Vectors for NDA

Vectors are one of the most reliable scoring chapters in NDA maths. The syllabus is short: vectors in two and three dimensions, magnitude and direction, unit and null vectors, addition and scalar multiplication, the dot and cross products, and applications to work done, moment of a force and geometry. The questions are mostly short computations, and almost all of them use five or six results.

Vectors also sit next to three-dimensional geometry. Direction cosines, the angle between lines and the distance formula appear in both, so time spent here pays twice.

Typical question types are:

• magnitude and unit vector along a given vector;
• the angle between two vectors, or a condition for them to be perpendicular or parallel;
• the value of a constant that makes vectors perpendicular or collinear;
• the projection of one vector on another;
• the area of a triangle or parallelogram;
• work done by a force, or moment of a force;
• identity-based questions: given |a|, |b| and one more fact, find |a + b| or a · b.

Basics

Write a vector as a = a₁i + a₂j + a₃k, where i, j and k are unit vectors along the axes.

• Magnitude: |a| = √(a₁² + a₂² + a₃²).
• Unit vector along a: a/|a|. A unit vector has magnitude 1; the null vector has magnitude 0 and no fixed direction.
• Position vector: for points A and B with position vectors a and b, AB = b − a (head minus tail).
• Section formula: the point dividing AB internally in the ratio m : n has position vector (mb + na)/(m + n). The midpoint is (a + b)/2.
• Collinear (parallel) vectors: a and b are parallel when b = λa, that is, when a₁/b₁ = a₂/b₂ = a₃/b₃.
• Direction cosines of a are a₁/|a|, a₂/|a|, a₃/|a|, and the sum of their squares is 1.

Dot (scalar) product

a · b = |a||b| cos θ = a₁b₁ + a₂b₂ + a₃b₃, where θ is the angle between them (0 ≤ θ ≤ π). The result is a number.

• Angle: cos θ = (a · b)/(|a||b|).
• Perpendicular: a · b = 0 (for non-zero vectors).
• Projection of a on b: (a · b)/|b|. Note which vector is divided out.
• i · i = j · j = k · k = 1 and i · j = j · k = k · i = 0.
• Work done by a constant force F over a displacement d is W = F · d.

Cross (vector) product

|a × b| = |a||b| sin θ. The direction of a × b is perpendicular to both a and b, given by the right-hand rule. The result is a vector.

In components, a × b is the determinant with rows (i, j, k), (a₁, a₂, a₃) and (b₁, b₂, b₃), which expands to (a₂b₃ − a₃b₂)i − (a₁b₃ − a₃b₁)j + (a₁b₂ − a₂b₁)k.

• i × j = k, j × k = i, k × i = j, and reversing the order changes the sign: j × i = −k.
• i × i = j × j = k × k = 0.
• a × b = −(b × a). The cross product is not commutative.
• Parallel vectors: a × b = 0.
• Area of a parallelogram with adjacent sides a and b = |a × b|. With diagonals d₁ and d₂, the area is ½|d₁ × d₂|.
• Area of a triangle ABC = ½|AB × AC|.
• Moment (torque) of a force F acting at a point with position vector r, about the origin, is r × F.

Identities that turn MCQs into one-liners

• a · a = |a|². Use it whenever you need to square a vector.
• |a + b|² = |a|² + |b|² + 2(a · b) gives the length of a sum.
• |a − b|² = |a|² + |b|² − 2(a · b) gives the length of a difference.
• (a + b) · (a − b) = |a|² − |b|². This settles "if (x − a) · (x + a) = 8" questions.
• |a × b|² + (a · b)² = |a|²|b|² links the two products (sin²θ + cos²θ = 1 in disguise).
• |a + b| = |a − b| exactly when a · b = 0, a quick perpendicularity test.

Scalar triple product (occasionally asked)

[a b c] = a · (b × c) equals the determinant of the three component rows. Its absolute value is the volume of the parallelepiped with edges a, b and c. Three vectors are coplanar exactly when [a b c] = 0. Also i · (j × k) = 1.

Worked NDA-style MCQs

Q1. The unit vector along 2i − j + 2k is:
(a) (2i − j + 2k)/9 (b) (2i − j + 2k)/3 (c) (2i − j + 2k)/√5 (d) 2i − j + 2k

|a| = √(4 + 1 + 4) = 3, so the unit vector is a/3. Answer: (b).

Q2. The angle between i + j and j + k is:
(a) 30° (b) 45° (c) 60° (d) 90°

The dot product is 0 + 1 + 0 = 1, and each magnitude is √2. cos θ = 1/2, so θ = 60°. Answer: (c).

Q3. The projection of a = 2i + 3j + 2k on b = i + 2j + k is:
(a) 10/√6 (b) 10/√17 (c) 10 (d) √6

a · b = 2 + 6 + 2 = 10 and |b| = √6. The projection is 10/√6. Option (b) divides by |a| instead. Answer: (a).

Q4. The area of the triangle with vertices A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) is:
(a) √21 (b) √21/2 (c) 21/2 (d) √14/2

AB = j + 2k and AC = i + 2j. The cross product AB × AC = (1 × 0 − 2 × 2)i − (0 × 0 − 2 × 1)j + (0 × 2 − 1 × 1)k = −4i + 2j − k, whose magnitude is √(16 + 4 + 1) = √21. The triangle's area is half of that. Answer: (b).

Q5. A force F = 2i + j − k moves a particle from (1, 2, 3) to (3, 4, 5). The work done is:
(a) 2 units (b) 4 units (c) 6 units (d) 8 units

The displacement is d = 2i + 2j + 2k, so W = F · d = 4 + 2 − 2 = 4. Answer: (b).

Q6. If |a| = 3, |b| = 4 and |a + b| = 5, then |a − b| is:
(a) 1 (b) 7 (c) 5 (d) √7

25 = 9 + 16 + 2a · b, so a · b = 0. Then |a − b|² = 9 + 16 − 0 = 25. Answer: (c).

Common mistakes

• Dividing by the wrong magnitude in a projection. The projection of a on b divides by |b|.
• Treating the cross product as commutative. b × a = −(a × b).
• Forgetting the ½ for a triangle, or using sides instead of diagonals in the parallelogram formula.
• Using the dot product for area or the cross product for work. Work is a scalar (dot); area and moment come from the cross product.
• Writing AB as a − b. It is always head minus tail: b − a.

Practice set

1. If |a| = 2, |b| = 3 and a · b = 3, the angle between a and b is: (a) 30° (b) 45° (c) 60° (d) 90°
2. The value of λ for which 2i + λj + k and i − 2j + 3k are perpendicular is: (a) 5/2 (b) −5/2 (c) 2 (d) 3/2
3. i · (j × k) + j · (k × i) + k · (i × j) equals: (a) 0 (b) 1 (c) 3 (d) −3
4. If |a × b| = 4 and a · b = 3, then |a||b| equals: (a) 7 (b) 12 (c) 5 (d) 25
5. The area of the parallelogram whose diagonals are 3i + j − 2k and i − 3j + 4k is: (a) 10√3 (b) 5√3 (c) 8√3 (d) 15
6. 2i − 3j + 4k and −4i + 6j + λk are collinear when λ is: (a) 8 (b) −8 (c) 2 (d) −2
7. The direction cosines of i + j + k are: (a) 1, 1, 1 (b) 1/3 each (c) 1/√3 each (d) √3 each
8. If a is a unit vector and (x − a) · (x + a) = 8, then |x| is: (a) 3 (b) 2√2 (c) 9 (d) √7

Answers:

1. (c). cos θ = 3/(2 × 3) = 1/2.
2. (a). 2 − 2λ + 3 = 0.
3. (c). Each term equals 1.
4. (c). 16 + 9 = |a|²|b|² = 25.
5. (b). The cross product is −2i − 14j − 10k, of magnitude √300 = 10√3; half of that is 5√3.
6. (b). The ratio −4/2 = 6/(−3) = −2, so λ = 4 × (−2) = −8.
7. (c). Each component 1 divided by the magnitude √3.
8. (a). |x|² − |a|² = 8, so |x|² = 9.

What to do next

• Write the identities list from memory and prove the |a + b|² line once by expanding (a + b) · (a + b).
• Practise the 3 × 3 cross-product expansion until it takes under 30 seconds; it decides the area questions.
• Carry on to three-dimensional geometry for lines, planes and direction ratios, and revise determinants in matrices and determinants.