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Class 11 · Tamil Nadu State Board

Vector Algebra Exercise Solutions | KalviGavani

TN Class 11 Maths Ch8 Vector Algebra exercise solutions — magnitude, dot product, cross product, direction cosines with worked examples.

Updated 15 June 202611 min read

1. How to Use This Page

This page contains complete exercise solutions for Chapter 8 — Vector Algebra of the Tamil Nadu State Board Class 11 Mathematics textbook. Attempt every problem yourself first, then compare your approach and final answer with the solutions below. Each section includes worked numerical examples with full step-by-step working.


2. Key Concepts (Quick Recap)

Types of Vectors

TypeDefinition
Zero vectorAll components are 0; has no direction
Unit vectorMagnitude = 1; â = a/a
Position vectorVector from origin O to point P
Co-initial vectorsVectors starting from the same point
Collinear vectorsVectors along the same or parallel lines
Equal vectorsSame magnitude and direction regardless of position

Key Formulas

Magnitude of a vector:

|a| = √(a₁² + a₂² + a₃²)

Unit vector:

â = a/|a|

Dot (Scalar) Product:

a · b = a₁b₁ + a₂b₂ + a₃b₃ a · b = |a||b|cos θ

Cross (Vector) Product:

a × b = |i j k | |a₁ a₂ a₃| |b₁ b₂ b₃| = i(a₂b₃ − a₃b₂) − j(a₁b₃ − a₃b₁) + k(a₁b₂ − a₂b₁)

Direction Cosines:

l = cos α = a₁/|a|, m = cos β = a₂/|a|, n = cos γ = a₃/|a| l² + m² + n² = 1

Angle between two vectors:

cos θ = (a · b)/(|a||b|)

Projection of a on b:

proj_b(a) = (a · b)/|b|

3. I. Choose the Best Answer (MCQs)

Q1. If a = (1, 2, 3) and b = (4, −8, 4), then the angle between them is:

(a) 0°   (b) 45°   (c) 90°   (d) 180°

Answer(c) 90°
Explanation
a · b = (1)(4) + (2)(−8) + (3)(4) = 4 − 16 + 12 = 0 Since dot product = 0, vectors are perpendicular. Angle = 90°

Q2. The magnitude of vector a = (3, −4, 12) is:

(a) 5   (b) 7   (c) 13   (d) 19

Answer(c) 13
Explanation
|a| = √(9 + 16 + 144) = √169 = 13

Q3. If a × b = 0, then:

(a) a ⊥ b   (b) a ∥ b   (c)a=b  (d) a = b
Answer(b) a ∥ b
ExplanationCross product is zero when vectors are parallel (or one is the zero vector).

Q4. The direction cosines of the vector (1, 1, 1) are:

(a) (1/3, 1/3, 1/3)   (b) (1/√3, 1/√3, 1/√3)   (c) (√3, √3, √3)   (d) (1, 1, 1)

Answer(b) (1/√3, 1/√3, 1/√3)
Explanation
|v| = √(1 + 1 + 1) = √3 Direction cosines = (1/√3, 1/√3, 1/√3) Check: (1/√3)² + (1/√3)² + (1/√3)² = 1/3 + 1/3 + 1/3 = 1 ✓

Q5. a · (b × c) is called:

(a) Cross product   (b) Dot product   (c) Scalar triple product   (d) Vector triple product

Answer(c) Scalar triple product
ExplanationThe dot product of a vector with a cross product gives a scalar, hence "scalar triple product" [a b c].

4. II. Fill in the Blanks

1. The magnitude of the unit vector is 1.

2. If a · b = 0 and a, b are non-zero vectors, then the angle between them is 90°.

3. The unit vector in the direction of (3, 4, 0) is (3/5, 4/5, 0).

4. If a = (2, −1, 3), thena= √14.

5. The direction cosines of the x-axis are (1, 0, 0).

6. a × b = −(b × a) — this property is called anti-commutative.

7. The cross product of two parallel vectors is the zero vector.

8. The projection of vector a on vector b is given by **(a · b)/b**.

9. If l, m, n are direction cosines, then l² + m² + n² = 1.

10. The scalar triple product [a b c] gives the volume of the parallelepiped formed by the three vectors.

11. For any vector a, a × a = 0 (zero vector).

12.a × b= **absin θ** where θ is the angle between a and b.

5. III. True or False (with Corrections)

1. Statement: The dot product of two vectors is always a scalar. Answer: True.

2. Statement: The cross product of two vectors is commutative: a × b = b × a. Answer: False. a × b = −(b × a). The cross product is anti-commutative.

3. Statement: Direction cosines of a vector can be greater than 1. Answer: False. Direction cosines are cosines of angles, so they lie in [−1, 1].

4. Statement: If two vectors are collinear, their cross product is the zero vector.

Answer: True. Parallel (collinear) vectors have sin θ = 0, soa × b= 0.

5. Statement: The dot product obeys the distributive law: a · (b + c) = a · b + a · c. Answer: True.


6. IV. Match the Following

Column AColumn B
1. a · b = 0 (non-zero vectors)a. Parallel vectors
2. a × b = 0 (non-zero vectors)b. Perpendicular vectors
3. a · a =a²c. Area of parallelogram
4.a × bd. Scalar triple product
5. a · (b × c)e. Self dot product
6. â = a/af. Unit vector
Answer1→b, 2→a, 3→e, 4→c, 5→d, 6→f

7. V. Answer Briefly (2-Mark Questions)

Q1. Find the unit vector in the direction of a = (2, −1, 2).

|a| = √(4 + 1 + 4) = √9 = 3 â = a/|a| = (2/3, −1/3, 2/3)

Q2. Find the dot product of a = (1, 2, −3) and b = (2, −1, 4).

a · b = (1)(2) + (2)(−1) + (−3)(4) = 2 − 2 − 12 = −12

Q3. Find the angle between a = (1, 0, 0) and b = (0, 1, 0).

a · b = 0 + 0 + 0 = 0 Since a · b = 0 and both are non-zero, angle = 90°

Q4. Find the direction cosines of the vector (2, 3, 6).

|r| = √(4 + 9 + 36) = √49 = 7 Direction cosines: l = 2/7, m = 3/7, n = 6/7 Check: (2/7)² + (3/7)² + (6/7)² = 4/49 + 9/49 + 36/49 = 49/49 = 1 ✓

8. VI. Answer in Detail (5-Mark Questions)

Q1. Find the angle between the vectors a = (3, −1, 2) and b = (1, 2, −1).

Step 1: Compute dot product a · b = (3)(1) + (−1)(2) + (2)(−1) = 3 − 2 − 2 = −1 Step 2: Compute magnitudes |a| = √(9 + 1 + 4) = √14 |b| = √(1 + 4 + 1) = √6 Step 3: Use the formula cos θ = (a · b)/(|a||b|) cos θ = (−1)/(√14 × √6) cos θ = −1/√84 cos θ = −1/(2√21) θ = cos⁻¹(−1/(2√21))

Q2. Find a × b where a = (2, 1, 3) and b = (1, −1, 2). Also finda × band verify thata × b² + (a · b)² =a²b².
Step 1: Compute a × b using determinant a × b = |i j k | |2 1 3 | |1 −1 2 | = i[(1)(2) − (3)(−1)] − j[(2)(2) − (3)(1)] + k[(2)(−1) − (1)(1)] = i[2 + 3] − j[4 − 3] + k[−2 − 1] = 5i − 1j − 3k a × b = (5, −1, −3) Step 2: Find |a × b| |a × b| = √(25 + 1 + 9) = √35 Step 3: Compute a · b a · b = (2)(1) + (1)(−1) + (3)(2) = 2 − 1 + 6 = 7 Step 4: Compute |a|² and |b|² |a|² = 4 + 1 + 9 = 14 |b|² = 1 + 1 + 4 = 6 Step 5: Verify identity |a × b|² + (a · b)² = 35 + 49 = 84 |a|²|b|² = 14 × 6 = 84 84 = 84 ✓ (Identity verified)

Q3. Find the projection of a = (4, −3, 5) on b = (2, 2, −1). Also find a vector perpendicular to both a and b.

Step 1: Find projection of a on b a · b = (4)(2) + (−3)(2) + (5)(−1) = 8 − 6 − 5 = −3 |b| = √(4 + 4 + 1) = √9 = 3 Projection = (a · b)/|b| = −3/3 = −1 Step 2: Find vector perpendicular to both (cross product) a × b = |i j k | |4 −3 5 | |2 2 −1 | = i[(−3)(−1) − (5)(2)] − j[(4)(−1) − (5)(2)] + k[(4)(2) − (−3)(2)] = i[3 − 10] − j[−4 − 10] + k[8 + 6] = (−7, 14, 14) This vector (−7, 14, 14) is perpendicular to both a and b. Step 3: Verify perpendicularity a · (a × b) = (4)(−7) + (−3)(14) + (5)(14) = −28 − 42 + 70 = 0 ✓ b · (a × b) = (2)(−7) + (2)(14) + (−1)(14) = −14 + 28 − 14 = 0 ✓

Q4. Find the direction cosines and direction ratios of the vector joining points A(1, 2, 3) and B(4, 6, 12).

Step 1: Find the vector AB AB = B − A = (4−1, 6−2, 12−3) = (3, 4, 9) Step 2: Direction ratios Direction ratios of AB = 3 : 4 : 9 Step 3: Find magnitude |AB| = √(9 + 16 + 81) = √106 Step 4: Direction cosines l = 3/√106 m = 4/√106 n = 9/√106 Verification: l² + m² + n² = 9/106 + 16/106 + 81/106 = 106/106 = 1 ✓

9. Common Mistakes to Avoid

1. Confusing dot and cross product: Dot product gives a scalar; cross product gives a vector. Never write a × b as a scalar.

2. Sign errors in cross product: The middle term in the determinant expansion has a negative sign: a × b = i(...) − j(...) + k(...). The minus before j is easy to forget.

3. Forgetting to normalize for unit vectors: The unit vector is a/a, not just a. Always divide by the magnitude.
4. Direction cosines vs. direction ratios: Direction ratios are the components; direction cosines are components divided by magnitude. They are different unlessv= 1.
5. Using sin θ for dot product: Remember a · b =abcos θ (cosine, not sine). The sine formula is fora × b.

6. Assuming commutativity of cross product: a × b ≠ b × a. In fact a × b = −(b × a). Order matters.


10. Exam Checklist

  • [ ] Can compute magnitude and unit vector for any 3D vector
  • [ ] Know dot product formula in both algebraic and geometric forms
  • [ ] Can find angle between two vectors using cos θ = (a · b)/(|a||b|)
  • [ ] Can compute cross product using the determinant method
  • [ ] Understand difference between dot product (scalar) and cross product (vector)
  • [ ] Can find direction cosines and verify l² + m² + n² = 1
  • [ ] Can find projection of one vector on another
  • [ ] Know conditions: a · b = 0 ⟹ perpendicular; a × b = 0 ⟹ parallel
  • [ ] Can verify identity |a × b|² + (a · b)² = |a|²|b|²
  • [ ] Remember cross product is anti-commutative and not associative

Frequently Asked Questions

If the dot product of two non-zero vectors is zero, the vectors are perpendicular (orthogonal) to each other.
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