Scalar and Vector Products — Dot & Cross (Syllabus Supplement)
1. What Are the Dot and Cross Products?
Two vectors can be "multiplied" in exactly two useful ways, and each answers a different geometric question. The scalar (dot) product A·B = AB cos θ asks: how much of B lies along A? — it is one magnitude times the projection of the other, and the output is a plain number. The vector (cross) product A×B of magnitude AB sin θ asks: how much do the two arrows "span an area"? — the output is a new vector perpendicular to both, direction fixed by the right-hand rule. Both formulas are explicit in the JEE Main and NEET syllabi for kinematics, and the rationalised NCERT defers the formal algebra to later chapters — which is precisely why this supplement page exists.
The division of labour across Class 11 physics is worth memorising on day one: work is a dot product (W = F·d, only the force along the displacement counts), while torque and angular momentum are cross products (τ = r×F, L = r×p, where only the part of r perpendicular to F matters). Every later chapter that "multiplies two vectors" is secretly this page.
2. Complete Theory: Component Rules and Identity Tests
Dot product in components. Using î·î = 1 and î·ĵ = 0 (unit vectors along different axes are perpendicular), the dot product becomes pure arithmetic: A·B = AxBx + AyBy (add +AzBz in 3-D). Three identities follow instantly: A·A = A² (magnitude squared), A·B = B·A (commutative), and A·B = 0 for nonzero vectors ⟺ A ⊥ B. The angle between vectors comes from cos θ = A·B/(AB) — Example 2 works one fully.
Cross product in components. With the cyclic right-handed relations î×ĵ = k̂, ĵ×k̂ = î, k̂×î = ĵ (and sign-flip when the order reverses), a 2-D cross product collapses to one term: A×B = (AxBy − AyBx)k̂. The scalar AxBy − AyBx is the signed area of the parallelogram spanned by the vectors. The antisymmetry B×A = −A×B is a property, not a mistake: reverse the rotation, flip the vector. |A×B| = 0 for nonzero vectors ⟺ parallel — and "parallel" includes antiparallel, since sin 180° also vanishes.
Geometric payoffs. The parallelogram spanned by A and B has area |A×B| = AB sin θ; the triangle formed by the two vectors has area ½|A×B|. The dot product, meanwhile, is the area-free "shadow" tool: A·B = (magnitude of A) × (component of B along A). One measures area, the other measures alignment — keeping these two pictures separate prevents the classic exam confusion of mixing cos θ into cross-product questions.
3. Visualising Both Products
Figure 3.5 — A·B is the shadow; |A×B| is the area
The orange dashed leg is B's projection on A (dot product, up to the factor A). The shaded parallelogram is the cross product's magnitude.
Exam read-out: same two vectors, two different questions. For A = 4 and B = 3 at 60°: dot = 4×3×0.5 = 6 (a number); cross magnitude = 4×3×0.866 ≈ 10.4 (a vector out of the page by the right-hand rule). No formula mixing survives this picture.
4. Solved Examples
Example 1 — Full workup of A = 2î + 3ĵ, B = 4î − ĵ
Find A·B, the angle between them, and A×B.
Solution (step by step): (i) A·B = (2)(4) + (3)(−1) = 8 − 3 = 5. (ii) |A| = √13 ≈ 3.606, |B| = √17 ≈ 4.123, so cos θ = 5/√221 ≈ 5/14.866 ≈ 0.336 → θ ≈ 70.4°. (iii) A×B = (AxBy − AyBx)k̂ = (2×(−1) − 3×4)k̂ = −14k̂, magnitude 14 — pointing into the page (negative ẑ), because rotating A toward B is a clockwise rotation. Consistency check: AB sin θ = 14.87 × sin 70.4° ≈ 14.87 × 0.942 ≈ 14.0 ✓.
Example 2 — Angle between î + ĵ and ĵ + k̂
Find the angle between A = î + ĵ and B = ĵ + k̂.
Solution: A·B = (1)(0) + (1)(1) + (0)(1) = 1. |A| = √2, |B| = √2. cos θ = 1/(√2·√2) = 1/2 → θ = 60°. The cross-check is instructive: A×B = î×ĵ + ĵ×ĵ + ĵ×k̂... computing component-wise gives (1, −1, 1) up to sign, |A×B| = √3 = AB sin θ = 2 sin 60° ✓. Unit-vector-free problems like this are stock NEET items — the entire solution is three lines if the component rules are automatic.
5. Practice Questions
Q1. A force F = (3î + 4ĵ) N displaces a body by d = 5î m. Find the work done by the force.
ANSWER: W = F·d = (3)(5) + (4)(0) = 15 J. The ĵ-component of the force does no work because the displacement is purely along x — the dot product's projection meaning made visible.
Q2 (MCQ). Which is always TRUE? (a) A·B = B·A (b) A×B = B×A (c) |A·B| = AB (d) |A×B| = AB for any θ
ANSWER: (a) only — the dot product is commutative; the cross product is anti-commutative. (c) holds only for parallel vectors (cos θ = 1); (d) only for perpendicular ones (sin θ = 1).
Q3. Find the area of the parallelogram whose adjacent sides are A = 3î and B = 4ĵ.
ANSWER: Area = |A×B| = |12k̂| = 12 square units (rectangle here, since A ⊥ B). The triangle with the same sides has area ½|A×B| = 6.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| A·B = AB cos θ = AxBx + AyBy | Scalar; commutative; A·A = A² |
| A·B = 0 ⟺ A ⊥ B | Nonzero vectors; cos 90° = 0 |
| cos θ = A·B/(AB) | Angle between any two vectors |
| |A×B| = AB sin θ; direction: right-hand rule | Vector ⊥ to both; anti-commutative |
| A×B = (AxBy − AyBx)k̂ | 2-D shortcut; signed parallelogram area |
| |A×B| = 0 ⟺ A ∥ B | Includes antiparallel; sin θ vanishes |
| î×ĵ = k̂, ĵ×k̂ = î, k̂×î = ĵ | Cyclic; reversed order flips sign |
| W = F·d; τ = r×F; L = r×p | Flagship uses later in the syllabus |
7. Frequently Asked Questions
What is the scalar (dot) product of two vectors and what does it mean geometrically?
The dot product A·B = AB cos θ is a scalar — the product of one vector's magnitude with the projection of the other on it. In components A·B = A_xB_x + A_yB_y (+A_zB_z in 3-D). It is commutative, A·B = B·A, and A·A = A². It equals zero exactly when the vectors are perpendicular (θ = 90°). Physically, work W = F·d is the flagship example: only the force component along the displacement does work.
What is the vector (cross) product and how is its direction found?
The cross product A×B is a vector of magnitude AB sin θ, perpendicular to both A and B, with direction given by the right-hand rule: rotate A toward B through the smaller angle and the thumb points along A×B. It is anti-commutative (B×A = −A×B) and zero for parallel vectors (θ = 0° or 180°). The cyclic relations î×ĵ = k̂, ĵ×k̂ = î, k̂×î = ĵ fix all sign questions, and in 2-D, A×B = (A_xB_y − A_yB_x)k̂.
When are two vectors perpendicular and when are they parallel, by product tests?
For nonzero vectors: A·B = 0 if and only if they are perpendicular, because cos 90° = 0; and |A×B| = 0 (equivalently A×B = 0) if and only if they are parallel, because sin 0° = sin 180° = 0. Combining both tests, two vectors are parallel when every component ratio matches, e.g., B = λA. NTA regularly hides 'perpendicularity' inside a dot product equal to zero and 'parallelism' inside a vanishing cross product.
What is the geometric meaning of |A × B|?
The magnitude |A×B| = AB sin θ equals the area of the parallelogram whose adjacent sides are A and B, and half of it, ½|A×B|, is the area of the triangle formed by the two vectors. For example, vectors 3î and 4ĵ form a rectangle of area |3î × 4ĵ| = |12k̂| = 12 square units. This is why cross products later compute torque (r×F) and angular momentum (r×p).
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