Vectors: Components, Dot & Cross Products

A rower who cannot win

A river is 150 m wide and flows at 3.0 m/s. A rower can only manage 2.0 m/s relative to the water. She is slower than the river, so no matter which way she points the bow, she is going downstream. The question is not whether she drifts. It is how little she can drift if she chooses her heading well.

You cannot answer that by thinking about speeds alone. Speed is one number, and this problem has two independent directions in it: across the river and along it. The whole trick is to stop treating the motion as one thing and start treating it as two numbers that evolve separately. That is what a vector is, not an arrow you draw, but a small bundle of independent numbers, one per direction. The previous lesson had you checking that the units of an expression made sense before trusting it. This one gives you the other habit that runs under every later unit: when a quantity has a direction, break it into pieces along axes you chose on purpose.

Splitting a vector into components

Put down a pair of perpendicular axes, $x$ and $y$. Draw your vector $\vec A$ starting at the origin, at an angle $\theta$ measured counterclockwise from the $+x$ axis. Now drop a perpendicular line from the tip of $\vec A$ down to the $x$ axis. You have made a right triangle whose hypotenuse is $\vec A$, whose horizontal leg runs along $x$, and whose vertical leg runs along $y$.

That triangle is the whole story. The hypotenuse has length $A$ (the magnitude, a plain positive number with units). Right-triangle trigonometry says the side adjacent to $\theta$ is $A\cos\theta$ and the side opposite is $A\sin\theta$. So

$A_x = A\cos\theta, \qquad A_y = A\sin\theta.$

$A_x$ is the $x$-component: how far the vector reaches in the $x$ direction. It can be negative, and a negative component just means the vector reaches backwards along that axis. Going the other way, the two components rebuild the magnitude and the angle by Pythagoras:

$A = \sqrt{A_x^2 + A_y^2}, \qquad \tan\theta = \frac{A_y}{A_x}.$

Which function goes with which leg depends on where the angle is measured from, not on some rule that sine is always vertical. If a problem gives you the angle from the vertical, then the vertical piece is the adjacent one and gets the cosine. Draw the triangle every single time and read it off. That habit costs five seconds and removes an entire category of mistake.

Worked example 1. A child pulls a sled with a rope at $35^\circ$ above the horizontal. The tension is $240$ N. How big is the horizontal pull?

The angle is measured from the horizontal, so the horizontal leg is adjacent, and it gets the cosine:

$F_x = (240\ \text{N})\cos 35^\circ = (240)(0.8192) = 197\ \text{N}.$

The vertical piece comes along free: $F_y = (240)(\sin 35^\circ) = (240)(0.5736) = 138$ N, upward. That vertical piece is doing nothing to move the sled forward, but it is lifting some weight off the ground, which is exactly why pulling at an angle reduces friction. One vector, two jobs, and you only see both once you split it.

Adding vectors means adding components

Two vectors add tip to tail: slide the second one so its tail sits at the first one's tip, and the sum runs from the first tail to the last tip. Geometrically that is a triangle. In components it is even simpler, because the $x$ journeys and the $y$ journeys never interfere:

$\vec A + \vec B = (A_x + B_x,\; A_y + B_y).$

A hiker who walks 4.0 km east then 3.0 km north has displacement $(4.0, 3.0)$ km, and the net distance from the start is $\sqrt{4.0^2 + 3.0^2} = 5.0$ km, not 7.0 km, because the two legs were never pointing the same way. Magnitudes do not add. Components do.

Subtraction works the same way with a sign flip, and this is where people slip. A car going 12 m/s east turns and ends up going 12 m/s north at the same speed. Its change in velocity is $\Delta\vec v = \vec v_f - \vec v_i = (0, 12) - (12, 0) = (-12, 12)$ m/s, whose magnitude is $12\sqrt2 = 17$ m/s. The speed never changed, and yet the velocity changed by a lot. That difference is the seed of everything in circular motion.

Relative velocity is just this addition applied to two observers. The rule is that velocities add through the middle man: your velocity relative to the ground equals your velocity relative to the water plus the water's velocity relative to the ground. For the rower, with $x$ downstream and $y$ across:

$\vec v_{\text{ground}} = \vec v_{\text{water}} + \vec v_{\text{current}}.$

Here is the key structural fact. The across-river component decides how long the crossing takes, and the downstream component decides how far she drifts during that time. They are independent. If a swimmer moving at 1.6 m/s points straight across a 40 m river, her crossing time is set by the 1.6 alone, $t = 40/1.6 = 25$ s, and the 1.2 m/s current carries her $(1.2)(25) = 30$ m downstream. The current never slows the crossing.

Worked example 2. The rower from the opening: $v_b = 2.0$ m/s, current $v_c = 3.0$ m/s, width $L = 150$ m. Point the bow at an angle $\theta$ upstream of straight across. What is the smallest possible drift?

Write the two components of her ground velocity. Across: $v_y = v_b\cos\theta$. Downstream: $v_x = v_c - v_b\sin\theta$. The crossing time is $t = L/(v_b\cos\theta)$, so the drift is

$d = v_x t = \frac{L\,(v_c - v_b\sin\theta)}{v_b\cos\theta}.$

Minimise it by setting $dd/d\theta = 0$. Differentiating the quotient and clearing the common $\cos^2\theta$ gives $-v_b\cos^2\theta + (v_c - v_b\sin\theta)\sin\theta = 0$, which expands to $-v_b\cos^2\theta - v_b\sin^2 \theta + v_c\sin\theta = 0$, and since $\cos^2 + \sin^2 = 1$,

$\sin\theta = \frac{v_b}{v_c} = \frac{2.0}{3.0} = 0.6667.$

So $\theta = 41.8^\circ$ and $\cos\theta = \sqrt{1 - 0.4444} = 0.7454$. Now put the numbers in. Across-speed $v_y = (2.0)(0.7454) = 1.491$ m/s, so $t = 150/1.491 = 100.6$ s. Downstream speed $v_x = 3.0 - (2.0)(0.6667) = 3.0 - 1.333 = 1.667$ m/s. Drift $d = (1.667)(100.6) = 168$ m.

Notice what made this tractable: never once did you think about the resultant arrow. You wrote two components, built the time out of one and the distance out of the other, and then did ordinary one-variable calculus.

The dot product: how much of this lies along that

Often you do not want a whole vector, only the part of it pointing some particular way. The horizontal pull on the sled was exactly that: the part of $\vec F$ along the direction of motion. Generalise it. Take any two vectors $\vec A$ and $\vec B$ with angle $\theta$ between them. The length of $\vec A$'s shadow on the line of $\vec B$ is $A\cos\theta$. Multiply that shadow by the length of $\vec B$ and you have defined the dot product:

$\vec A\cdot\vec B = AB\cos\theta.$

It is a plain number, not a vector. It is positive when the two roughly agree in direction, zero at exactly $90^\circ$, and negative when they oppose. In components it is even easier to compute, because each axis contributes separately:

$\vec A\cdot\vec B = A_xB_x + A_yB_y + A_zB_z.$

Having both forms is the point. The component form gets you a number from raw coordinates; the $AB\cos\theta$ form turns that number into an angle. Set them equal and you can find the angle between any two vectors: $\cos\theta = (\vec A\cdot\vec B)/(AB)$.

Three things it buys you. Work: $W = \vec F\cdot\vec d$, since only the part of the force along the displacement does anything. For $\vec F = (12, -5)$ N and $\vec d = (3, 4)$ m, $W = (12)(3) + (-5)(4) = 36 - 20 = 16$ J. A perpendicularity test: $\vec A\cdot\vec B = 0$ with neither vector zero means they are at right angles, which is how you find the $k$ that makes $(3, k, -2)$ perpendicular to $(4, 5, 1)$: set $12 + 5k - 2 = 0$, so $k = -2$. And projection: the component of $\vec F$ along a direction is $\vec F\cdot\hat u$, where $\hat u$ is a unit vector, meaning a vector of length exactly 1 pointing that way. You build one by dividing any vector by its own magnitude: $\hat u = \vec d/|\vec d|$.

Worked example 3. A bead is threaded on a straight rod pointing along $\vec d = (1, 2, 2)$. A force $\vec F = (3, 4, 0)$ N acts on the bead. How large is the part of $\vec F$ perpendicular to the rod?

The direct route (finding the perpendicular piece geometrically) is painful in three dimensions. The cheap route is to find the parallel piece with a dot product and let Pythagoras give you the rest, since the parallel and perpendicular parts are the two legs of a right triangle with $\vec F$ as hypotenuse.

First make the unit vector along the rod. $|\vec d| = \sqrt{1 + 4 + 4} = \sqrt9 = 3$, so $\hat d = (1/3,\, 2/3,\, 2/3)$. Then

$F_\parallel = \vec F\cdot\hat d = (3)\tfrac13 + (4)\tfrac23 + (0)\tfrac23 = 1 + 2.667 = 3.667\ \text{N}.$

The full magnitude is $|\vec F| = \sqrt{9 + 16} = 5$ N. So

$F_\perp = \sqrt{5^2 - 3.667^2} = \sqrt{25 - 13.44} = \sqrt{11.56} = 3.40\ \text{N}.$

That 3.40 N is what the rod has to supply to keep the bead on it. Splitting a vector into "along" and "across" some awkward direction, rather than along $x$ and $y$, is one of the most reusable moves you will learn.

The cross product: how much of this lies across that

The dot product throws away the perpendicular part. Sometimes the perpendicular part is the whole point. A wrench does nothing if you push along the handle; only the sideways push turns the bolt. That calls for a second product, built from $\sin\theta$ instead of $\cos\theta$:

$|\vec A\times\vec B| = AB\sin\theta.$

Unlike the dot product, this one is itself a vector. Its magnitude is the area of the parallelogram with $\vec A$ and $\vec B$ as sides: base $A$, height $B \sin\theta$. Its direction is perpendicular to both, given by the right-hand rule: point your fingers along $\vec A$, curl them toward $\vec B$, and your thumb points along $\vec A\times\vec B$. Swapping the order reverses it, so $\vec B\times\vec A = -\vec A\times\vec B$. The cross product of a vector with itself, or with any parallel vector, is zero, because $\sin 0 = 0$.

In components, with $\hat\imath$, $\hat\jmath$, $\hat k$ the unit vectors along $x$, $y$, $z$:

$\vec A\times\vec B = (A_yB_z - A_zB_y,\; A_zB_x - A_xB_z,\; A_xB_y - A_yB_x).$

The pattern is cyclic: the $x$ entry uses $y$ and $z$, the $y$ entry uses $z$ and $x$, the $z$ entry uses $x$ and $y$. Write the cycle $x \to y \to z \to x$ in the margin and you will never misremember a sign.

A wrench pushed with 80 N at 0.25 m from the bolt, at $60^\circ$ to the handle, gives torque $\tau = rF\sin\theta = (0.25)(80)(0.866) = 17.3$ N·m. Same formula, different name: torque is $\vec r\times\vec F$, angular momentum is $\vec r \times \vec p$, and magnetic force is $q\vec v\times\vec B$.

Worked example 4. A proton, $q = 1.6\times10^{-19}$ C, moves with $\vec v = (2.0, -1.0, 3.0)\times10^6$ m/s through $\vec B = (0.050, 0.10, -0.020)$ T. Find the magnitude of the magnetic force.

Nothing here is perpendicular to anything, so the angle route is hopeless and you go straight to components. Drop the $10^6$ and restore it at the end. Taking the three entries of $\vec v\times\vec B$ in turn:

$x: (-1.0)(-0.020) - (3.0)(0.10) = 0.020 - 0.300 = -0.280$ $y: (3.0)(0.050) - (2.0)(-0.020) = 0.150 + 0.040 = 0.190$

Careful here: the middle entry of the cyclic formula is $A_zB_x - A_xB_z$, which is what was computed, so $y = +0.190$.

$z: (2.0)(0.10) - (-1.0)(0.050) = 0.200 + 0.050 = 0.250$

Magnitude: $\sqrt{0.280^2 + 0.190^2 + 0.250^2} = \sqrt{0.0784 + 0.0361 + 0.0625} = \sqrt{0.1770} = 0.4207$, so $|\vec v\times\vec B| = 4.207\times10^5$ T·m/s. Then

$F = q|\vec v\times\vec B| = (1.6\times10^{-19})(4.207\times10^5) = 6.73\times10^{-14}\ \text{N}.$

Choosing the axes

Nothing forces $x$ to be horizontal. Axes are yours to place, and the single highest-return decision in a mechanics problem is where to point them. On a ramp, tilt the axes so one runs along the slope. Then the acceleration lies entirely on one axis, the normal force entirely on the other, and only gravity needs splitting: into $mg\sin\alpha$ down the slope and $mg\cos\alpha$ into it.

Try it. A 40 kg crate sits on a frictionless $25^\circ$ ramp while a worker pushes horizontally into the hill with 200 N. Along the slope, gravity pulls down-slope with $mg\sin 25^\circ = (40)(9.8)(0.4226) = 165.6$ N. The horizontal push makes angle $25^\circ$ with the slope surface, so its along-slope (up-slope) component is $(200)\cos 25^\circ = (200)(0.9063) = 181.3$ N. The net along-slope force is $181.3 - 165.6 = 15.6$ N up the slope. Two lines, because the axes were chosen to make the normal force irrelevant.

Recognising when this is the tool

Some signals to watch for in a problem statement.

Any mention of an angle with a force, a velocity, or a displacement means components. Write $A\cos\theta$ and $A\sin\theta$ before you do anything else.

Two motions combining (a boat and a current, a plane and a wind, a passenger walking on a moving train) means relative velocity, which means adding vectors in components. Ask which component controls the time and which controls the distance; they are usually different, and that observation is the solution.

The words "component along", "how much of the force does work", "perpendicular to", "at what angle", or "for what value of $k$ are these at right angles" are all dot product. So is finding the perpendicular leftover: get the parallel part with a dot product, then use Pythagoras.

Torque, angular momentum, magnetic force, the area of a parallelogram or triangle, the perpendicular distance from a point to a line, and any phrase like "lever arm" or "impact parameter" are cross product. That last one is worth keeping: the closest approach of a particle travelling in a straight line, with position $\vec r$ and velocity $\vec v$, is $|\vec r\times\vec v|/|\vec v|$, because the cross product already contains the perpendicular distance times the speed. Two ships on straight courses, an alpha particle flying past a nucleus: same formula both times. And when a problem hands you three cables or three supports in 3D, you are being asked for three component equations in three unknown tensions, each cable's pull written as its tension times a unit vector along the cable.

Once a quantity can be split into pieces this way, the next natural question is how each piece changes over time. That is where derivatives come in.