Resolution of Vectors

Why we resolve vectors in the first place

One-dimensional kinematics is comfortable because every quantity lives on a single line — speeding up and slowing down are nothing but a positive or negative sign. The moment a particle moves in a plane, that comfort vanishes. A velocity vector pointing $37°$ above the horizontal cannot be added to a wind blowing east using ordinary arithmetic; you need a geometric rule. NCERT records two such rules earlier in the chapter — the parallelogram and triangle laws — but both are tedious for anything beyond two vectors. The way out is to choose two perpendicular axes, split each vector into components along those axes, and add the components separately. That trick is resolution of vectors, and NCERT introduces it in §3.5 as the bridge to every analytical method that follows.

The payoff is enormous. Once a vector $\mathbf{A}$ is replaced by the pair $(A_x, A_y)$, addition becomes addition of pairs, subtraction becomes subtraction of pairs, and motion in a plane reduces — as NCERT §3.7 states — to "two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions". Projectile motion, inclined-plane mechanics, and uniform circular motion all rest on this single decomposition. Get resolution right, and the rest of the chapter follows; get it wrong, and every downstream answer is wrong.

Think of resolution as the universal first move. Whenever a NEET problem hands you a force at an angle or a velocity tilted from the axes, the next step on your rough sheet should read "resolve into components". Newton's second law, once vector-valued, becomes two ordinary scalar equations, one for each axis. The two equations can be solved independently and the answers reassembled at the end. This idea is sometimes called the principle of independence of perpendicular motions; Galileo first articulated it for falling bodies, NCERT §3.9 invokes it for projectiles, and it powers every analytical kinematic result in the syllabus.

"It is convenient to resolve a general vector along the axes of a rectangular coordinate system using vectors of unit magnitude."

NCERT Class 11 Physics, §3.5

Unit vectors $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$ and the right-handed system

A unit vector is a vector of magnitude exactly one. It carries no unit and no dimension — it exists only to label a direction. NCERT writes unit vectors with a caret: $\hat{\mathbf{n}}$. The three Cartesian unit vectors are $\hat{\mathbf{i}}$ along the positive $x$-axis, $\hat{\mathbf{j}}$ along the positive $y$-axis and $\hat{\mathbf{k}}$ along the positive $z$-axis. Each has magnitude one:

The three are mutually perpendicular and obey the right-hand convention: if the fingers of the right hand curl from $\hat{\mathbf{i}}$ to $\hat{\mathbf{j}}$, the thumb points along $\hat{\mathbf{k}}$. NEET routinely tests this orientation through cross-product questions in rotational mechanics — once you accept the right-handed system here, the sign of $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$ stops being mysterious.

Multiplying a unit vector by a scalar produces a vector of arbitrary length but the same direction: $\lambda \hat{\mathbf{n}}$ has magnitude $|\lambda|$ and direction $\hat{\mathbf{n}}$ (or opposite, if $\lambda$ is negative). So any vector $\mathbf{A}$ can be written as a magnitude times a unit vector in the same direction:

This separation — magnitude in a number, direction in a hat — is exactly what allows resolution to work. We re-express the single direction $\hat{\mathbf{n}}$ as a combination of $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$, then collect the numbers. The choice of which directions to call $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ is yours, and a clever choice — for example, aligning the axes with an incline — can collapse a problem from four equations to one.

Rectangular components — $A_x = A\cos\theta$, $A_y = A\sin\theta$

Place a vector $\mathbf{A}$ in the $xy$-plane with its tail at the origin, pointing at an angle $\theta$ measured anticlockwise from the positive $x$-axis. Drop perpendiculars from the head of $\mathbf{A}$ to each axis. The foot on the $x$-axis is at distance $A\cos\theta$ from the origin; the foot on the $y$-axis is at distance $A\sin\theta$. These two numbers are the rectangular components of $\mathbf{A}$:

The vector itself is the sum of two perpendicular vector pieces — one along $\hat{\mathbf{i}}$, one along $\hat{\mathbf{j}}$:

NCERT §3.5 derives these from a parallelogram construction; the trigonometry is the same right-triangle exercise you have done since school. A few features worth fixing in mind. First, $\theta$ is always measured from the positive $x$-axis. If a question gives the angle from a different reference (the vertical, or the $y$-axis), you must convert before applying these identities. Second, $A_x$ and $A_y$ are signed — a vector in the second quadrant has $A_x < 0$ and $A_y > 0$, because $\cos\theta$ is negative for $90° < \theta < 180°$. Third, the sum of the squares recovers the magnitude:

That Pythagorean identity is the heart of the reconstruction rule we develop next. The mnemonic students find most useful is drawn from the right triangle: the side adjacent to the angle is the cosine piece; the side opposite to the angle is the sine piece. As long as you know which axis the angle is measured from, the labels follow. The sign of a component carries information that the magnitude does not — when you solve a problem and arrive at $A_x = -8$ m/s, the negative sign is telling you the vector points in the $-x$ direction; carrying the sign honestly through every step is the single most reliable habit in this section.

Reconstruction — magnitude, direction, and the quadrant trap

If a vector is presented through its components $(A_x, A_y)$, NCERT Eqs. (3.14) and (3.15) tell us how to get the magnitude back and how to recover the angle:

The magnitude formula is unambiguous. The angle formula is the source of the most common NEET trap. The function $\tan^{-1}$ on a calculator returns a single value between $-90°$ and $+90°$. But $\tan\theta$ has the same value at $\theta$ and at $\theta + 180°$ — so two different vectors, pointing in opposite directions, give the same $\tan^{-1}$ output. To choose the right one, inspect the signs of $A_x$ and $A_y$ and decide which quadrant the vector lies in. The fix is a two-step procedure: compute the reference angle $\alpha = \tan^{-1}|A_y/A_x|$, then add $0°$, $180°$ or $360°$ depending on the quadrant.

Worked example — resolving a force at $37°$

Worked example — velocity at $53°$ west of north

Three-dimensional resolution and direction cosines

A vector in space — not just in a plane — needs three components. NCERT Eq. (3.16b) writes this as

and the magnitude generalises directly:

The angles the vector makes with the three axes are no longer coplanar — they live in space, and NCERT calls them $\alpha$, $\beta$, $\gamma$ with the $x$-, $y$- and $z$-axes respectively. The components are then $A_x = A\cos\alpha$, $A_y = A\cos\beta$, $A_z = A\cos\gamma$. The three numbers $\cos\alpha$, $\cos\beta$, $\cos\gamma$ are called the direction cosines of $\mathbf{A}$.

Because $A_x^2 + A_y^2 + A_z^2 = A^2$, dividing through by $A^2$ gives a striking identity that every NEET-3D problem leans on:

This is the three-dimensional Pythagorean check. Squared direction cosines must sum to one; if they do not, the data is inconsistent. The body-diagonal of a cube makes equal angles with all three axes, each $\cos^{-1}(1/\sqrt{3}) \approx 54.7°$.

Resolution along non-perpendicular axes

NCERT §3.5 opens with a more general construction. Given any two non-collinear vectors $\mathbf{a}$ and $\mathbf{b}$ in a plane, any third vector $\mathbf{A}$ in the same plane can be written as $\mathbf{A} = \lambda\,\mathbf{a} + \mu\,\mathbf{b}$ for real numbers $\lambda$ and $\mu$. The rectangular case is just the special instance where $\mathbf{a} = \hat{\mathbf{i}}$ and $\mathbf{b} = \hat{\mathbf{j}}$ and the two basis vectors are perpendicular unit vectors. NEET rarely tests non-orthogonal resolution numerically, but knowing it exists explains why the rectangular split is a convenience-driven choice: perpendicular axes make $\cos$ and $\sin$ the only functions you need, and make the components independent.

Sanity checks

Three checks catch most errors. Magnitude: $\sqrt{A_x^2 + A_y^2}$ must equal $A$. Limiting angles: at $\theta = 0°$, $A_y$ vanishes; at $\theta = 90°$, $A_x$ vanishes. Units: components carry the same unit as the original vector.

The inclined-plane split — $mg\sin\theta$ along, $mg\cos\theta$ perpendicular

Of all the standard NEET set-ups, none uses resolution more relentlessly than a block on an inclined plane. A block of mass $m$ sits on a smooth incline tilted at angle $\theta$ to the horizontal. Gravity pulls vertically downward with force $mg$. The natural axes for this problem are not horizontal and vertical — they are along the incline and perpendicular to the incline, because the block can only move along the incline.

Resolving $mg$ along these tilted axes:

• The component along the incline (pointing downhill) is $mg\sin\theta$. This is the unbalanced force that accelerates the block down the slope.
• The component perpendicular to the incline (pressing into the surface) is $mg\cos\theta$. This is balanced by the normal reaction $N$, so $N = mg\cos\theta$.

The acceleration of the block down the smooth incline is therefore $a = g\sin\theta$ — independent of mass. With $g = 10$ m s$^{-2}$ and $\theta = 30°$, the slope acceleration is $5$ m s$^{-2}$. Add friction, and the equation gains a $-\mu mg\cos\theta$ term, giving $a = g(\sin\theta - \mu\cos\theta)$. The geometric reason the slope angle reappears between $mg$ and the perpendicular-to-incline direction: the slope is tilted $\theta$ above the horizontal, so the normal to the slope is tilted $\theta$ from the vertical, and since $mg$ acts vertically, the angle between $mg$ and the normal-to-incline direction is exactly $\theta$.

Resolution in projectile motion — a preview

Projectile motion is the cleanest demonstration of why resolution matters. A stone thrown with initial speed $u$ at angle $\theta$ above the horizontal becomes, after resolution, two one-dimensional problems: a horizontal motion at constant velocity $u\cos\theta$ (no horizontal force, no horizontal acceleration) and a vertical motion that starts at $u\sin\theta$ and decelerates at $g$. NEET 2022 turned this directly into a question — a ball thrown at $10$ m/s at $60°$ from the vertical has horizontal component $10\cos 30° = 5\sqrt{3}$ m/s, which is also its speed at the highest point. Students who misread $60°$ as the angle from the horizontal wrote $10\cos 60° = 5$ m/s and lost the mark. The sibling article on projectile motion develops the time of flight, range and maximum height from this resolved-velocity starting point.

Components are signed scalars, not vectors

A subtle but heavily tested fact: the symbol $A_x$ (and likewise $A_y$, $A_z$) denotes a real number, not a vector. It is a signed scalar. Only the product $A_x\,\hat{\mathbf{i}}$ — the component multiplied by its unit vector — is a vector. NCERT §3.5 is explicit: "Note that $A_x$ is itself not a vector, but $A_x\,\hat{\mathbf{i}}$ is a vector, and so is $A_y\,\hat{\mathbf{j}}$."

NCERT Exercise 3.4(f) — adding a component of a vector to the same vector — is flagged as meaningless. NCERT Exercise 3.5(b), which asks whether "each component of a vector is always a scalar", is answered true: only the unit-vector-attached object is itself a vector.

Adding vectors by components

Once two vectors are resolved, addition becomes addition of pairs. If $\mathbf{A} = A_x\hat{\mathbf{i}} + A_y\hat{\mathbf{j}}$ and $\mathbf{B} = B_x\hat{\mathbf{i}} + B_y\hat{\mathbf{j}}$, then $\mathbf{R} = (A_x + B_x)\hat{\mathbf{i}} + (A_y + B_y)\hat{\mathbf{j}}$. Subtraction works identically. The sibling article on vector addition works out the full machinery, including the law-of-cosines magnitude formula $R^2 = A^2 + B^2 + 2AB\cos\theta$.