Product of
Two Vectors
The two most powerful vector operations — Scalar (Dot) Product and Vector (Cross) Product — with complete theory, formulas, properties, and solved examples.
The scalar product (dot product) of two vectors a and b is defined as a scalar equal to the product of their magnitudes and the cosine of the angle between them.
where θ is the angle between a and b (0° ≤ θ ≤ 180°)
Component form: if a = a₁î+a₂ĵ+a₃k̂, b = b₁î+b₂ĵ+b₃k̂:
a · b = a₁b₁ + a₂b₂ + a₃b₃
a = 2î + 3ĵ − k̂, b = î − 2ĵ + 3k̂
a·b = (2)(1) + (3)(−2) + (−1)(3) = 2 − 6 − 3 = −7
a = î + ĵ, b = ĵ + k̂. Find angle θ between them.
a·b = 0+1+0 = 1
|a| = √2, |b| = √2
cosθ = 1/(√2·√2) = 1/2 → θ = 60°
2. Distributive: a·(b+c) = a·b + a·c
3. Scalar factor: (λa)·b = λ(a·b) = a·(λb)
4. Self dot product: a·a = |a|²
5. Zero vector: 0·a= 0
î·î = ĵ·ĵ = k̂·k̂ = 1
(Each unit vector dotted with itself = 1)
î·ĵ = ĵ·k̂ = k̂·î = 0
(Perpendicular unit vectors dot to 0)
a ⊥ b ⟺ a·b = 0
(cosθ = cos90° = 0)
The result a·a = |a|² means |a| = √(a·a). This is used to find magnitudes. Also: |a + b|² = |a|² + 2(a·b) + |b|² — expanding like algebraic identity using dot product.
The scalar projection of b onto a is the component of b in the direction of a. The vector projection is that component as a vector.
= a·b / |a| = |b| cosθ
Vector projection of b on a:
= (a·b / |a|²) · a = (a·b / |a|) · â
Find projection of b = 2î + 3ĵ + 2k̂ on a = î + 2ĵ + 2k̂
a·b = 2+6+4 = 12
|a| = √(1+4+4) = 3
Scalar projection = 12/3 = 4
Projection problems are a staple 4-mark question in CBSE boards. Always write the formula first, then substitute — you get step marks even with a calculation error.
θ = cos⁻¹ [ (a₁b₁ + a₂b₂ + a₃b₃) / (|a| · |b|) ]
| Angle θ | cos θ | Dot Product | Relation |
|---|---|---|---|
| 0° | 1 | |a||b| | Same direction (parallel) |
| 90° | 0 | 0 | Perpendicular |
| 180° | −1 | −|a||b| | Antiparallel |
| 60° | 1/2 | |a||b|/2 | Acute angle |
| 120° | −1/2 | −|a||b|/2 | Obtuse angle |
The vector product (cross product) of two vectors a and b produces a new vector that is perpendicular to both a and b, with magnitude = |a| |b| sin θ.
where θ is the angle between a and b (0° ≤ θ ≤ 180°)
and n̂ is the unit vector ⊥ to both a and b (Right-Hand Rule)
Magnitude only: |a × b| = |a| |b| sin θ
For vectors in component form, the cross product is calculated using a 3×3 determinant:
a₁ a₂ a₃
b₁ b₂ b₃
Find a × b where a = 2î + 3ĵ − k̂ and b = î − 2ĵ + 3k̂
(3)(3) − (−1)(−2) = 9 − 2 = 7
−[(2)(3) − (−1)(1)] = −[6 + 1] = −7
(2)(−2) − (3)(1) = −4 − 3 = −7
a × b = 7î − 7ĵ − 7k̂
|a × b| = √(49+49+49) = 7√3
Cover the î column → multiply diagonally for î coefficient. Cover the ĵ column → multiply (but negate!). Cover the k̂ column → multiply for k̂. The sign pattern is + , − , + for î, ĵ, k̂.
2. Distributive: a × (b + c) = a×b + a×c
3. Scalar factor: (λa) × b = λ(a × b)
4. Self cross product: a × a = 0⃗ (any vector)
5. Zero vector: a × 0⃗ = 0⃗
î×ĵ = k̂ , ĵ×k̂ = î , k̂×î = ĵ
ĵ×î = −k̂ , k̂×ĵ = −î , î×k̂ = −ĵ
î×î = ĵ×ĵ = k̂×k̂ = 0⃗
a ∥ b ⟺ a × b = 0⃗
(sinθ = sin0° or sin180° = 0)
This is an alternative collinearity test using the cross product.
Cross product is NOT commutative: a×b ≠ b×a. In fact a×b = −b×a. Never swap without changing the sign. This is the most common error in board and JEE answers.
The magnitude of the cross product gives the area of the parallelogram formed by the two vectors as adjacent sides. Half of that is the area of a triangle.
Area = |a × b|
Area of triangle with sides a and b:
Area = (1/2)|a × b|
Unit normal vector to the plane of a and b:
n̂ = (a × b) / |a × b|
Find area of triangle with vertices A(1,1,1), B(2,3,4), C(3,2,5).
AB = B−A = î+2ĵ+3k̂
AC = C−A = 2î+ĵ+4k̂
AB×AC = î(8−3) − ĵ(4−6) + k̂(1−4) = 5î+2ĵ−3k̂
|AB×AC| = √(25+4+9) = √38
Area = (1/2)√38 = (√38)/2 sq. units
Area of triangle / parallelogram using cross product is a frequently asked 5-mark question. Always label your steps clearly and find AB and AC from the given vertices before computing the cross product.
| Property | Dot Product (a·b) | Cross Product (a×b) | ||||||
|---|---|---|---|---|---|---|---|---|
| Result | Scalar | Vector | ||||||
| Formula | |a||b|cosθ | |a||b|sinθ n̂ | ||||||
| Commutative | Yes: a·b = b·a | No: a×b = −b×a | ||||||
| Parallel vectors | a·b = ±|a||b| | a×b = 0⃗ | ||||||
| Perpendicular vectors | a·b = 0 | |a×b| = |a ||<.span style='color:var(--pink)'>b|
| î, ĵ cross | î·ĵ = 0 | î×ĵ = k̂ | Application | Angle, work, projection | Area, torque, normal | |
The scalar triple product of vectors a, b, c is defined as a·(b×c). It gives the volume of the parallelepiped formed by the three vectors.
Computed as 3×3 determinant:
|a₁ a₂ a₃|
|b₁ b₂ b₃|
|c₁ c₂ c₃|
Coplanarity: [a b c] = 0 ⟺ vectors are coplanar
Volume of parallelepiped = |[a b c]|
Volume of tetrahedron = (1/6)|[a b c]|
The scalar triple product is tested heavily in JEE Advanced. Key facts:
• [a b c] = [b c a] = [c a b] (cyclic permutation = same value)
• [a b c] = −[a c b] (swapping two = sign change)
• Vectors coplanar ⟺ box product = 0
Unit vector: â = a/|a|
Direction cosines: l²+m²+n² = 1
Section formula (internal): r = (mb+na)/(m+n)
Dot product: a·b = a₁b₁+a₂b₂+a₃b₃ = |a||b|cosθ
Angle: cosθ = (a·b)/(|a||b|)
Projection of b on a: (a·b)/|a|
Cross product mag: |a×b| = |a||b|sinθ
Parallelogram area: |a×b|
Triangle area: (1/2)|a×b|
Coplanarity: a·(b×c) = 0
Parallelepiped vol: |a·(b×c)|