Vectors and unit-vector notation

A vector is a quantity with a magnitude and a direction that combines with others of its kind by the tip-to-tail rule: put the tail of B at the tip of A, and A + B is the arrow from the first tail to the last tip. Displacement is the model — walk 40 m east and then 30 m north and you stand 50 m from the start, not 70 m — and force, velocity and acceleration earn the name because experiment shows they combine the same way. Drawing the arrows shows the shape of an answer, but only to the precision of a ruler and a protractor. Unit vectors remove that limit. î, ĵ and k̂ are dimensionless arrows of length one along the axes, so any vector can be written A = Aₓ î + A_y ĵ + A_z k̂, and the geometric rule collapses into adding signed numbers axis by axis; magnitude and direction are rebuilt afterwards, with a quadrant check that a calculator's arctan will not make for you. The model has edges. A size and a direction are not enough: a finite three-dimensional rotation has both, yet a quarter-turn about x followed by one about y leaves a book in a different orientation from the same turns taken in reverse order, so rotations do not add like arrows. And components belong to a frame — turn the axes and the components generally change while A itself does not.

Something counts as a vector when it has both a size and a direction and combines with others tip to tail; in unit-vector notation it is written as signed components multiplying î, ĵ and k̂, the dimensionless unit vectors along the x, y and z axes.

F = (6.0 î − 8.0 ĵ) N has magnitude √(6.0² + 8.0²) = 10.0 N and direction F̂ = 0.60 î − 0.80 ĵ, which points 53.1° below the +x axis; the minus sign, not the size, is what sends it downward.

A vector is specified by two things, and each is easy to state carelessly. The magnitude |A| is a non-negative number with a unit: a displacement of 250 m, a force of 12 N. The direction needs a reference, or it says nothing. “30°” is incomplete; “30° north of east”, “143° counterclockwise from +x” and “along −ĵ” are complete, and so is a compass bearing measured clockwise from north. Where an arrow is drawn does not matter: two arrows with the same length and the same direction are the same vector, which is what lets you slide B across the page to put its tail on the tip of A. Reversing the arrow gives −A, with the same magnitude pointing the opposite way, and multiplying by a scalar c stretches the arrow by |c|, flipping it when c is negative. Scalars such as mass, time and energy carry no direction and add as ordinary numbers. A vector's magnitude is a scalar too, which is exactly why it cannot carry the direction on its own.

To add B to A, slide B until its tail sits on the tip of A; the sum R = A + B runs from the tail of A to the tip of B. Draw B first and A second and you trace the other half of a parallelogram with the same diagonal, so A + B = B + A. With three or more vectors keep chaining tips to tails; the grouping does not matter, (A + B) + C = A + (B + C), and the resultant always closes the chain. Subtraction needs no new rule: A − B = A + (−B), so reverse B and chain it on. Change of velocity is the everyday case. A puck moving at v₁ = 12 m s⁻¹ east leaves a collision at v₂ = 5.0 m s⁻¹ north. Δv = v₂ − v₁ is v₂ followed by a 12 m s⁻¹ arrow pointing west, a right triangle with hypotenuse √(12² + 5.0²) = 13 m s⁻¹ directed 22.6° north of west. The speed fell by only 7 m s⁻¹, yet the velocity changed by 13 m s⁻¹, and it is the 13, not the 7, that sets how hard the collision had to push.

Pick a right-handed set of axes and lay an arrow of length one along each: î along +x, ĵ along +y, k̂ along +z. They are dimensionless, so once a vector is written on them, every unit and every sign lives in its components; Aₓ = −8.0 m is a full statement, eight metres in the −x direction. The same idea works along any line, not just the axes. Divide a vector by its own magnitude and what survives is pure direction, the unit vector  = A/|A|. A drone at P = (2.0 m, −1.0 m) is sent toward Q = (−6.0 m, 14.0 m). Its displacement is d = Q − P = (−8.0 î + 15.0 ĵ) m, of length √(8.0² + 15.0²) = 17.0 m, so d̂ = −0.471 î + 0.882 ĵ. Any vector along that line is now a magnitude times d̂: flying at 34 m s⁻¹ from P toward Q means v = (34 m s⁻¹) d̂ = (−16 î + 30 ĵ) m s⁻¹. Check a unit vector by squaring its components: 0.471² + 0.882² = 1.00.

The magnitude of a sum is not the sum of the magnitudes. Two ropes pulling a sled with 250 N and 180 N, 60° apart, do not deliver 430 N. Put the first along +x: the second is (90 î + 155.9 ĵ) N, the sum is (340 î + 155.9 ĵ) N, and its magnitude is 374 N. |A + B| = |A| + |B| holds only when the two vectors point the same way; at any other angle part of each vector runs across the other, and pulling in opposite directions the same two ropes would give just 70 N. The slip has a component-form twin: |R| is √(Rₓ² + R_y²), never |Rₓ| + |R_y|, which here would claim 496 N. Add the vectors first, as arrows or as components, and take the magnitude last.