Vector Addition and Subtraction — Graphical Method (NCERT 3.3, 3.4)
1. What Is Graphical Vector Addition?
Because vectors carry direction, they cannot be added like grocery bills; the geometry of the arrows decides the sum. The graphical method turns addition into drawing: represent each vector as an arrow whose length (to scale) is the magnitude and whose orientation is the direction. Two equivalent constructions exist — the triangle (head-to-tail) law and the parallelogram law — and both produce the identical resultant. Mastering the drawing is not optional book-work: many NEET questions are answered in five seconds by a correct sketch, and the diagrams later generate the analytical formula of the analytical method page.
Subtraction needs no separate rule. To find A − B, reverse B to get −B (same length, opposite direction) and add: A − B = A + (−B). Since the angle between A and −B is 180° − θ (where θ is the angle between A and B), the subtraction diagonal is different from the addition diagonal — this single observation spawns a family of NTA MCQs on |A + B| versus |A − B|.
2. Complete Theory: Triangle Law, Parallelogram Law, Properties
Triangle law. To add B to A, place the tail of B at the head of A. The resultant R = A + B runs from the tail of A to the head of B, closing the triangle. The construction is order-insensitive: adding A to B by the same rule gives the same R — the algebraic statement is the commutative law A + B = B + A. For three vectors, the head-to-tail chain A → B → C closes with R from the first tail to the last head, and the associative law (A + B) + C = A + (B + C) guarantees any grouping works.
Parallelogram law. If two vectors are drawn from a common point as the two adjacent sides of a parallelogram, the resultant is the diagonal of that parallelogram passing through the common point. Its magnitude follows from the law of cosines, R = √(A² + B² + 2AB cos θ), where θ is the angle between the vectors (tails together). Two boundary cases anchor every estimate: θ = 0° (parallel) gives R = A + B, the maximum; θ = 180° (antiparallel) gives R = |A − B|, the minimum. For any intermediate angle, |A − B| ≤ R ≤ A + B.
Zero resultant. Two vectors cancel only if they are equal in magnitude and antiparallel. Three vectors can cancel with unequal magnitudes if they can form a closed triangle taken head-to-tail — each magnitude must then be smaller than the sum of the other two (triangle inequality). A four-vector closure requires a closed polygon; NTA tests exactly this "closed polygon ⇒ zero resultant" idea in statement questions.
3. Visualising the Two Laws
Figure 3.2 — Triangle law (left) and parallelogram law (right)
Follow the arrows head-to-tail; the green arrow is always the resultant closing the figure.
Triangle law: B at the head of A; R closes from tail of A to head of B
Parallelogram law: A, B from one point; R is the diagonal through it
Exam read-out: both panels encode R = A + B with the same green arrow. The dashed light arrows in the right panel complete the parallelogram; the other diagonal (not drawn) would be A − B. Whenever a question says "the two diagonals of the parallelogram are 10 and 6", translate instantly to |A + B| = 10, |A − B| = 6.
4. Solved Examples
Example 1 — Perpendicular forces, the 6-8-10 case
A force of 6 N acts due east and a force of 8 N acts due north on a body. Find the resultant by the parallelogram law.
Solution (step by step): The forces are perpendicular, θ = 90°, so cos θ = 0 and R = √(6² + 8²) = √(36 + 64) = √100 = 10 N. Direction: tan α = 8/6 = 4/3 → α ≈ 53.1° north of east (α measured from the 6 N force). This is the 3-4-5 triangle again — NTA recycles 6, 8, 10 and its 37°/53° siblings relentlessly, so the angle bank on the chapter hub pays for itself here.
Example 2 — Maximum, minimum and a target resultant
Two forces have magnitudes 7 N and 5 N. Find (i) the maximum and minimum possible resultants, (ii) whether a resultant of exactly 3 N is possible.
Solution: (i) Maximum at θ = 0°: R = 7 + 5 = 12 N; minimum at θ = 180°: R = |7 − 5| = 2 N. (ii) The resultant varies continuously from 2 N to 12 N as θ goes from 180° to 0°, and every value between the limits is attained — so yes, 3 N is possible (at some obtuse angle). The general statement "the resultant of two vectors can take any value between |A − B| and A + B" is a standard assertion–reason item; the justification is the continuous dependence of R on θ through cos θ.
5. Practice Questions
Q1. A walker goes 4 km east, then 3 km north. Using the triangle law, find the displacement's magnitude and direction.
ANSWER: R = √(4² + 3²) = 5 km, at tan⁻¹(3/4) ≈ 36.9° north of east (from the eastward leg). Distance walked 7 km, displacement 5 km — the Ch 2 inequality in action.
Q2 (MCQ). Two vectors of magnitudes 4 and 9 are added. Which resultant is impossible? (a) 5 (b) 9 (c) 13 (d) 14
ANSWER: (d) 14 — the resultant cannot exceed A + B = 13. The limits are |9 − 4| = 5 to 13, so 5, 9 and 13 are all attainable.
Q3. Vectors of 10 units and 10 units act at 60° to each other. Predict the resultant's magnitude graphically and verify: is it more, less, or equal to 10?
ANSWER: More than 10. Equal vectors at θ < 120° always give R > each vector; here the analytical check (next topic's method) gives R = 2 × 10 × cos 30° = 17.3 units along the bisector. Graphically the closing diagonal is visibly longer than either side.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| R = A + B (triangle law) | Head-to-tail; R from tail of A to head of B |
| A + B = B + A; (A+B)+C = A+(B+C) | Commutative and associative — order/grouping free |
| R = √(A²+B²+2AB cos θ) | Parallelogram diagonal; θ between tails |
| |A − B| = √(A²+B²−2AB cos θ) | Second diagonal; subtract via −B |
| Rmax = A+B (θ=0°); Rmin = |A−B| (θ=180°) | Every intermediate value is attainable |
| |A+B| = |A−B| ⟺ θ = 90° | MCQ favourite; also A·B = 0 there |
7. Frequently Asked Questions
State the triangle law of vector addition.
Place the tail of vector B at the head of vector A. The resultant R = A + B is drawn from the tail of A to the head of B, closing the triangle. The order does not matter: A + B = B + A (commutative law). Geometrically the two vectors and their resultant form the three sides of a triangle.
When is the resultant of two vectors maximum and when minimum?
The resultant magnitude R = √(A² + B² + 2AB cos θ) is maximum when the vectors are parallel (θ = 0°), giving R_max = A + B, and minimum when they are antiparallel (θ = 180°), giving R_min = |A − B|. For any other angle the resultant lies strictly between these two limits. For A = 7 and B = 5, R runs from 12 down to 2.
Can two vectors of unequal magnitudes add up to zero? Can three vectors do it?
Two vectors of unequal magnitudes can never add to zero, because the smallest possible resultant |A − B| is nonzero when A ≠ B. Three vectors CAN add to zero even with unequal magnitudes, provided they can be represented by the three sides of a triangle taken in order — the magnitudes must satisfy the triangle inequality, each being less than the sum of the other two.
How is subtraction of vectors performed graphically?
Subtraction is addition of the reversed vector: A − B = A + (−B). Reverse the direction of B (keeping its magnitude), then apply the triangle or parallelogram law exactly as for addition. The magnitude follows |A − B| = √(A² + B² − 2AB cos θ), where θ is the angle between A and B. Note that |A − B| is also the diagonal of the parallelogram formed by A and B, drawn from B's head region toward A's head.
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