Magnitude of a Vector (Length)

Square each coordinate, add the squares and take the root: |v⃗$| = \sqrt{(v_{1}^{2} + v_{2}^{2} + v_{3}^{2})}$. Example: v⃗ = (2|3|6) has length $\sqrt{(4 + 9 + 36)} = \sqrt{49} = 7$. In the plane the third coordinate simply drops: $|(3|4)| = \sqrt{(9 + 16)} = 5$. 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.

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 $d = \sqrt{(9 + 16 + 0)} = 5 LU$. 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$| = |BA$⃗|. The distance is the basis for triangle side lengths, perimeters and sphere equations in space.

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 $(\frac{2}{3}|-\frac{\frac{1}{3}|2}{3})$; check: $\frac{4}{9} +\frac{ 1}{9} +\frac{ 4}{9} = 1$ ✓. 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.

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$| = \sqrt{(9 + 16)} = 5$. 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: $|(0|0|-5)| = 5$.

The core identity is |v⃗$|^{2} = v$⃗·v⃗: the dot product of a vector with itself is the square of its length, since $v_{1}\cdot{}v_{1} + v_{2}\cdot{}v_{2} + v_{3}\cdot{}v_{3}$ 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$|\cdot{}|b$⃗|); without the magnitudes in the denominator the cosine would not be normalized. Third, the identity yields the rule $|k\cdot{}v$⃗$| = |k|\cdot{}|v$⃗| for scalings. Example: v⃗ = (2|3|6): v⃗·v⃗ = 4 + 9 + 36 = 49 = 7², consistent with |v⃗$| = 7$.