Push a lawnmower across the lawn and something quietly frustrating happens. You shove along the handle, which slants down toward the ground, so your 200\ \text{N} of effort does not all go into moving the mower forward. Part of it drives the mower ahead — the useful part — and part of it merely presses the mower harder into the grass, doing nothing but tiring you out. Tilt the handle steeper and you press down more and push forward less; hold it flatter and almost all your effort goes into motion. The same force, aimed differently, does a different job.

That is the whole reason physics needs vectors. A force isn't just "how hard" — it is "how hard and which way", and to predict what it actually accomplishes you must split it into directions and recombine the pieces. On this page we take the physical quantities that carry a direction — displacement, velocity, force — and learn the three moves that make them useful: resolving a vector into components, adding vectors, and the two products that hide inside real physics laws — the dot product (which gives you work) and the cross product (which gives you torque).

Scalars carry a size; vectors carry a size and a direction

Some physical quantities are fully described by a single number with a unit. Your mass is 70\ \text{kg}; the room is at 21\,^\circ\text{C}; the trip took 3\ \text{hours}. There is no "direction of 70 kilograms". These are scalars — mass, temperature, time, energy, speed, electric charge.

Other quantities are meaningless until you also say which way. "I walked 5\ \text{km}" doesn't tell you where I ended up; "I walked 5\ \text{km} north-east" does. Quantities that need a magnitude and a direction are vectors — displacement, velocity, acceleration, force, momentum. We write them in bold, \vec{F}, and draw them as arrows: the length is the magnitude |\vec{F}| = F, the way the arrow points is the direction.

Resolving a vector into components

To do anything quantitative with a slanted vector, we break it into pieces that point along the axes — a horizontal part and a vertical part. Drop a perpendicular from the tip of the arrow to the x-axis and you make a right triangle whose hypotenuse is the vector itself. If the force \vec{F} has magnitude F and makes an angle \theta with the horizontal, plain trigonometry on that triangle gives the two components:

Reveal the figure step by step: first the slanted force, then the horizontal leg it casts onto the x-axis, then the vertical leg, and finally the angle that ties them together.

The components are the vector's shadow on each axis. Going the other way, if you know F_x and F_y you rebuild the magnitude with Pythagoras and the direction with the tangent:

Notice the magnitude is not F_x + F_y — a point two blocks east and two blocks north is only 2\sqrt{2}\approx 2.83 blocks away, not four.

Adding vectors: add the components

Two forces act on the same object; what single force do they amount to? Geometrically you slide one arrow so its tail sits on the other's head, and the resultant runs from the very first tail to the very last head — the "head-to-tail" rule. But the arithmetic is where the magic is: once every vector is resolved into components, you just add the components separately.

The awkward geometry of adding slanted arrows collapses into two easy sums, one per axis. Then Pythagoras rebuilds the magnitude of the resultant and the tangent rebuilds its direction. This "resolve, add component-wise, recombine" recipe is the single most-used technique in first-year mechanics — every free-body-diagram problem is really this.

The dot product: how much a force works

Back to the lawnmower. You push with force \vec{F} along the handle while the mower moves through a displacement \vec{d} along the ground. The work you do — the energy you transfer — depends only on the part of the force that lies along the motion. That "amount of one vector along another" is exactly what the dot product computes:

The \cos\theta is the whole story. When the force points along the motion (\theta = 0, \cos\theta = 1) you get maximum work, W = Fd. When the force is perpendicular to the motion (\theta = 90^\circ, \cos\theta = 0) you do zero work, no matter how hard you push. And when the force opposes the motion (\theta = 180^\circ, \cos\theta = -1) the work is negative — friction, for instance, drains energy away. The graph below plots W = Fd\cos\theta against the angle for a fixed F and d; watch it cross zero at exactly 90^\circ.

The cross product: how much a force turns

Loosen a stubborn bolt with a wrench. Pushing straight along the handle does nothing; pushing perpendicular to it, as far out as you can reach, turns it best. The turning effect — the torque — grows with how far out you push (r), how hard (F), and how squarely (the angle between them). That is the fingerprint of the cross product:

Notice the beautiful contrast. The dot product uses \cos\theta and peaks when the vectors are parallel; the cross product uses \sin\theta and peaks when they are perpendicular. Work cares about pushing along; torque cares about pushing across. One returns a scalar (energy), the other a vector (an axis of spin). Between them they capture most of what forces actually do.

Worked examples

Example 1 — resolve a force. A child pulls a sled with a rope, exerting F = 50\ \text{N} at \theta = 37^\circ above the horizontal. Split it into components (using \cos 37^\circ \approx 0.80, \sin 37^\circ \approx 0.60):

Only the 40\ \text{N} horizontal part drags the sled forward; the 30\ \text{N} vertical part lifts it, easing the load on the ground.

Example 2 — work done. That same sled is pulled d = 10\ \text{m} along level ground by the 50\ \text{N} rope at 37^\circ. The work done by the rope is

Equivalently, only the 40\ \text{N} horizontal component acts over the 10\ \text{m}: 40 \times 10 = 400\ \text{J} — the same answer, as it must be.

Example 3 — torque on a wrench. You push with F = 80\ \text{N} at the end of a r = 0.25\ \text{m} wrench, at 90^\circ to the handle. The torque magnitude is

Push at only 30^\circ to the handle instead and \sin 30^\circ = 0.5, so the torque halves to 10\ \text{N·m} — same force, half the turning.

Example 4 — add two forces. Two ropes pull a boat: \vec{A} = 30\ \text{N} due east and \vec{B} = 40\ \text{N} due north. Their components add axis by axis: R_x = 30, R_y = 40, so the resultant has magnitude

at an angle \theta = \tan^{-1}(40/30) \approx 53^\circ north of east. Note 50 \ne 30 + 40: vectors don't add like plain numbers.