Vector components

Physics 12 lives in two dimensions, so we have to be careful about direction. A scalar is just a number with a unit — mass, time, temperature, energy. A vector carries a magnitude and a direction — displacement, velocity, acceleration, force. We draw a vector as an arrow: its length is the magnitude and the way it points is the direction. The whole game of 2D kinematics is handling those directions correctly, and the single most powerful tool for that is breaking a vector into components.

You could add vectors by carefully drawing them tip-to-tail and measuring the result with a ruler and protractor — but that is slow and imprecise. Instead we notice that any vector can be replaced by two perpendicular 'shadow' vectors: one along the x-axis and one along the y-axis. These are its components. Because x and y are independent, we can handle horizontal and vertical motion separately — and that trick unlocks projectiles, forces on ramps, and everything that follows.

Suppose a vector $\vec{A}$ has magnitude $A$ and points at an angle $\theta$ measured from the positive x-axis. Drop perpendiculars to each axis and a right triangle appears, with $\vec{A}$ as the hypotenuse. Straight from trigonometry, the adjacent side is $A_x = A\cos\theta$ and the opposite side is $A_y = A\sin\theta$. That is the entire recipe: x gets the cosine, y gets the sine — provided $\theta$ is measured from the x-axis.

Going the other way is just Pythagoras and a tangent. Given the components, the magnitude is $A = \sqrt{A_x^{2} + A_y^{2}}$ and the direction is $\theta = \tan^{-1}\!\left(\dfrac{A_y}{A_x}\right)$. The components and the magnitude–direction pair are two languages for the same vector — you convert freely between them depending on which is easier for the problem in front of you.

Here is why components are worth the effort. To add several vectors, resolve each into x and y, then simply add the x-components together and add the y-components together: $R_x = A_x + B_x + \dots$ and $R_y = A_y + B_y + \dots$. The sums $R_x$ and $R_y$ are the components of the resultant $\vec{R}$, which you then rebuild with $R = \sqrt{R_x^{2} + R_y^{2}}$ and $\theta = \tan^{-1}(R_y/R_x)$. No rulers, no protractors — just arithmetic.

Components carry signs. A vector pointing left has a negative $A_x$; one pointing down has a negative $A_y$. When you finish with $\tan^{-1}$, your calculator only returns angles between $-90^\circ$ and $+90^\circ$, so you must look at the *signs of $R_x$ and $R_y$* to decide which quadrant the resultant truly lives in. A resultant with $R_x < 0$ and $R_y > 0$ is in the second quadrant, so add $180^\circ$ to the calculator's answer. Skipping this is the classic 2D blunder.

For students ready to push further: two vectors can be multiplied to give a scalar through the dot product, $\vec{A}\cdot\vec{B} = AB\cos\theta = A_xB_x + A_yB_y$, where $\theta$ is the angle between them. It measures how much the two vectors point the same way: it is maximum when they are parallel, zero when they are perpendicular, and negative when they oppose. This is exactly the machinery behind $W = \vec{F}\cdot\vec{d}$ for work — a force does no work when it acts at right angles to the motion.

Components always work, but when you add just two vectors separated by an angle $\phi$ there is a direct shortcut from the law of cosines: the resultant magnitude is $R = \sqrt{A^{2} + B^{2} + 2AB\cos\phi}$, and its direction relative to $\vec{A}$ is $\alpha = \tan^{-1}\!\dfrac{B\sin\phi}{A + B\cos\phi}$. Always sanity-check the limits: when $\phi = 0$ (parallel) it gives $R = A+B$; when $\phi = 90^\circ$, $R = \sqrt{A^2+B^2}$; when $\phi = 180^\circ$ (opposite), $R = |A-B|$. Watch the sign — the $+2AB\cos\phi$ term automatically becomes a subtraction whenever $\phi$ is obtuse, because $\cos\phi$ turns negative.

Physicists write any 2D vector compactly as $\vec{A} = A_x\,\hat{i} + A_y\,\hat{j}$, where $\hat{i}$ and $\hat{j}$ are unit vectors (length exactly 1) pointing along $x$ and $y$. Addition then needs no diagram at all — you just collect like terms: $\vec{A} + \vec{B} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}$. The dot product drops out the same way, since $\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = 1$ and $\hat{i}\cdot\hat{j} = 0$, giving $\vec{A}\cdot\vec{B} = A_xB_x + A_yB_y$. The notation extends to 3D by simply adding a $\hat{k}$ term for the $z$-direction.

Almost every vector question yields to the same recipe, and building the habit now pays off all year. (1) Sketch the vectors tip-to-tail and choose $x$–$y$ axes (often along the strongest force or the direction of motion). (2) Resolve each vector into components, watching the signs. (3) Sum separately: $R_x = \sum A_x$ and $R_y = \sum A_y$. (4) Rebuild $$R = \sqrt{R_x^2 + R_y^2}$$ and $\theta = \tan^{-1}(R_y/R_x)$, then check the quadrant from the signs of $R_x$ and $R_y$. For an object in equilibrium the net force is zero, so you instead set $\sum F_x = 0$ and $\sum F_y = 0$ and solve the two equations together — exactly how you crack tension-in-two-ropes and hanging-sign problems.

You are never forced to keep $x$ horizontal and $y$ vertical — the axes are yours to choose, and choosing them well can turn a messy problem into a clean one. On a ramp or inclined plane, tilt the whole coordinate system so that $x$ runs along the slope and $y$ runs perpendicular to it. Then the weight $mg$, which points straight down, splits into $mg\sin\theta$ down the slope and $mg\cos\theta$ pressing into the surface — and any sliding motion stays entirely along your new $x$-axis. The working rule of thumb is simple: point one axis along the direction of motion or the strongest force, and most of your components will collapse to zero, leaving far less arithmetic to carry.

Let us add $\vec{A} = 40$ units at $20^\circ$, $\vec{B} = 30$ units at $110^\circ$, and $\vec{C} = 25$ units at $250^\circ$, all measured from the positive $x$-axis. Resolve each one in turn, carrying two decimals: $A_x = 40\cos 20^\circ = 37.59$ and $A_y = 40\sin 20^\circ = 13.68$; $B_x = 30\cos 110^\circ = -10.26$ and $B_y = 30\sin 110^\circ = 28.19$; $C_x = 25\cos 250^\circ = -8.55$ and $C_y = 25\sin 250^\circ = -23.49$. Now sum the columns: $R_x = 37.59 - 10.26 - 8.55 = 18.78$ and $R_y = 13.68 + 28.19 - 23.49 = 18.38$. Rebuild the resultant: $R = \sqrt{18.78^{2} + 18.38^{2}} = 26.3$ at $\theta = \tan^{-1}(18.38/18.78) = 44.4^\circ$. Notice how the signs from each quadrant took care of themselves the moment every angle was measured consistently from the $x$-axis.

Subtracting one vector from another sounds like a new operation, but it is nothing more than addition in disguise: $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$, where $-\vec{B}$ is simply $\vec{B}$ turned around by $180^\circ$, so every one of its components flips sign. In component form it could not be tidier — $(\vec{A} - \vec{B})_x = A_x - B_x$ and $(\vec{A} - \vec{B})_y = A_y - B_y$. This is exactly what you compute to find a change in a vector quantity, such as $\Delta\vec{v} = \vec{v}_f - \vec{v}_i$ when working out an acceleration, or the relative velocity of one object as seen from another. Geometrically, $\vec{A} - \vec{B}$ is the arrow that points from the tip of $\vec{B}$ to the tip of $\vec{A}$ when both are drawn from the same origin.

A recurring instinct — and a wrong one — is to add magnitudes directly. If $\vec{A}$ is $6$ units and $\vec{B}$ is $8$ units, the resultant is $14$ only if they point the very same way. Set them at right angles and you get $\sqrt{6^{2} + 8^{2}} = 10$; point them in opposite directions and you get merely $2$. In general the resultant magnitude obeys the triangle inequality, $|A - B| \le |\vec{A} + \vec{B}| \le A + B$: the sum can never exceed the total length, nor fall below their difference. Keeping this bracket in mind is a fast sanity check — if your computed resultant lands outside it, you have made an arithmetic slip somewhere and should hunt it down before moving on.

Multiplying a vector by a scalar $k$ scales its length by $|k|$ and, when $k$ is negative, reverses its direction, while every component simply multiplies through: $k\vec{A} = (kA_x,\ kA_y)$. Doubling a velocity doubles both components and so doubles the speed; multiplying by $-1$ produces the anti-parallel vector we used in subtraction. This is precisely why $\vec{F} = m\vec{a}$ works component by component — the scalar mass stretches the acceleration vector into the force vector without turning it at all. A unit vector is the special case $\hat{A} = \vec{A}/A$, which keeps the direction but rescales the length to exactly $1$, handy whenever you need a pure 'pointer' carrying no magnitude of its own.

Two angle conventions collide in real problems, and mixing them is a classic trap. In mathematics an angle is measured counter-clockwise from the positive $x$-axis, so east is $0^\circ$ and north is $90^\circ$. In navigation a bearing is measured clockwise from north, so north is $000^\circ$ and east is $090^\circ$. A heading of 'N $30^\circ$ E' means $30^\circ$ east of due north, which is a standard angle of $60^\circ$. Before resolving anything, convert every direction into one consistent system — usually standard angles — and then apply $A_x = A\cos\theta$ and $A_y = A\sin\theta$. A surprising fraction of all component errors trace back to a bearing that was never converted.

An object sitting still, or gliding at constant velocity, is in equilibrium — the net force on it is zero. In component language that single vector statement splits into two separate conditions that must hold at once: $\sum F_x = 0$ and $\sum F_y = 0$. A street lamp hung from two cables, a sign held out by a strut, a block resting on a ramp — each becomes a pair of simultaneous equations the instant you resolve every force into $x$ and $y$. The real power of the method is exactly this reduction: a single vector balance ('the forces cancel') becomes two ordinary algebra equations, which you solve together for two unknowns such as the tensions in two supporting ropes.

Sometimes you need to know how much of one vector lies along a second vector and how much lies across it — the parallel and perpendicular parts. If $\phi$ is the angle between them, the parallel piece has magnitude $A\cos\phi$, which is exactly what the dot product extracts through $\vec{A}\cdot\hat{B} = A\cos\phi$, while the leftover perpendicular piece has magnitude $A\sin\phi$. This decomposition is the engine behind resolving weight on a ramp, finding the effective driving force along a direction of travel, and computing work, where only the component of force parallel to the displacement counts toward $W = Fd\cos\phi$. Learn to see any vector as 'so much along' plus 'so much across' and a great many problems simplify at a glance.

Before you trust a vector result, run three fast checks. First, the magnitude bracket: does the resultant sit between $|A - B|$ and $A + B$? Second, the quadrant: do the signs of $R_x$ and $R_y$ agree with the direction your sketch predicted? Third, a limiting case: push a variable to an extreme — set an angle to $0^\circ$ or $90^\circ$ — and confirm the formula gives the obvious answer. These habits cost only seconds and catch the overwhelming majority of sign slips, swapped sines and cosines, and forgotten quadrant corrections. Building them into your routine now is what separates a confident solver from someone merely hoping the calculator was kind.

Everything you have built extends to three dimensions with almost no new ideas. A vector simply gains a third component, $\vec{A} = (A_x,\ A_y,\ A_z) = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$; its magnitude becomes $A = \sqrt{A_x^{2} + A_y^{2} + A_z^{2}}$; and addition still means summing like components. The dot product just gains a term, $\vec{A}\cdot\vec{B} = A_xB_x + A_yB_y + A_zB_z$. Direction now needs two angles rather than one, but the philosophy is unchanged — break the problem into independent axes, work each separately, and reassemble at the end. Master the two-dimensional machinery and the jump to 3D, and later to velocity and acceleration vectors, is a genuinely short step.

It is worth pausing on why all this vector machinery matters so much in kinematics: the three headline quantities — displacement, velocity, and acceleration — are every one of them vectors, so each obeys exactly the component rules of this lesson. Average velocity is the displacement vector divided by the time, $\vec{v}_{avg} = \Delta\vec{r}/\Delta t$, and points the same way as the displacement. Acceleration is the change in the velocity vector divided by time, $\vec{a} = \Delta\vec{v}/\Delta t$, and because it comes from a vector difference it can point in a wholly different direction from the velocity itself — which is precisely how an object rounding a curve accelerates even at constant speed. Every 2D motion problem from here on is really this component bookkeeping applied to $\vec{r}$, $\vec{v}$, and $\vec{a}$.