Rectangular components

Every vector problem in two dimensions becomes easy once you stop thinking of a vector as one arrow at an awkward angle and start thinking of it as two arrows at right angles — one along $x$ , one along $y$ . That splitting is resolution into rectangular components, and it is the single most-used trick in Motion in a Plane: it sits right after vector addition and right before the dot and cross products in Unit 2, "Kinematics," and it is what makes projectile motion, inclined-plane problems, and multi-force equilibrium solvable by plain algebra instead of geometry. Once a vector is written as $\vec A = A_x\hat\imath + A_y\hat\jmath$ , adding several vectors, however many and however tilted, becomes nothing more than adding their $x$ -parts and adding their $y$ -parts. This article builds that idea from the definitions up, works two full numeric examples, and flags the sign and quadrant mistakes that trip up almost every beginner.

The core insight

Any vector splits into perpendicular $x$ - and $y$ -parts, $A_x = A\cos\theta$ and $A_y = A\sin\theta$ , and vectors add by adding these parts separately, which turns vector addition into ordinary arithmetic.

The core idea

A vector $\vec A$ lying in the $xy$ -plane can always be replaced by two perpendicular vectors, one along the x-axis and one along the y-axis, whose combined effect is identical to $\vec A$ itself. These two pieces are the rectangular components of $\vec A$ . If $\vec A$ makes angle $\theta$ with the positive x-axis, simple right-triangle trigonometry gives the components directly:

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

where: $A$ is the magnitude of the vector, $\theta$ is measured from the positive x-axis, and $A_x, A_y$ are the (signed) lengths of the projections onto the two axes. Once the components are known, the vector is written compactly using the unit vectors $\hat\imath$ (length 1, along +x) and $\hat\jmath$ (length 1, along +y):

\vec A = A_x\hat\imath + A_y\hat\jmath.

In three dimensions a third unit vector $\hat k$ along +z is added, so $\vec A = A_x\hat\imath + A_y\hat\jmath + A_z\hat k$ ; almost every Class 11 kinematics problem, projectile motion included, stays in a single plane and needs only $\hat\imath$ and $\hat\jmath$ . The reverse operation — going from components back to magnitude and direction — uses Pythagoras and the tangent ratio:

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

where: the positive square root gives the magnitude, and $\theta$ must be placed in the correct quadrant using the individual signs of $A_x$ and $A_y$ , since a calculator's inverse tangent alone cannot distinguish, say, $45°$ from $225°$ . The real payoff of this scheme is that vector addition becomes component-wise addition. For $\vec A = A_x\hat\imath + A_y\hat\jmath$ and $\vec B = B_x\hat\imath + B_y\hat\jmath$ ,

\vec A + \vec B = (A_x+B_x)\hat\imath + (A_y+B_y)\hat\jmath,

which completely replaces the parallelogram-law cosine formula and scales effortlessly to any number of vectors: resolve every vector onto the same pair of axes, sum the x-parts, sum the y-parts, and recombine only at the end if a single magnitude and direction is wanted. A further trick that recurs throughout mechanics is choosing the axes wisely — along and perpendicular to an inclined plane, or horizontal and vertical for a projectile — so that one component of some force or velocity comes out to be exactly zero, cutting the algebra in half.

Figure

A vector at angle theta from the x axis with dashed projections onto the x axis and y axis labelled A cos theta and A sin theta

Derivation AComponent-wise vector addition

Show that when several vectors are each resolved into rectangular components along the same fixed x-axis and y-axis, the components of their vector sum are simply the algebraic sum of the individual x-components and the algebraic sum of the individual y-components, so that the resultant's magnitude and direction can be recovered afterward from those two sums alone.

Step 1. Write each vector in unit-vector form using its own angle from the x-axis: for vectors $\vec A$ and $\vec B$ , $\vec A = A\cos\theta_A\,\hat\imath + A\sin\theta_A\,\hat\jmath$ and $\vec B = B\cos\theta_B\,\hat\imath + B\sin\theta_B\,\hat\jmath$ , i.e. $\vec A = A_x\hat\imath + A_y\hat\jmath$ and $\vec B = B_x\hat\imath + B_y\hat\jmath$ .

Step 2. Add the two vectors directly, treating $\hat\imath$ and $\hat\jmath$ as ordinary algebraic symbols that cannot mix with each other (they are perpendicular, independent directions): $\vec A + \vec B = (A_x\hat\imath + A_y\hat\jmath) + (B_x\hat\imath + B_y\hat\jmath)$ .

Step 3. Collect the $\hat\imath$ terms together and the $\hat\jmath$ terms together, since ordinary addition is commutative and associative: $\vec A + \vec B = (A_x+B_x)\hat\imath + (A_y+B_y)\hat\jmath$ . The resultant's own x-component is therefore $R_x = A_x+B_x$ and its y-component is $R_y = A_y+B_y$ — nothing more than adding numbers.

Step 4. Recover the resultant's magnitude and direction from $R_x, R_y$ exactly as for any single vector: $R = \sqrt{R_x^2+R_y^2}$ and $\tan\theta_R = R_y/R_x$ , with the quadrant fixed by the signs of $R_x$ and $R_y$ . The argument extends unchanged to any number of vectors — add all the x-components, add all the y-components — which is exactly why component resolution, not the geometric parallelogram law, is the standard method whenever more than two vectors must be combined.

Worked example

Part 1. A displacement vector has magnitude $50\ \text{m}$ and makes an angle of $37°$ with the x-axis (use $\sin 37° \approx 0.6$ , $\cos 37° \approx 0.8$ ). Find its rectangular components.

Step 1. Apply the component formulas directly: $A_x = A\cos\theta = 50 \times 0.8 = 40\ \text{m}$ and $A_y = A\sin\theta = 50 \times 0.6 = 30\ \text{m}$ .

Step 2. Write the vector in unit-vector form: $\vec A = 40\hat\imath + 30\hat\jmath\ \text{m}$ .

Step 3. Check by recombining: $\sqrt{40^2+30^2} = \sqrt{1600+900} = \sqrt{2500} = 50\ \text{m}$ , matching the given magnitude, so the resolution is consistent.

Part 2 (component addition). Three forces act on a particle: $\vec F_1 = 30\ \text{N}$ at $0°$ , $\vec F_2 = 20\ \text{N}$ at $90°$ (along +y), and $\vec F_3 = 10\ \text{N}$ at $180°$ (along $-x$ ). Find the resultant using rectangular components.

Step 1. Resolve each force: $\vec F_1 = 30\hat\imath + 0\hat\jmath$ , since $\cos0°=1,\sin0°=0$ ; $\vec F_2 = 0\hat\imath + 20\hat\jmath$ , since $\cos90°=0,\sin90°=1$ ; $\vec F_3 = -10\hat\imath + 0\hat\jmath$ , since $\cos180°=-1,\sin180°=0$ .

Step 2. Add the x-components and y-components separately: $F_x = 30+0-10 = 20\ \text{N}$ , $F_y = 0+20+0 = 20\ \text{N}$ , so $\vec F_{\text{net}} = 20\hat\imath + 20\hat\jmath\ \text{N}$ .

Step 3. Recover magnitude and direction: $F = \sqrt{20^2+20^2} = \sqrt{800} \approx 28.3\ \text{N}$ , and $\tan\theta = 20/20 = 1$ , so $\theta = 45°$ from the x-axis. The resultant force is about 28.3 N, directed at 45° above the x-axis.

Competitive stretch

If a fourth force $\vec F_4 = 10\sqrt2\ \text{N}$ were added at $225°$ (i.e. into the third quadrant), its components would be $F_{4x} = 10\sqrt2\cos225° = -10\ \text{N}$ and $F_{4y} = 10\sqrt2\sin225° = -10\ \text{N}$ , so the new totals become $F_x = 20-10=10\ \text{N}$ and $F_y = 20-10=10\ \text{N}$ — the resultant shrinks to $\sqrt{200}\approx 14.1\ \text{N}$ but stays at $45°$ , because $\vec F_4$ happened to point exactly opposite to the original resultant.

Common traps

Swapping the formulas — using $\sin\theta$ for $A_x$ and $\cos\theta$ for $A_y$ . Always anchor to the definition: $\theta$ is measured from the x-axis, so the x-component carries the cosine.
Dropping the sign of a component when the vector lies in the second, third, or fourth quadrant; a negative $A_x$ or $A_y$ is real information and must be carried through the addition.
Adding raw magnitudes of different vectors instead of their like components; component addition only works once every vector is resolved onto the same pair of axes.
Reading $\tan^{-1}(A_y/A_x)$ off a calculator without checking the quadrant from the signs of $A_x$ and $A_y$ — the inverse tangent function alone cannot tell $45°$ from $225°$ .

Ask the tutor: if resolving vectors along a tilted incline still feels confusing, ask the AI tutor to walk through choosing the axes step by step.

Why do we even bother splitting a vector into components instead of just using the parallelogram law directly?

The parallelogram law works fine for two vectors but gets messy fast — you need a new cosine-rule calculation for every fresh pair, and it gets much harder with three or more vectors at different angles. Components turn every addition into simple arithmetic: resolve once, add the x-parts, add the y-parts, done, no matter how many vectors are involved.

Does the choice of x-axis and y-axis matter, or can I resolve a vector along any two perpendicular directions?

Any pair of mutually perpendicular axes works — the physics does not care which way you draw them. What changes is convenience: on an inclined plane, axes along and perpendicular to the slope often make gravity's component along one axis zero, while horizontal/vertical axes are natural for projectile motion. Pick whichever pair makes the most components vanish.

If Ax and Ay come out negative, did I make a mistake?

Not necessarily — a negative component is completely normal and simply means the vector points partly in the negative x or negative y direction. A vector at $200°$ , for instance, has both components negative because it lies in the third quadrant. The mistake would be dropping that sign when you add it to other vectors.

Rectangular components

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads