Resolution of Vectors — Rectangular Components (NCERT 3.5)
1. What Is Resolution of a Vector?
Resolution is the reverse of addition: instead of combining two perpendicular vectors into one resultant, we split one vector into two perpendicular parts (conventionally along the x and y axes) that reproduce it exactly. The two parts are called the rectangular components of the vector. If a vector A of magnitude A makes an angle θ with the +x axis, its components are Ax = A cos θ and Ay = A sin θ, and the vector is reconstructed as A = Axî + Ayĵ. Geometrically, A is the diagonal of the rectangle whose sides are Ax and Ay.
Why bother? Because components turn vector problems into arithmetic. Once every vector in a problem is written in the î–ĵ language, adding vectors, adding forces, or writing Newton's second law becomes adding numbers column-wise. This one skill powers projectile motion (separate horizontal and vertical worlds), inclined planes, and every free-body diagram you will ever draw in Laws of Motion. NTA builds roughly half of this chapter's numericals on a clean resolution followed by nothing more than Pythagoras and a tangent.
2. Complete Theory: Components, Unit Vectors, Reconstruction
Unit vectors as direction carriers. The symbols î, ĵ, k̂ are unit vectors along +x, +y, +z of a right-handed system; each has magnitude 1 and no dimension. Any vector becomes a triple of numbers plus standard directions: A = Axî + Ayĵ + Ayk̂ in 3-D (here we work mainly in the plane). Because î·î = 1 and î·ĵ = 0, the component numbers extract physical information effortlessly — this is the machinery the dot and cross product page exploits.
Reconstruction (the inverse problem). Given components, the magnitude follows from Pythagoras, A = √(Ax² + Ay²), and the direction from tan θ = Ay/Ax. The catch is the quadrant: the arctangent returns a reference angle only, so the signs of Ax and Ay must fix the true quadrant — quadrant I (+, +), II (−, +), III (−, −), IV (+, −). Resolution in the "bad" direction gives zero: the component of A perpendicular to A itself is A cos 90° = 0, a fact NTA uses in "what is the component of 6 î along ĵ?" style questions (answer: zero).
Choosing smart axes. Axes are your choice, and a smart choice minimises trigonometry: align one axis with the motion or with the surface. On an incline, resolving weight into "along the slope" (mg sin θ) and "into the slope" (mg cos θ) kills two components to zero and is the single most reused resolution in all of mechanics.
3. Visualising Components
Figure 3.3 — Resolving A = 50 units at θ = 37°
The orange dashed lines drop from the arrowhead to the axes; they are the components, and A is their rectangle's diagonal.
Exam read-out: the 3-4-5 triangle again: 50 → (40, 30). Check by Pythagoras: 40² + 30² = 1600 + 900 = 2500 = 50². Whenever NTA quotes 37° or 53°, expect integer components — that is exactly why those angles are chosen.
4. Solved Examples
Example 1 — Force at 37° above the horizontal
A rope pulls a crate with F = 50 N at 37° above the horizontal. Find the horizontal and vertical components and their units.
Solution (step by step): Fx = F cos θ = 50 × cos 37° = 50 × 0.8 = 40 N (horizontal); Fy = F sin θ = 50 × 0.6 = 30 N (vertical). Physics meaning: only the 40 N part drags the crate forward; the 30 N part fights gravity (reducing the normal reaction). Units survive resolution — components carry newtons, and the unit vectors carry directions, which is why  itself is dimensionless.
Example 2 — The quadrant-II case, done correctly
A vector of magnitude 10 makes 120° with the +x axis. Find its components and verify the reconstruction.
Solution: Ax = 10 cos 120° = 10 × (−0.5) = −5; Ay = 10 sin 120° = 10 × (√3/2) ≈ +8.66. Signs (−, +) correctly place the arrow in quadrant II. Reconstruction: √((−5)² + (8.66)²) = √(25 + 75) = √100 = 10 ✓. Note how the calculator's tan⁻¹(8.66/−5) = tan⁻¹(−1.732) = −60° would have misdirected you by 180° — the sign check rescues the answer, giving 180° − 60° = 120°.
5. Practice Questions
Q1. A body is displaced 10 m along a direction 53° above the horizontal. Find the horizontal and vertical displacements.
ANSWER: Δx = 10 cos 53° = 10 × 0.6 = 6 m; Δy = 10 sin 53° = 10 × 0.8 = 8 m — the 6-8-10 triangle. Resultant check: √(36 + 64) = 10 ✓.
Q2. Write the unit vector along B = 4î − 3ĵ + 12k̂.
ANSWER: |B| = √(16 + 9 + 144) = √169 = 13, so B̂ = (4î − 3ĵ + 12k̂)/13. In 3-D the same Pythagoras extends to three squares under the root.
Q3 (MCQ). The component of the vector 6î along the y-axis is: (a) 6 (b) 0 (c) 6√2 (d) 3
ANSWER: (b) 0 — î is perpendicular to ĵ, and the component along a perpendicular direction is A cos 90° = 0.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| Ax = A cos θ; Ay = A sin θ | θ measured anticlockwise from +x; signs follow quadrant |
| A = Axî + Ayĵ | Reconstruction; A is the components' rectangle diagonal |
| A = √(Ax² + Ay²) | Pythagoras; extends to +Az² in 3-D |
| tan θ = Ay/Ax | Reference angle only — fix quadrant from component signs |
| Component ⊥ to A is zero | A cos 90° = 0; e.g., component of 6î along ĵ is 0 |
| Smart axes | Align an axis with the motion/surface to kill a component |
7. Frequently Asked Questions
What is the resolution of a vector into rectangular components?
Resolution replaces one vector by two perpendicular vectors (usually along the x and y axes) that add up to the original. If A makes an angle θ with the +x axis, the components are A_x = A cos θ along x and A_y = A sin θ along y, and A = A_x î + A_y ĵ. The original vector is exactly the diagonal of the rectangle formed by its components. Resolution is the reverse of addition by the parallelogram law with θ = 90°.
How do you find the magnitude and direction of a vector from its components?
The magnitude is A = √(A_x² + A_y²) by the Pythagorean theorem. The direction follows from tan θ = A_y/A_x, but the arctangent returns only a reference angle between −90° and +90°, so you must place the angle in the correct quadrant by inspecting the signs of A_x and A_y: both positive → quadrant I, A_x negative → quadrant II, A_y negative → quadrant III or IV.
Why must signs be checked when using tan θ = A_y/A_x?
Because tan θ repeats for angles separated by 180°: tan 30° and tan 210° are equal, yet the vectors point in opposite directions. A component pair A_x = −4, A_y = 3 gives tan θ = −0.75 and the calculator reports −36.9°, but the vector actually lies in quadrant II at 143.1° from +x. Checking the signs of the components against a quick sketch is the only reliable way to fix the quadrant.
What are î, ĵ, k̂ and why are unit vectors useful?
î, ĵ and k̂ are unit vectors (magnitude 1, dimensionless) pointing along the +x, +y and +z axes of a right-handed system. They act as standard direction carriers: any vector can be written as A = A_x î + A_y ĵ + A_z k̂, where the numbers A_x, A_y, A_z carry the physics and the î ĵ k̂ carry the geometry. Adding vectors then reduces to adding numbers component by component, which is how the analytical method of vector addition works.
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