Resolution of vectors

This topic sits directly inside the prescribed CBSE Class 11 Physics scope for Motion in a Plane: "resolution into rectangular components" is named explicitly alongside vector addition/subtraction and unit vectors. It follows the graphical triangle and parallelogram laws of vector addition and feeds straight into the analytical method of vector addition, which is nothing but resolution applied to two vectors and added component-wise. Everything below stays within algebra and right-triangle trigonometry in two dimensions — no 3-D geometry, no vector triple products, both of which are explicitly out of scope for this board.

Picture a vector $\vec A$ drawn from the origin of an $xy$-plane, making an angle $\theta$ with the positive $x$-axis. Drop a perpendicular from the tip of $\vec A$ onto the $x$-axis; the foot of that perpendicular marks off a length $A_x$ along the $x$-axis, and the perpendicular itself has length $A_y$ along the $y$-axis. Together $A_x$ and $A_y$ are the two legs of a right triangle whose hypotenuse is $\vec A$, so ordinary trigonometry gives them directly:

where: $A$ is the magnitude (length) of the vector, $\theta$ is the angle it makes with the chosen $x$-axis (measured counter-clockwise), $A_x$ is the component along the $x$-axis (the projection of $\vec A$ onto the $x$-axis), and $A_y$ is the component along the $y$-axis. Physically, $A_x$ and $A_y$ are what the vector is "worth" along each axis on its own — for a force, $A_x$ is the part of the force pulling along $x$ and $A_y$ the part pulling along $y$, and each behaves in every calculation exactly like an independent one-dimensional vector along its own axis. Once you have the two components, the original vector can be rebuilt exactly using the unit vectors $\hat i$ (a vector of length 1 along $x$) and $\hat j$ (a vector of length 1 along $y$):

This is the standard component form of a vector, and it is the form used for essentially every calculation from here through projectile motion and circular motion. It is worth being clear about what $A_x\hat i$ and $A_y\hat j$ actually are: they are two genuine, separate vectors, one lying entirely along the $x$-axis and one entirely along the $y$-axis, whose vector sum reconstructs $\vec A$ by the very same triangle law used for adding any two vectors. Nothing new is being invented here — resolution is just vector addition run in reverse. The process also runs backward. Given the components $A_x$ and $A_y$ of any vector, its magnitude and direction are recovered by:

where: the magnitude formula is simply the Pythagoras theorem applied to the right triangle formed by $A_x$, $A_y$ and $\vec A$, and $\theta$ is the angle whose tangent is the ratio of the opposite side ($A_y$) to the adjacent side ($A_x$). This inverse-tangent step is where most marks are lost in exams, because a calculator's $\tan^{-1}$ always returns a value strictly between $-90^\circ$ and $+90^\circ$, regardless of which quadrant the vector actually lies in — you must look at the signs of $A_x$ and $A_y$ separately to place the angle correctly, a point developed fully in Worked example 2 below.

Resolution also makes vector addition mechanical: if $\vec A = A_x\hat i+A_y\hat j$ and $\vec B = B_x\hat i+B_y\hat j$, adding them simply means adding the like components,

which is exactly the analytical method of vector addition covered in the next article — resolution is the tool that makes that method possible.

Board level - resolving a force at a given angle

A force of magnitude 50 N acts at an angle of $37^\circ$ above the horizontal. Resolve it into horizontal and vertical components. (Use $\sin 37^\circ \approx 0.6$, $\cos 37^\circ \approx 0.8$.)

Step 1. Choose axes matching the problem: horizontal as the $x$-axis, vertical as the $y$-axis, with $\theta = 37^\circ$ measured from the horizontal.

Step 2. Apply $F_x = F\cos\theta$: $F_x = 50\times 0.8 = 40\ \text{N}$.

Step 3. Apply $F_y = F\sin\theta$: $F_y = 50\times 0.6 = 30\ \text{N}$.

Step 4. Write the component form and verify: $\vec F = 40\hat i + 30\hat j\ \text{N}$, and $\sqrt{40^2+30^2} = \sqrt{1600+900} = \sqrt{2500} = 50\ \text{N}$, matching the given magnitude exactly.

The horizontal component is 40 N and the vertical component is 30 N.

Competitive level - recovering magnitude and direction, with correct quadrant

A displacement vector has components $\Delta x = -3\ \text{m}$ and $\Delta y = 4\ \text{m}$. Find its magnitude and the angle it makes with the positive $x$-axis.

Step 1. Apply the magnitude formula: $|\Delta\vec r| = \sqrt{(-3)^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5\ \text{m}$.

Step 2. Find the reference angle using magnitudes only, to avoid sign confusion: $\tan\theta_{ref} = |4/(-3)| = 4/3 \implies \theta_{ref} \approx 53.1^\circ$.

Step 3. Fix the quadrant from the signs: $\Delta x$ is negative and $\Delta y$ is positive, so the vector points up and to the left — the second quadrant.

Step 4. Convert the reference angle to the true angle measured counter-clockwise from the positive $x$-axis, for the second quadrant that means $\theta = 180^\circ - \theta_{ref} = 180^\circ - 53.1^\circ = 126.9^\circ$.

The displacement is 5 m at 126.9° from the positive x-axis. Timing note: in an exam, budget about 30 seconds to compute the reference angle with $\tan^{-1}$ of the magnitude ratio, then a further 10 seconds just to look at the two signs and place the angle in the right quadrant — skipping that second step is the single most common way this question loses marks, even when the arithmetic is otherwise perfect.

• Applying $\theta = \tan^{-1}(A_y/A_x)$ mechanically without checking the quadrant from the signs of $A_x$ and $A_y$ — a calculator's $\tan^{-1}$ only ever returns an angle between $-90^\circ$ and $90^\circ$, so answers in the 2nd or 3rd quadrant need a manual $180^\circ$ correction, and answers in the 4th quadrant need a $360^\circ$ correction if angles are being measured as positive values.
• Swapping sine and cosine — remember $A_x = A\cos\theta$ (the side adjacent to the angle) and $A_y = A\sin\theta$ (the side opposite the angle), always relative to the axis $\theta$ is measured from.
• Adding magnitudes instead of using Pythagoras when recovering the resultant magnitude — a frequent slip is writing $A = A_x + A_y$ instead of the correct $A=\sqrt{A_x^2+A_y^2}$.
• Mixing degrees and radians mid-calculation, especially when a calculator's mode has been left on the wrong setting from a previous question — always confirm the mode before evaluating $\sin$, $\cos$ or $\tan^{-1}$.
• Choosing axes that fight the geometry of the problem instead of aligning with it — any pair of perpendicular axes is mathematically valid, but axes chosen along an incline or along a projectile's initial velocity usually turn a two-line calculation into a much longer, error-prone one.

• Resolve a given vector into rectangular components along chosen perpendicular axes, using $A_x=A\cos\theta$ and $A_y=A\sin\theta$.
• Write any 2-D vector in unit-vector component form $A_x\hat i+A_y\hat j$, and explain what $A_x\hat i$ and $A_y\hat j$ physically represent.
• Recover a vector's magnitude and direction from its components using $A=\sqrt{A_x^2+A_y^2}$ and $\theta=\tan^{-1}(A_y/A_x)$, fixing the quadrant correctly from the signs of $A_x$ and $A_y$.
• Verify a resolution by checking that $\sqrt{A_x^2+A_y^2}$ reproduces the original magnitude.
• Add two vectors by adding their components separately, anticipating the analytical method of the next article.
• Justify why choosing axes aligned with a problem's geometry (an incline, an initial velocity direction) simplifies the resulting algebra.

Ask the tutor: Not sure which quadrant your resolved vector falls in, or how to pick smart axes for a tricky incline or projectile diagram — ask the AI tutor to walk through your specific vector step by step.