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Vector Algebra Exercise Solutions | KalviGavani

Samacheer Kalvi 11th Business Maths Vector Algebra exercise solutions — scalars, vectors, addition, scalar product, vector product, direction cosines, worked examples, MCQs, full answers.

Updated 15 June 202617 min read

How to Use This Page

This page contains complete exercise solutions for Chapter 6 — Vector Algebra from the Samacheer Kalvi 11th Business Maths textbook (Tamil Nadu State Board, English Medium, Commerce stream).

Work through each section in order:

1. Key Concepts — quick revision tables before you start solving. 2. MCQs, Fill in the Blanks, True or False, Match the Following — one-mark practice. 3. Answer Briefly — two-mark short answers. 4. Answer in Detail — five-mark problems with fully worked numerical examples. 5. Common Mistakes and Exam Checklist — final review before your exam.

Every answer follows the TN Board marking scheme. Code blocks show step-by-step calculations so you can follow the working.


Key Concepts (Quick Recap)

Scalar vs Vector

PropertyScalarVector
DefinitionHas only magnitudeHas both magnitude and direction
ExamplesMass, temperature, speedDisplacement, velocity, force
NotationOrdinary letter (m, t)Bold (a) or with arrow (a⃗)
AdditionSimple arithmeticTriangle or parallelogram law
RepresentationSingle numberComponents along axes

Types of Vectors

TypeDefinition
Zero vector (0⃗)Magnitude = 0; no specific direction
Unit vectorMagnitude = 1; denoted â = a/a
Equal vectorsSame magnitude and same direction
Negative vectorSame magnitude, opposite direction
Collinear vectorsParallel to the same line (may differ in magnitude)
Coplanar vectorsLie in the same plane
Position vectorVector from origin O to point P; OP⃗ = xî + yĵ + zk̂

Vector Addition

LawStatement
Triangle lawIf two sides of a triangle represent two vectors, the third side (from tail of first to head of second) gives the resultant: a⃗ + b⃗
Parallelogram lawIf two vectors are drawn from the same point as adjacent sides of a parallelogram, the diagonal from that point gives the resultant

Scalar Multiplication

If k is a scalar and a⃗ = a₁î + a₂ĵ + a₃k̂, then k · a⃗ = (k a₁)î + (k a₂)ĵ + (k a₃)k̂ |k · a⃗| = |k| · |a⃗| If k > 0 → same direction If k < 0 → opposite direction

Magnitude of a Vector

For a⃗ = a₁î + a₂ĵ + a₃k̂: |a⃗| = √(a₁² + a₂² + a₃²)

Dot (Scalar) Product

a⃗ · b⃗ = |a⃗||b⃗| cos θ Component form: a⃗ · b⃗ = a₁b₁ + a₂b₂ + a₃b₃ Properties: a⃗ · a⃗ = |a⃗|² a⃗ · b⃗ = 0 ⟺ a⃗ ⊥ b⃗ î · î = ĵ · ĵ = k̂ · k̂ = 1 î · ĵ = ĵ · k̂ = k̂ · î = 0

Cross (Vector) Product

|a⃗ × b⃗| = |a⃗||b⃗| sin θ Determinant form: | î ĵ k̂ | a⃗ × b⃗ = | a₁ a₂ a₃ | | b₁ b₂ b₃ | = î(a₂b₃ − a₃b₂) − ĵ(a₁b₃ − a₃b₁) + k̂(a₁b₂ − a₂b₁) Properties: a⃗ × a⃗ = 0⃗ a⃗ × b⃗ = −(b⃗ × a⃗) a⃗ × b⃗ = 0⃗ ⟺ a⃗ ∥ b⃗ î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ

Direction Cosines and Ratios

If a⃗ makes angles α, β, γ with x, y, z axes: Direction cosines: l = cos α, m = cos β, n = cos γ Fundamental relation: l² + m² + n² = 1 Direction ratios (a, b, c) relate to direction cosines: l = a/√(a²+b²+c²), m = b/√(a²+b²+c²), n = c/√(a²+b²+c²)

Projection of a Vector

Projection of a⃗ on b⃗ = (a⃗ · b⃗) / |b⃗| = |a⃗| cos θ

Angle Between Two Vectors

cos θ = (a⃗ · b⃗) / (|a⃗||b⃗|) θ = cos⁻¹ [(a⃗ · b⃗) / (|a⃗||b⃗|)]

I. Choose the Best Answer (MCQs)

1. If a⃗ = 2î + 3ĵ − k̂, thena⃗is:
|a⃗| = √(2² + 3² + (−1)²) = √(4 + 9 + 1) = √14
Answer(b) √14

2. If a⃗ · b⃗ = 0, then the angle between a⃗ and b⃗ is:

a⃗ · b⃗ = |a⃗||b⃗| cos θ = 0 Since |a⃗| ≠ 0 and |b⃗| ≠ 0 (assuming non-zero vectors), cos θ = 0 ⟹ θ = 90°
Answer(c) 90°

3. The direction cosines of the vector î + ĵ + k̂ are:

|a⃗| = √(1² + 1² + 1²) = √3 Direction cosines: l = 1/√3, m = 1/√3, n = 1/√3 Check: l² + m² + n² = 1/3 + 1/3 + 1/3 = 1 ✓
Answer(a) 1/√3, 1/√3, 1/√3

4. If a⃗ × b⃗ = 0⃗, then:

|a⃗ × b⃗| = |a⃗||b⃗| sin θ = 0 Since |a⃗| ≠ 0 and |b⃗| ≠ 0, sin θ = 0 ⟹ θ = 0° or 180° Vectors are parallel (or anti-parallel).
Answer(a) a⃗ and b⃗ are parallel

5. The unit vector in the direction of a⃗ = 3î − 4ĵ + 12k̂ is:

|a⃗| = √(9 + 16 + 144) = √169 = 13 â = a⃗/|a⃗| = (3/13)î − (4/13)ĵ + (12/13)k̂
Answer(d) (3/13)î − (4/13)ĵ + (12/13)k̂

6. î × ĵ equals:

From the cross product of unit vectors: î × ĵ = k̂
Answer(c) k̂

7. The projection of a⃗ on b⃗ is:

Projection = (a⃗ · b⃗) / |b⃗|
**Answer: (b) (a⃗ · b⃗) /b⃗**

II. Fill in the Blanks

1. If a⃗ = a₁î + a₂ĵ + a₃k̂, thena⃗= √(a₁² + a₂² + a₃²)

2. The direction cosines l, m, n satisfy the relation l² + m² + n² = 1


3. If a⃗ ⊥ b⃗, then a⃗ · b⃗ = 0


4. a⃗ × b⃗ = −(b⃗ × a⃗) (anti-commutative property)


5. The unit vector in the direction of a⃗ is **a⃗/a⃗**

6. If a⃗ = 2î − ĵ + 3k̂, then 2a⃗ = 4î − 2ĵ + 6k̂

2a⃗ = 2(2î − ĵ + 3k̂) = 4î − 2ĵ + 6k̂

7. The dot product of parallel vectors equals **a⃗b⃗** (since cos 0° = 1)

8. The cross product of two parallel vectors is 0⃗ (zero vector)


9. If the position vector of point P is 3î − 2ĵ + 5k̂, then the coordinates of P are (3, −2, 5)


10. î · k̂ = 0 (since î and k̂ are perpendicular)


11. The magnitude of the vector −6î + 8ĵ is 10

|a⃗| = √((−6)² + 8² + 0²) = √(36 + 64) = √100 = 10

12. If l = 1/2 and m = 1/2 are direction cosines, then n = 1/√2

l² + m² + n² = 1 1/4 + 1/4 + n² = 1 n² = 1 − 1/2 = 1/2 n = 1/√2

III. True or False

1. "The dot product of two vectors is always a vector."

False. The dot product a⃗ · b⃗ =a⃗b⃗cos θ is always a scalar quantity, not a vector.

2. "If the direction cosines of a vector are l, m, n, then l² + m² + n² = 1."

True. This is the fundamental relation satisfied by direction cosines.


3. "Two vectors are perpendicular if their dot product equals 1."

False. Two non-zero vectors are perpendicular if their dot product equals 0, not 1. The correction: a⃗ · b⃗ = 0 ⟺ a⃗ ⊥ b⃗.


4. "The cross product of a vector with itself is the zero vector."

True. a⃗ × a⃗ = 0⃗ because the angle between a⃗ and itself is 0°, and sin 0° = 0.


5. "The magnitude of a vector is always negative."

False. The magnitudea⃗= √(a₁² + a₂² + a₃²) is always non-negative (≥ 0). It is zero only for the zero vector.

IV. Match the Following

Column AColumn B
1. a⃗ · b⃗ = 0(c) Vectors are perpendicular
2. a⃗ × b⃗ = 0⃗(e) Vectors are parallel
3. a⃗ · a⃗(a) \a⃗\²
4. l² + m² + n²(d) 1
5. a⃗/a⃗(f) Unit vector
6. î × k̂(b) −ĵ

Explanations:

1. a⃗ · b⃗ = 0 ⟹ cos θ = 0 ⟹ θ = 90°, so vectors are perpendicular. 2. a⃗ × b⃗ = 0⃗ ⟹ sin θ = 0 ⟹ θ = 0° or 180°, so vectors are parallel.

4. Direction cosines always satisfy l² + m² + n² = 1. 5. Dividing a vector by its magnitude gives a unit vector. 6. î × k̂ = −(k̂ × î) = −ĵ.

3. a⃗ · a⃗ =a⃗² cos 0° =a⃗².

V. Answer Briefly

1. Distinguish between scalar and vector quantities with two examples each.

ScalarVector
DefinitionHas only magnitudeHas both magnitude and direction
Example 1Mass (5 kg)Displacement (5 m east)
Example 2Temperature (37 °C)Velocity (20 m/s north)

A scalar is fully described by its magnitude alone. A vector requires both magnitude and direction for complete specification. Scalars follow ordinary algebra; vectors follow vector algebra (triangle/parallelogram laws).


2. Write the properties of the dot product.

1. Commutative: a⃗ · b⃗ = b⃗ · a⃗ 2. Distributive: a⃗ · (b⃗ + c⃗) = a⃗ · b⃗ + a⃗ · c⃗ 3. Scalar multiply: (k a⃗) · b⃗ = k(a⃗ · b⃗) 4. Self dot product: a⃗ · a⃗ = |a⃗|² 5. Perpendicularity: a⃗ · b⃗ = 0 ⟺ a⃗ ⊥ b⃗ (a⃗, b⃗ ≠ 0⃗) 6. Maximum value: a⃗ · b⃗ = |a⃗||b⃗| when θ = 0° (parallel, same direction) 7. Minimum value: a⃗ · b⃗ = −|a⃗||b⃗| when θ = 180° (opposite direction)

3. What are direction cosines and direction ratios? How are they related?

Direction cosines: If a vector a⃗ makes angles α, β, γ with the positive x, y, z axes respectively, then cos α, cos β, cos γ are called direction cosines, denoted l, m, n. They satisfy:

l² + m² + n² = 1

Direction ratios: Three numbers a, b, c proportional to the direction cosines are called direction ratios. The relation between them:

l = a / √(a² + b² + c²) m = b / √(a² + b² + c²) n = c / √(a² + b² + c²)

Direction ratios are not unique (any proportional set works), but direction cosines are unique (they satisfy l² + m² + n² = 1).


4. Find the direction cosines of the vector 2î + 3ĵ + 6k̂.

Given: a⃗ = 2î + 3ĵ + 6k̂ Step 1: Find magnitude |a⃗| = √(2² + 3² + 6²) = √(4 + 9 + 36) = √49 = 7 Step 2: Find direction cosines l = 2/7, m = 3/7, n = 6/7 Verification: l² + m² + n² = 4/49 + 9/49 + 36/49 = 49/49 = 1 ✓

Direction cosines: 2/7, 3/7, 6/7


VI. Answer in Detail

1. Vector Addition and Magnitude Calculation

Explain the triangle law and parallelogram law of vector addition. Find the magnitude and direction cosines of the resultant vector when a⃗ = î + 2ĵ + 3k̂ and b⃗ = 4î − 5ĵ + 6k̂.

Triangle Law of Vector Addition:

If two vectors are represented as two sides of a triangle taken in order, then the third side taken in the opposite order represents the resultant vector. If OA⃗ = a⃗ and AB⃗ = b⃗, then OB⃗ = a⃗ + b⃗.

Parallelogram Law of Vector Addition:

If two vectors are represented as adjacent sides of a parallelogram drawn from the same point, then the diagonal from that point represents the resultant vector. If OA⃗ = a⃗ and OB⃗ = b⃗, then OC⃗ (diagonal) = a⃗ + b⃗.

Worked Example:

Given: a⃗ = î + 2ĵ + 3k̂ b⃗ = 4î − 5ĵ + 6k̂ Step 1: Find the resultant R⃗ = a⃗ + b⃗ R⃗ = (1+4)î + (2−5)ĵ + (3+6)k̂ R⃗ = 5î − 3ĵ + 9k̂ Step 2: Find the magnitude of R⃗ |R⃗| = √(5² + (−3)² + 9²) = √(25 + 9 + 81) = √115 Step 3: Find direction cosines of R⃗ l = 5/√115, m = −3/√115, n = 9/√115 Verification: l² + m² + n² = 25/115 + 9/115 + 81/115 = 115/115 = 1 ✓ Result: |R⃗| = √115 Direction cosines: 5/√115, −3/√115, 9/√115

2. Dot Product and Angle Between Vectors

Explain the scalar (dot) product of two vectors and its properties. Find the angle between the vectors a⃗ = î + 2ĵ + 3k̂ and b⃗ = 2î − ĵ + 4k̂.

Definition: The dot product of two vectors a⃗ and b⃗ is defined as:

a⃗ · b⃗ = |a⃗||b⃗| cos θ Component form: a⃗ · b⃗ = a₁b₁ + a₂b₂ + a₃b₃

Properties:

1. Commutative: a⃗ · b⃗ = b⃗ · a⃗ 2. Distributive: a⃗ · (b⃗ + c⃗) = a⃗ · b⃗ + a⃗ · c⃗ 3. Scalar multiplier: (ka⃗) · b⃗ = k(a⃗ · b⃗) 4. Self product: a⃗ · a⃗ = |a⃗|² 5. Orthogonality: a⃗ · b⃗ = 0 ⟺ a⃗ ⊥ b⃗

Worked Example:

Given: a⃗ = î + 2ĵ + 3k̂ → a₁=1, a₂=2, a₃=3 b⃗ = 2î − ĵ + 4k̂ → b₁=2, b₂=−1, b₃=4 Step 1: Compute dot product a⃗ · b⃗ = (1)(2) + (2)(−1) + (3)(4) = 2 − 2 + 12 = 12 Step 2: Find |a⃗| |a⃗| = √(1² + 2² + 3²) = √(1 + 4 + 9) = √14 Step 3: Find |b⃗| |b⃗| = √(2² + (−1)² + 4²) = √(4 + 1 + 16) = √21 Step 4: Find cos θ cos θ = (a⃗ · b⃗) / (|a⃗||b⃗|) = 12 / (√14 × √21) = 12 / √294 = 12 / (√294) Simplify: √294 = √(49 × 6) = 7√6 cos θ = 12 / (7√6) = 12√6 / 42 = 2√6 / 7 Step 5: Find θ θ = cos⁻¹(2√6 / 7) Numerical approximation: 2√6 ≈ 2 × 2.449 = 4.899 cos θ ≈ 4.899 / 7 ≈ 0.6999 θ ≈ 45.6°

The angle between a⃗ and b⃗ is cos⁻¹(2√6/7) ≈ 45.6°


3. Cross Product and Its Applications

Explain the vector (cross) product of two vectors and its properties. Find the cross product and the area of the parallelogram formed by a⃗ = 3î − ĵ + 2k̂ and b⃗ = î + 2ĵ − k̂.

Definition: The cross product of two vectors a⃗ and b⃗ is defined as:

a⃗ × b⃗ = |a⃗||b⃗| sin θ n̂ where n̂ is the unit vector perpendicular to both a⃗ and b⃗, following the right-hand rule.

Determinant form:

| î ĵ k̂ | a⃗ × b⃗ = | a₁ a₂ a₃ | | b₁ b₂ b₃ |

Properties:

1. Anti-commutative: a⃗ × b⃗ = −(b⃗ × a⃗) 2. Distributive: a⃗ × (b⃗ + c⃗) = a⃗ × b⃗ + a⃗ × c⃗ 3. Self product: a⃗ × a⃗ = 0⃗ 4. Parallel vectors: a⃗ × b⃗ = 0⃗ ⟺ a⃗ ∥ b⃗ 5. Geometric meaning: |a⃗ × b⃗| = area of parallelogram 6. Triangle area: (1/2)|a⃗ × b⃗| = area of triangle

Worked Example:

Given: a⃗ = 3î − ĵ + 2k̂ → a₁=3, a₂=−1, a₃=2 b⃗ = î + 2ĵ − k̂ → b₁=1, b₂=2, b₃=−1 Step 1: Compute cross product using determinant | î ĵ k̂ | a⃗ × b⃗ = | 3 −1 2 | | 1 2 −1 | = î[(−1)(−1) − (2)(2)] − ĵ[(3)(−1) − (2)(1)] + k̂[(3)(2) − (−1)(1)] = î[1 − 4] − ĵ[−3 − 2] + k̂[6 + 1] = î(−3) − ĵ(−5) + k̂(7) = −3î + 5ĵ + 7k̂ Step 2: Find magnitude of cross product |a⃗ × b⃗| = √((−3)² + 5² + 7²) = √(9 + 25 + 49) = √83 Step 3: Area of parallelogram Area of parallelogram = |a⃗ × b⃗| = √83 square units Step 4: Verify — compute |a⃗| and |b⃗| |a⃗| = √(9 + 1 + 4) = √14 |b⃗| = √(1 + 4 + 1) = √6 Verification check: |a⃗ × b⃗| = |a⃗||b⃗| sin θ √83 = √14 × √6 × sin θ √83 = √84 × sin θ sin θ = √83/√84 ≈ 0.994 θ ≈ 83.8° (vectors are nearly perpendicular)

Cross product: −3î + 5ĵ + 7k̂; Area of parallelogram: √83 sq. units


Common Mistakes to Avoid

1. Confusing dot and cross products — The dot product gives a scalar; the cross product gives a vector. Never write a⃗ · b⃗ as a vector or a⃗ × b⃗ as a scalar.

2. Forgetting the negative sign in cross product determinant — When expanding the determinant, the ĵ component has a minus sign: −ĵ(a₁b₃ − a₃b₁). Many students forget this.

3. Applying dot product rules to cross product — The cross product is anti-commutative (a⃗ × b⃗ = −b⃗ × a⃗), not commutative. This is different from the dot product.

4. Direction cosines vs direction ratios — Direction cosines satisfy l² + m² + n² = 1; direction ratios do not. Do not assume a² + b² + c² = 1 for direction ratios.

5. Magnitude must be positive —a⃗= √(a₁² + a₂² + a₃²) is always ≥ 0. Never write a negative magnitude.
6. Unit vector formula — â = a⃗/a⃗. Some students write â =a⃗/a⃗ or forget to divide each component by the magnitude.
7. Angle between vectors — Use cos θ = (a⃗ · b⃗)/(a⃗b⃗), NOT sin θ with the dot product. The sine formula uses the cross product: sin θ =a⃗ × b⃗/(a⃗b⃗).

8. Zero vector vs scalar zero — The zero vector 0⃗ = 0î + 0ĵ + 0k̂ is a vector, not the scalar 0. Writing just "0" in vector equations can cause confusion.


Exam Checklist

Use this checklist before your exam. Tick each item once you are confident.

  • [ ] I can distinguish between scalars and vectors with examples
  • [ ] I know the types of vectors: zero, unit, equal, negative, collinear, coplanar
  • [ ] I can find the position vector of a point and write components
  • [ ] I can add vectors using both triangle and parallelogram laws
  • [ ] I can compute the magnitude |a⃗| = √(a₁² + a₂² + a₃²)
  • [ ] I can compute the dot product a⃗ · b⃗ = a₁b₁ + a₂b₂ + a₃b₃ and use it to find angles
  • [ ] I can compute the cross product using the determinant method
  • [ ] I know that |a⃗ × b⃗| gives the area of a parallelogram and (1/2)|a⃗ × b⃗| gives the area of a triangle
  • [ ] I can find direction cosines from a vector and verify l² + m² + n² = 1
  • [ ] I can find the projection of one vector on another: proj = (a⃗ · b⃗)/|b⃗|

Frequently Asked Questions

a · b = |a||b| cos θ = a₁b₁ + a₂b₂ + a₃b₃. It gives a scalar. If a · b = 0, the vectors are perpendicular.
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