JEE Main Mathematics · Chapter 16

Vector Algebra

Vector operations, dot and cross products, projections, scalar triple product and coplanarity.

NCERT Class 12JEE Main: 1–2 Qs per shift 7 min read

1. Chapter at a Glance

A vector is a quantity with both magnitude and direction, such as displacement, velocity or force. Vector algebra gives us the dot product (which measures how much two vectors point the same way) and the cross product (which gives a vector perpendicular to both, with magnitude equal to an area). This chapter is the engine of 3D geometry and also appears in physics. JEE Main asks one or two questions, mostly on dot/cross product, projection and scalar triple product.

What JEE asks most
  • Angle between vectors; projection of one vector on another.
  • Magnitude of a × b from |a|, |b| and a · b; area of triangles and parallelograms.
  • Coplanarity of three vectors and volume of a parallelepiped.
  • Vector triple product expansions.

2. Basics of Vectors

  • A vector is written a = a1i + a2j + a3k, with magnitude |a| = √(a1² + a2² + a3²).
  • Unit vector in the direction of a: â = a/|a|.
  • Position vector of a point P(x, y, z) is xi + yj + zk; the vector AB = (position vector of B) − (position vector of A).
  • Addition follows the triangle and parallelogram laws; it is commutative and associative.
  • Two vectors are collinear (parallel) if one is a scalar multiple of the other.
  • Section formula: the point dividing AB internally in the ratio m : n has position vector (mb + na)/(m + n).
aba + b Area = |a × b| = |a||b| sin θ θ
Parallelogram law of addition; the area of the parallelogram formed by a and b is |a × b|.

3. Scalar (Dot) Product

a · b = |a||b| cos θ = a1b1 + a2b2 + a3b3
  • a · b = 0 ⇔ a ⊥ b (for non-zero vectors). a · a = |a|².
  • i · i = j · j = k · k = 1 and i · j = j · k = k · i = 0.
  • Projection of a on b: (a · b)/|b|. Vector projection: ((a · b)/|b|²) b.
  • |a + b|² = |a|² + |b|² + 2a · b; so |a + b| = |a − b| ⇔ a ⊥ b.
  • Work done by force F in displacement d: W = F · d.

4. Vector (Cross) Product

a × b = |a||b| sin θ n̂,   with a × b = |i j k; a1 a2 a3; b1 b2 b3|
  • n̂ is perpendicular to both a and b (right-hand rule).
  • a × b = −(b × a) — the cross product is not commutative. a × a = 0; a × b = 0 ⇔ a ∥ b.
  • i × j = k, j × k = i, k × i = j (cyclic order).
  • Area of parallelogram with adjacent sides a, b: |a × b|. Area of triangle: ½|a × b|; with diagonals d1, d2: ½|d1 × d2|.
  • Lagrange's identity: |a × b|² + (a · b)² = |a|²|b|².
  • Unit vector perpendicular to a and b: ±(a × b)/|a × b|.

5. Triple Products

Scalar triple product

[a b c] = a · (b × c) = |a1 a2 a3; b1 b2 b3; c1 c2 c3|
  • It equals the volume of the parallelepiped with edges a, b, c (in absolute value); the tetrahedron has volume (1/6)|[a b c]|.
  • Three vectors are coplanar ⇔ [a b c] = 0.
  • Cyclic shifts do not change the value: [a b c] = [b c a] = [c a b]; swapping two vectors changes the sign.
  • [a a b] = 0; [a + b, b + c, c + a] = 2[a b c].

Vector triple product

a × (b × c) = (a · c) b − (a · b) c

Memory aid: "outer-far times middle minus outer-near times far". The result lies in the plane of b and c. In general (a × b) × c ≠ a × (b × c).

6. Formula Master Table

ResultFormula
Magnitude|a| = √(a1² + a2² + a3²)
Angle between vectorscos θ = a·b/(|a||b|)
Projection of a on ba·b/|b|
Area of parallelogram|a × b|
Area of triangle ABC½|AB × AC|
Lagrange's identity|a × b|² = |a|²|b|² − (a·b)²
Volume of parallelepiped|[a b c]|
Volume of tetrahedron(1/6)|[a b c]|
Coplanar condition[a b c] = 0
Vector triple producta × (b × c) = (a·c)b − (a·b)c

7. Mind Map

Vector Algebra
Basics
  • Magnitude, unit vector
  • Position vectors
  • Collinearity
Dot product
  • |a||b| cos θ
  • Perpendicular ⇔ 0
  • Projection, work
Cross product
  • |a||b| sin θ n̂
  • Area
  • Lagrange identity
Triple products
  • [a b c] = volume
  • Coplanar ⇔ 0
  • a × (b × c) expansion

8. Tips and Tricks

Shortcuts
  • Given |a|, |b| and a · b, find |a × b| instantly with Lagrange's identity — no need to find the angle.
  • If a + b + c = 0, then a × b = b × c = c × a; also squaring gives relations between dot products.
  • For coplanarity with an unknown λ, expand the 3 × 3 determinant and set it to 0.
  • To find a vector of given length along a direction, multiply the unit vector by that length.
Mistakes students make
  • Writing b × a = a × b — the order matters (they are negatives).
  • Forgetting the ½ for the area of a triangle.
  • Confusing the scalar projection (a·b/|b|) with the vector projection.

9. Important One-liners

  • The zero vector has no definite direction.
  • a · b is a scalar; a × b is a vector.
  • If a · b = 0 and a × b = 0, then a = 0 or b = 0.
  • The diagonals of a rhombus are perpendicular, which can be proved using (a + b) · (a − b) = |a|² − |b|² = 0.
  • The magnitude of a × b is maximum when a ⊥ b.
  • Three points A, B, C are collinear if AB × AC = 0.
  • Four points are coplanar if the scalar triple product of the three edge vectors from one of them is 0.

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Medium

If |a| = 3, |b| = 4 and a · b = 6, then |a × b| equals:

  1. 6√3
  2. 12
  3. 6
  4. 3√3
Show solution

|a × b|² = |a|²|b|² − (a · b)² = 144 − 36 = 108, so |a × b| = √108 = 6√3.

Answer: (A)

Q2Easy

The projection of a = i + 2j + 2k on b = 2i − j + 2k is:

  1. 4/3
  2. 4/9
  3. 4
  4. 2/3
Show solution

a · b = 2 − 2 + 4 = 4 and |b| = √(4 + 1 + 4) = 3. Projection = 4/3.

Answer: (A)

Q3Numerical

Find λ if the vectors i + j + k, i + 2j + 3k and 2i + 3j + λk are coplanar.

Show solution

[a b c] = |1 1 1; 1 2 3; 2 3 λ| = 1(2λ − 9) − 1(λ − 6) + 1(3 − 4) = λ − 4.

Coplanar ⇒ λ − 4 = 0 ⇒ λ = 4.

Answer: 4

Q4Medium

A unit vector perpendicular to both i + j and j + k is:

  1. (i − j + k)/√3
  2. (i + j + k)/√3
  3. (i − k)/√2
  4. (i + j − k)/√3
Show solution

(i + j) × (j + k) = |i j k; 1 1 0; 0 1 1| = i(1 − 0) − j(1 − 0) + k(1 − 0) = i − j + k, with magnitude √3.

Answer: (A)

Q5Numerical

Find the volume of the parallelepiped whose edges are i, 2j and 3k.

Show solution

[a b c] = |1 0 0; 0 2 0; 0 0 3| = 6. Volume = 6 cubic units.

Answer: 6

Q6Easy

If |a + b| = |a − b| for non-zero vectors a and b, then:

  1. a ∥ b
  2. a ⊥ b
  3. |a| = |b|
  4. a = b
Show solution

Squaring: |a|² + |b|² + 2a·b = |a|² + |b|² − 2a·b ⇒ 4a·b = 0 ⇒ a ⊥ b.

Answer: (B)

Q7Hard

If a, b, c are unit vectors with a + b + c = 0, then a · b + b · c + c · a equals:

  1. 3/2
  2. −3/2
  3. 0
  4. −3
Show solution

Square both sides: |a + b + c|² = 0 ⇒ |a|² + |b|² + |c|² + 2(a·b + b·c + c·a) = 0.

3 + 2S = 0 ⇒ S = −3/2.

Answer: (B)

Q8Hard

The value of i × (i × j) is:

  1. j
  2. −j
  3. k
  4. 0
Show solution

Using a × (b × c) = (a·c)b − (a·b)c with a = i, b = i, c = j: (i·j)i − (i·i)j = 0 − j = −j.

(Directly: i × j = k and i × k = −j.)

Answer: (B)

11. Quick Revision Summary

Remember before the exam
  • a · b = |a||b| cos θ = Σaibi; zero means perpendicular.
  • |a × b| = |a||b| sin θ = area of parallelogram; zero means parallel.
  • |a × b|² + (a·b)² = |a|²|b|².
  • [a b c] = volume; zero means coplanar.
  • a × (b × c) = (a·c)b − (a·b)c.
All Mathematics chaptersJEE Main Mock Tests