Vector Addition — Analytical Method (NCERT 3.6)
1. What Is the Analytical Method?
Drawing arrows is fine for two vectors and rough estimates, but NTA wants numbers to two significant figures — and for that the geometry must become algebra. The analytical method converts the parallelogram construction of the graphical page into two compact formulas: one for the resultant's magnitude, R = √(A² + B² + 2AB cos θ), and one for its direction, tan α = B sin θ/(A + B cos θ). Every vector-addition MCQ in both exams is one of these two formulas plus a special-angle value, so the payoff for mastering this page is enormous relative to its length.
The method generalises beautifully. For more than two vectors, resolve each into components and add column-wise (shown in Example 2) — the same machinery the 2-D kinematics page uses on velocities and accelerations. In effect, analytical addition is resolution (previous topic) applied twice and Pythagoras applied once.
2. Complete Theory: Derivation, Direction, Special Cases
Derivation (know it — it is asked). Draw A and B tail-to-tail with angle θ between them and complete the parallelogram; the resultant R is the diagonal through the common point. Drop a perpendicular from the tip of B onto the line of A. The right-triangle legs are B sin θ (perpendicular to A) and A + B cos θ (along A). Pythagoras then gives R² = (A + B cos θ)² + (B sin θ)² = A² + 2AB cos θ + B²(cos²θ + sin²θ) = A² + B² + 2AB cos θ, using the identity cos²θ + sin²θ = 1 — the same identity Chapter 1's dimensional work treats as dimensionless bookkeeping. The direction follows from the same triangle: tan α = (opposite leg)/(adjacent leg) = B sin θ/(A + B cos θ), with α measured from A.
Special angles — memorise this row set. θ = 0°: R = A + B. θ = 90°: R = √(A² + B²) with tan α = B/A. θ = 180°: R = |A − B|. Equal vectors (A = B) at angle θ: R = 2A cos(θ/2) along the bisector — giving R = √3A at 60°, √2A at 90°, A at 120°, and 0 at 180°. Subtraction changes only the sign of the cosine term: |A − B| = √(A² + B² − 2AB cos θ), since subtracting is adding the antiparallel clone of B.
Angle bookkeeping that saves marks. The θ inside both formulas is the angle between the vectors' tails when drawn from one point. If a problem quotes "the angle between A and the extension of −B" or an interior angle of some polygon, first redraw the tails together. This single re-drawing habit eliminates the most common analytical-method error reported in exam reviews.
3. Visualising the Parallelogram Computation
Figure 3.4 — Where R² = A² + B² + 2AB cos θ comes from
The dashed legs are the two sides of the derivation's right triangle; the green diagonal is R.
Exam read-out: with A = B = 5 and θ = 60°, the legs are 5 + 2.5 = 7.5 and 5 × (√3/2) ≈ 4.33, so R = √(56.25 + 18.75) = √75 = 5√3 ≈ 8.66 — equal-vector shortcut 2A cos(θ/2) confirmed numerically.
4. Solved Examples
Example 1 — Equal forces at 60°
Two forces of 5 N each act at 60° to each other. Find the magnitude and direction of the resultant.
Solution (step by step): Since A = B = 5, use the equal-vector shortcut: R = 2A cos(θ/2) = 2 × 5 × cos 30° = 10 × 0.866 = 8.66 N. Direction: along the bisector, i.e., 30° from each force (equivalently tan α = 5 sin 60°/(5 + 5 cos 60°) = 4.33/7.5 = 0.577 → α = 30° ✓). Cross-check with the full formula: R = √(25 + 25 + 2·25·0.5) = √75 ≈ 8.66 ✓. A resultant larger than either force but smaller than their sum — exactly what the triangle inequality promised.
Example 2 — Component method for two arbitrary vectors
Add A = 3î + 4ĵ and B = 2î − ĵ (units arbitrary). Find R, its magnitude and direction.
Solution: Component addition: Rx = 3 + 2 = 5; Ry = 4 + (−1) = 3. So R = 5î + 3ĵ. Magnitude: |R| = √(25 + 9) = √34 ≈ 5.83. Direction: tan α = 3/5 = 0.6, α ≈ 31.0° with +x (both components positive → quadrant I, no correction needed). Notice no trigonometry of "the angle between A and B" was ever needed — the component method is angle-free, which is why it scales to any number of vectors.
5. Practice Questions
Q1. Forces of 3 N and 4 N act at 60° to each other. Find the resultant magnitude and its angle with the 3 N force.
ANSWER: R = √(9 + 16 + 2·3·4·0.5) = √(25 + 12) = √37 ≈ 6.08 N. tan α = 4 sin 60°/(3 + 4 cos 60°) = 3.464/5 = 0.693 → α ≈ 34.7° from the 3 N force.
Q2. Two forces of 10 N each act at 120°. Find the resultant.
ANSWER: R = 2 × 10 × cos 60° = 10 N along the bisector (60° from each force). Equal vectors at 120° reproduce either vector — the exam's favourite special case.
Q3 (MCQ). For what angle between A and B is |A + B| = |A − B|? (a) 0° (b) 45° (c) 90° (d) 180°
ANSWER: (c) 90°. Equating √(A²+B²+2AB cos θ) and √(A²+B²−2AB cos θ) forces cos θ = 0. This also means A·B = 0 — see the products page.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| R = √(A² + B² + 2AB cos θ) | θ between the tails; law of cosines |
| tan α = B sin θ/(A + B cos θ) | α from A; equal vectors → α = θ/2 |
| |A − B| = √(A² + B² − 2AB cos θ) | Subtraction diagonal; add −B |
| Equal vectors: R = 2A cos(θ/2) | Along bisector; A at 120°, √3A at 60°, √2A at 90° |
| Rx = ΣAx, Ry = ΣAy | Component method — scales to N vectors, angle-free |
| Rmin/Rmax = |A−B| / (A+B) | Range of possible resultants; all values attainable |
7. Frequently Asked Questions
Derive the magnitude of the resultant of two vectors using the analytical method.
Draw A and B from a common point at angle θ and complete the parallelogram; R is the diagonal through that point. Drop a perpendicular from the tip of B onto the extension of A. In the right triangle, the projection of B on A is B cos θ and the perpendicular part is B sin θ. Then R² = (A + B cos θ)² + (B sin θ)² = A² + 2AB cos θ + B², so R = √(A² + B² + 2AB cos θ). This is the law of cosines applied to the vector triangle.
What is the formula for the direction of the resultant?
If α is the angle the resultant makes with vector A, then tan α = B sin θ / (A + B cos θ). The numerator is the total component perpendicular to A, and the denominator is the total component along A. For equal vectors (A = B), α = θ/2 — the resultant bisects the angle between them, as the formula reduces to tan α = sin θ/(1 + cos θ) = tan(θ/2).
What is the resultant of two equal vectors acting at an angle θ?
For A = B, the general formula gives R = √(2A² + 2A² cos θ) = 2A cos(θ/2), directed along the bisector of the angle between them. Useful checks: θ = 0° gives R = 2A; θ = 60° gives R = √3 A; θ = 90° gives R = √2 A; θ = 120° gives R = A; and θ = 180° gives R = 0. The θ = 120° case (two equal vectors whose resultant equals either vector) is a recurring NTA MCQ.
How do you add more than two vectors analytically?
Use the component method: resolve every vector into x and y components, add all x components to get R_x = ΣA_x, add all y components to get R_y = ΣA_y, and then combine: R = √(R_x² + R_y²) with direction from tan α = R_y/R_x (quadrant checked from the signs of R_x and R_y). This method extends to any number of vectors and is exactly how the kinematics pages of this chapter handle two-dimensional motion.
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