2/3)² = 9/9 = 1 ✓.

Common mistakes

Adding the coordinates first, then squaring.

Square each coordinate individually, then sum, then take the root.

Carrying negative coordinates without squaring.

(−3)² = 9; squaring removes all signs.

Confusing the magnitude of a vector with the absolute value of a number.

|v⃗| is a length from all coordinates, not just dropping signs.

Exam context

  • Side lengths and distances in geometry tasks, normalizing for HNF and angles, magnitudes of physical vectors.

These mistakes cost points in real exams. The set drills them until they stick.

Worked example

v⃗ = (2|3|6): |v⃗| = √(4 + 9 + 36) = √49 = 7 LU. Distance of A(1|2|3) and B(4|6|3): AB⃗ = (3|4|0), |AB⃗| = √(9 + 16 + 0) = 5 LU.

Applications

Lengths and distances in vector geometry, unit vectors and normalization (Hesse normal form), magnitudes of velocity and force in physics, side lengths of triangles in space

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Curated exam set for "Magnitude of a Vector (Length)":

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How do you rearrange |v⃗| = √(v₁² + v₂² + v₃²) for Unit vector?

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Which common mistake happens with Magnitude of a Vector (Length)?

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Scientific sources

Common notations & search queries

|v|=sqrt(x^2+y^2+z^2)Betrag Vektor berechnenLänge Vektor FormelVektorbetragAbstand zweier Punkte VektorEinheitsvektor berechnenvector magnitude formulaNorm eines Vektors

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Frequently asked questions about Magnitude of a Vector (Length)

How do you calculate the length of a vector?+

Square each coordinate, add the squares and take the root: |v⃗| = √(v₁² + v₂² + v₃²). Example: v⃗ = (2|3|6) has length √(4 + 9 + 36) = √49 = 7. In the plane the third coordinate simply drops: |(3|4)| = √(9 + 16) = 5. Behind it is the Pythagorean theorem, applied twice in space (first the diagonal of the base, then with the height). The order matters: square first, then add, then take the root; the root of a sum must not be pulled apart. Negative coordinates automatically lose their sign when squared, the length is never negative.

How do you calculate the distance of two points with vectors?+

Form the connection vector and take its magnitude: d(A;B) = |AB⃗| with AB⃗ = b⃗ − a⃗ (tip minus tail, i.e. target point minus start point). Example: A(1|2|3) and B(4|6|3): AB⃗ = (3|4|0) and d = √(9 + 16 + 0) = 5 LU. That is exactly the distance formula of analytic geometry, just in vector notation. Frequent mistakes: squaring the point coordinates directly instead of forming the differences first, or swapping start and target; the latter is harmless for the length, since |AB⃗| = |BA⃗|. The distance is the basis for triangle side lengths, perimeters and sphere equations in space.

What is a unit vector and how do you form it?+

A unit vector has length 1 and stores only a direction. You get it by normalizing: v⃗⁰ = v⃗/|v⃗|, i.e. divide each coordinate by the length. Example: v⃗ = (2|−1|2) has |v⃗| = 3, the unit vector is (2/3|−1/3|2/3); check: 4/9 + 1/9 + 4/9 = 1 ✓. You need unit vectors wherever lengths would disturb: in the Hesse normal form (unit normal vector), for direction cosines, for points at a prescribed distance along a direction (P + k·v⃗⁰ lies exactly k LU from P) and in physics for the directions of forces. The zero vector is the only one that cannot be normalized.

Why can you not simply add the coordinates?+

Because length does not arise coordinate-wise but via the Pythagorean theorem. Counterexample: v⃗ = (3|4) would have the coordinate-wise "length" 3 + 4 = 7, but actually |v⃗| = √(9 + 16) = 5. The coordinates are perpendicular to each other, so the length is composed as a hypotenuse, not as the sum of the legs; the sum would be the detour along the axis directions. Even clearer with signs: (3|−4) would sum to −1, but lengths are never negative. Remember: square first (makes everything positive and weights correctly), then add, then the root. Only the special case where all but one coordinate are 0 allows direct reading: |(0|0|−5)| = 5.

What links the magnitude with the dot product?+

The core identity is |v⃗|² = v⃗·v⃗: the dot product of a vector with itself is the square of its length, since v₁·v₁ + v₂·v₂ + v₃·v₃ is exactly the sum of squares. This has practical consequences. First, in proofs and calculations you can work root-free by comparing lengths via their squares. Second, the magnitude sits inside the angle formula cos φ = (a⃗·b⃗)/(|a⃗|·|b⃗|); without the magnitudes in the denominator the cosine would not be normalized. Third, the identity yields the rule |k·v⃗| = |k|·|v⃗| for scalings. Example: v⃗ = (2|3|6): v⃗·v⃗ = 4 + 9 + 36 = 49 = 7², consistent with |v⃗| = 7.

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Create a curated FSRS exam set for |v⃗| = √(v₁² + v₂² + v₃²): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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How do you calculate with Magnitude of a Vector (Length)?

Here is how to work through a typical Magnitude of a Vector (Length) (|v⃗| = √(v₁² + v₂² + v₃²)) task step by step:

  1. 1

    Task

    Compute the distance of the points A(1|2|3) and B(4|6|3).

    Solution path

    AB⃗ = (3|4|0), |AB⃗| = √(9 + 16 + 0) = √25 = 5 LU.

  2. 2

    Task

    Determine the unit vector of v⃗ = (2|−1|2).

    Solution path

    |v⃗| = √(4 + 1 + 4) = 3, so v⃗⁰ = (2/3|−1/3|2/3). Check: (2/3)² + (1/3)² + (2/3)² = 9/9 = 1 ✓.