Class 12 Mathematics Notes Chapter 4 (Vector Algebra) – Mathematics Part-II Book

Detailed Notes with MCQs of Chapter 4: Vector Algebra from your NCERT Part-II book. This is a fundamental chapter, not just for your board exams but also crucial for many government exams that include mathematics, especially those related to engineering or technical fields. We'll break down the key concepts you need to master.

Vector Algebra: Key Concepts for Government Exams

1. Scalars and Vectors

• Scalar: A quantity having only magnitude. Examples: Length, mass, time, speed, temperature, work, density.
• Vector: A quantity having both magnitude and direction. Examples: Displacement, velocity, acceleration, force, momentum, electric field.
• Representation: A directed line segment. Let A be the initial point and B be the terminal point. The vector is denoted as vec(AB) or often by a single bold letter like a or vec(a).
• Magnitude: The length of the vector. The magnitude of vec(AB) is denoted by |vec(AB)| or AB. The magnitude of vec(a) is denoted by |vec(a)| or simply 'a'. Magnitude is always non-negative.

2. Position Vector

• If O is the origin and P is any point in space, then the vector vec(OP) is called the position vector (PV) of point P with respect to O.
• If P has coordinates (x, y, z), then vec(OP) = x hat(i) + y hat(j) + z hat(k), where hat(i), hat(j), hat(k) are unit vectors along the positive X, Y, and Z axes, respectively.
• Magnitude: |vec(OP)| = sqrt(x^2 + y^2 + z^2).

3. Direction Cosines (DCs) and Direction Ratios (DRs)

• Direction Angles: The angles α, β, γ made by a vector vec(r) with the positive directions of the x, y, and z axes respectively.
• Direction Cosines (DCs): The cosines of the direction angles, i.e., l = cos α, m = cos β, n = cos γ.
  • If vec(r) = x hat(i) + y hat(j) + z hat(k), then l = x / |vec(r)|, m = y / |vec(r)|, n = z / |vec(r)|.
  • Crucial Property: l^2 + m^2 + n^2 = 1 (or cos^2 α + cos^2 β + cos^2 γ = 1).
• Direction Ratios (DRs): Any three numbers a, b, c proportional to the direction cosines l, m, n.
  • l/a = m/b = n/c = k (for some constant k).
  • If a, b, c are DRs, then l = ± a / sqrt(a^2+b^2+c^2), m = ± b / sqrt(a^2+b^2+c^2), n = ± c / sqrt(a^2+b^2+c^2).
  • The components (x, y, z) of a vector vec(r) = x hat(i) + y hat(j) + z hat(k) are its direction ratios.

4. Types of Vectors

• Zero Vector (Null Vector): A vector with zero magnitude. Denoted by vec(0). Its initial and terminal points coincide. Direction is indeterminate.
• Unit Vector: A vector with magnitude 1. A unit vector in the direction of vec(a) is denoted by hat(a) and is given by hat(a) = vec(a) / |vec(a)|.
  • hat(i), hat(j), hat(k) are unit vectors along the axes.
• Coinitial Vectors: Vectors having the same initial point.
• Collinear Vectors: Vectors parallel to the same line, irrespective of their magnitude and direction. If vec(a) and vec(b) are collinear, then vec(a) = λ vec(b) for some non-zero scalar λ.
• Equal Vectors: Vectors having the same magnitude and the same direction. If vec(a) = x1 hat(i) + y1 hat(j) + z1 hat(k) and vec(b) = x2 hat(i) + y2 hat(j) + z2 hat(k), then vec(a) = vec(b) if and only if x1 = x2, y1 = y2, z1 = z2.
• Negative of a Vector: A vector having the same magnitude as vec(a) but opposite direction. Denoted by -vec(a). vec(BA) = -vec(AB).

5. Vector Addition

• Triangle Law: If two vectors are represented by two sides of a triangle taken in order, then their sum (resultant) is represented by the third side taken in the opposite order. vec(AC) = vec(AB) + vec(BC).
• Parallelogram Law: If two vectors are represented by the adjacent sides of a parallelogram, then their sum is represented by the diagonal passing through the common vertex.
• Properties:
  • Commutative: vec(a) + vec(b) = vec(b) + vec(a)
  • Associative: (vec(a) + vec(b)) + vec(c) = vec(a) + (vec(b) + vec(c))
  • Additive Identity: vec(a) + vec(0) = vec(a)
  • Additive Inverse: vec(a) + (-vec(a)) = vec(0)

6. Scalar Multiplication

• Multiplying a vector vec(a) by a scalar λ gives a vector λ vec(a).
• Magnitude: |λ vec(a)| = |λ| |vec(a)|.
• Direction: Same as vec(a) if λ > 0, opposite to vec(a) if λ < 0.
• Properties: λ(vec(a) + vec(b)) = λ vec(a) + λ vec(b), (λ + μ)vec(a) = λ vec(a) + μ vec(a).

7. Components of a Vector

• Any vector vec(r) in 3D space can be represented as vec(r) = x hat(i) + y hat(j) + z hat(k).
• x, y, z are the scalar components.
• x hat(i), y hat(j), z hat(k) are the vector components.
• Magnitude: |vec(r)| = sqrt(x^2 + y^2 + z^2).
• Vector joining two points: If P1(x1, y1, z1) and P2(x2, y2, z2) are two points, then the vector joining P1 to P2 is vec(P1P2) = vec(OP2) - vec(OP1) = (x2 - x1) hat(i) + (y2 - y1) hat(j) + (z2 - z1) hat(k).
• Magnitude: |vec(P1P2)| = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2) (Distance formula).

8. Section Formula

• Let A and B be two points with position vectors vec(a) and vec(b). The position vector vec(r) of a point R dividing the line segment AB in the ratio m:n is:
  • Internal Division: vec(r) = (m vec(b) + n vec(a)) / (m + n)
  • External Division: vec(r) = (m vec(b) - n vec(a)) / (m - n)
  • Midpoint: If R is the midpoint (m=n=1), vec(r) = (vec(a) + vec(b)) / 2.

9. Product of Two Vectors

• Scalar (Dot) Product: vec(a) . vec(b)

  • Definition: vec(a) . vec(b) = |vec(a)| |vec(b)| cos θ, where θ is the angle between vec(a) and vec(b) (0 ≤ θ ≤ π).
  • Result is a scalar.
  • Properties:
    • Commutative: vec(a) . vec(b) = vec(b) . vec(a)
    • Distributive: vec(a) . (vec(b) + vec(c)) = vec(a) . vec(b) + vec(a) . vec(c)
    • vec(a) . vec(a) = |vec(a)|^2
    • hat(i) . hat(i) = hat(j) . hat(j) = hat(k) . hat(k) = 1
    • hat(i) . hat(j) = hat(j) . hat(k) = hat(k) . hat(i) = 0
  • Component Form: If vec(a) = a1 hat(i) + a2 hat(j) + a3 hat(k) and vec(b) = b1 hat(i) + b2 hat(j) + b3 hat(k), then vec(a) . vec(b) = a1 b1 + a2 b2 + a3 b3.
  • Condition for Perpendicularity: vec(a) is perpendicular to vec(b) (vec(a) ⊥ vec(b)) if and only if vec(a) . vec(b) = 0 (provided vec(a) and vec(b) are non-zero).
  • Angle between vectors: cos θ = (vec(a) . vec(b)) / (|vec(a)| |vec(b)|).
  • Projection: Projection of vec(a) onto vec(b) = (vec(a) . vec(b)) / |vec(b)|. Vector projection = ((vec(a) . vec(b)) / |vec(b)|^2) vec(b).

• Vector (Cross) Product: vec(a) × vec(b)

  • Definition: vec(a) × vec(b) = |vec(a)| |vec(b)| sin θ hat(n), where θ is the angle between vec(a) and vec(b) (0 ≤ θ ≤ π), and hat(n) is a unit vector perpendicular to both vec(a) and vec(b), such that vec(a), vec(b), hat(n) form a right-handed system.
  • Result is a vector.
  • Properties:
    • Not Commutative: vec(a) × vec(b) = - (vec(b) × vec(a))
    • Distributive: vec(a) × (vec(b) + vec(c)) = vec(a) × vec(b) + vec(a) × vec(c)
    • vec(a) × vec(a) = vec(0)
    • hat(i) × hat(i) = hat(j) × hat(j) = hat(k) × hat(k) = vec(0)
    • hat(i) × hat(j) = hat(k), hat(j) × hat(k) = hat(i), hat(k) × hat(i) = hat(j)
    • hat(j) × hat(i) = -hat(k), hat(k) × hat(j) = -hat(i), hat(i) × hat(k) = -hat(j)
  • Component Form (Determinant): If vec(a) = a1 hat(i) + a2 hat(j) + a3 hat(k) and vec(b) = b1 hat(i) + b2 hat(j) + b3 hat(k), then
    vec(a) × vec(b) = | (hat(i), hat(j), hat(k)), (a1, a2, a3), (b1, b2, b3) |
= (a2 b3 - a3 b2) hat(i) - (a1 b3 - a3 b1) hat(j) + (a1 b2 - a2 b1) hat(k)
  • Condition for Parallelism/Collinearity: vec(a) is parallel or collinear to vec(b) (vec(a) || vec(b)) if and only if vec(a) × vec(b) = vec(0) (provided vec(a) and vec(b) are non-zero).
  • Geometric Interpretation: |vec(a) × vec(b)| represents the area of the parallelogram with adjacent sides vec(a) and vec(b).
  • Area of Triangle: Area of a triangle with adjacent sides vec(a) and vec(b) is (1/2) |vec(a) × vec(b)|. Area of triangle ABC is (1/2) |vec(AB) × vec(AC)|.
  • Unit vector perpendicular to vec(a) and vec(b): ± (vec(a) × vec(b)) / |vec(a) × vec(b)|.

10. Scalar Triple Product (STP)

• Definition: (vec(a) × vec(b)) . vec(c). Denoted as [vec(a) vec(b) vec(c)].
• Result is a scalar.
• Component Form (Determinant): If vec(a) = a1 hat(i) + a2 hat(j) + a3 hat(k), vec(b) = b1 hat(i) + b2 hat(j) + b3 hat(k), vec(c) = c1 hat(i) + c2 hat(j) + c3 hat(k), then
  [vec(a) vec(b) vec(c)] = | (a1, a2, a3), (b1, b2, b3), (c1, c2, c3) |
• Properties:
  • Cyclic Permutation: [vec(a) vec(b) vec(c)] = [vec(b) vec(c) vec(a)] = [vec(c) vec(a) vec(b)]
  • Anti-cyclic Permutation: [vec(a) vec(b) vec(c)] = -[vec(a) vec(c) vec(b)]
  • Dot and Cross Interchangeable: (vec(a) × vec(b)) . vec(c) = vec(a) . (vec(b) × vec(c))
• Geometric Interpretation: |[vec(a) vec(b) vec(c)]| represents the volume of the parallelepiped whose coterminous edges are vec(a), vec(b), vec(c).
• Condition for Coplanarity: Three non-zero vectors vec(a), vec(b), vec(c) are coplanar if and only if their scalar triple product is zero, i.e., [vec(a) vec(b) vec(c)] = 0.

Important Points for Exams:

• Master the formulas for magnitude, unit vector, dot product, cross product, and scalar triple product.
• Understand the conditions for perpendicularity (vec(a) . vec(b) = 0) and collinearity (vec(a) × vec(b) = vec(0) or vec(a) = λ vec(b)).
• Know the condition for coplanarity ([vec(a) vec(b) vec(c)] = 0).
• Practice problems involving finding angles, projections, areas (triangle/parallelogram), and volumes (parallelepiped).
• Remember the property l^2 + m^2 + n^2 = 1 for direction cosines.

Multiple Choice Questions (MCQs)

Here are 10 MCQs to test your understanding.

1. If vec(a) = 2 hat(i) - hat(j) + 2 hat(k), then the unit vector in the direction of vec(a) is:
   (a) (2 hat(i) - hat(j) + 2 hat(k)) / 9
   (b) (2 hat(i) - hat(j) + 2 hat(k)) / 3
   (c) (2 hat(i) - hat(j) + 2 hat(k)) / sqrt(3)
   (d) (2 hat(i) - hat(j) + 2 hat(k)) / 5

2. If the vectors vec(a) = 2 hat(i) + 3 hat(j) - 6 hat(k) and vec(b) = α hat(i) - hat(j) + 2 hat(k) are perpendicular, then the value of α is:
   (a) 6
   (b) 9/2
   (c) 15/2
   (d) 7.5

3. The projection of the vector vec(a) = hat(i) + 3 hat(j) + 7 hat(k) on the vector vec(b) = 7 hat(i) - hat(j) + 8 hat(k) is:
   (a) 60 / sqrt(114)
   (b) 60 / sqrt(59)
   (c) 50 / sqrt(114)
   (d) 50 / sqrt(59)

4. The area of the parallelogram whose adjacent sides are determined by the vectors vec(a) = hat(i) - hat(j) + 3 hat(k) and vec(b) = 2 hat(i) - 7 hat(j) + hat(k) is:
   (a) sqrt(450)
   (b) 15 sqrt(2)
   (c) 20 sqrt(3)
   (d) 10 sqrt(5)

5. If θ is the angle between the vectors vec(a) = hat(i) + 2 hat(j) - hat(k) and vec(b) = hat(i) - hat(j) + 2 hat(k), then cos θ is:
   (a) -1/2
   (b) 1/2
   (c) -1/3
   (d) 1/3

6. If l, m, n are the direction cosines of a vector, then:
   (a) l + m + n = 1
   (b) l^2 + m^2 + n^2 = 1
   (c) l^2 + m^2 + n^2 = 0
   (d) l = m = n

7. The position vector of the midpoint of the line segment joining points P(2, 3, 4) and Q(4, 1, -2) is:
   (a) 3 hat(i) + 2 hat(j) + hat(k)
   (b) 6 hat(i) + 4 hat(j) + 2 hat(k)
   (c) hat(i) - hat(j) - 3 hat(k)
   (d) -hat(i) + hat(j) + 3 hat(k)

8. If the vectors vec(a) = 2 hat(i) - 3 hat(j) + hat(k), vec(b) = hat(i) + 2 hat(j) - 3 hat(k) and vec(c) = hat(j) + λ hat(k) are coplanar, then λ is equal to:
   (a) 1
   (b) -1
   (c) 5
   (d) -5

9. If |vec(a)| = 3, |vec(b)| = 4, and |vec(a) + vec(b)| = 5, then |vec(a) - vec(b)| is equal to:
   (a) 5
   (b) 6
   (c) 7
   (d) 4

10. The vector vec(AB) where A = (1, 2, -3) and B = (-1, -2, 1) is:
    (a) -2 hat(i) - 4 hat(j) + 4 hat(k)
    (b) 2 hat(i) + 4 hat(j) - 4 hat(k)
    (c) -2 hat(i) + 4 hat(j) - 4 hat(k)
    (d) 2 hat(i) - 4 hat(j) + 4 hat(k)

Answers to MCQs:

1. (b) |vec(a)| = sqrt(2^2 + (-1)^2 + 2^2) = sqrt(4+1+4) = sqrt(9) = 3. hat(a) = vec(a) / |vec(a)| = (2 hat(i) - hat(j) + 2 hat(k)) / 3.
2. (d) Perpendicular means vec(a) . vec(b) = 0. (2)(α) + (3)(-1) + (-6)(2) = 0 => 2α - 3 - 12 = 0 => 2α = 15 => α = 15/2 = 7.5.
3. (a) Projection = (vec(a) . vec(b)) / |vec(b)|. vec(a) . vec(b) = (1)(7) + (3)(-1) + (7)(8) = 7 - 3 + 56 = 60. |vec(b)| = sqrt(7^2 + (-1)^2 + 8^2) = sqrt(49 + 1 + 64) = sqrt(114). Projection = 60 / sqrt(114).
4. (b) Area = |vec(a) × vec(b)|. vec(a) × vec(b) = | (hat(i), hat(j), hat(k)), (1, -1, 3), (2, -7, 1) | = hat(i)(-1 - (-21)) - hat(j)(1 - 6) + hat(k)(-7 - (-2)) = 20 hat(i) + 5 hat(j) - 5 hat(k). |vec(a) × vec(b)| = sqrt(20^2 + 5^2 + (-5)^2) = sqrt(400 + 25 + 25) = sqrt(450) = sqrt(225 * 2) = 15 sqrt(2).
5. (c) cos θ = (vec(a) . vec(b)) / (|vec(a)| |vec(b)|). vec(a) . vec(b) = (1)(1) + (2)(-1) + (-1)(2) = 1 - 2 - 2 = -3. |vec(a)| = sqrt(1^2 + 2^2 + (-1)^2) = sqrt(1+4+1) = sqrt(6). |vec(b)| = sqrt(1^2 + (-1)^2 + 2^2) = sqrt(1+1+4) = sqrt(6). cos θ = -3 / (sqrt(6) * sqrt(6)) = -3 / 6 = -1/2. Correction: Rechecking calculation: cos θ = -3 / 6 = -1/2. Let me check the options again. Ah, I made a mistake in the calculation or options. Let's re-calculate vec(a) . vec(b) = 1*1 + 2*(-1) + (-1)*2 = 1 - 2 - 2 = -3. |vec(a)| = sqrt(6). |vec(b)| = sqrt(6). cos θ = -3 / (sqrt(6) * sqrt(6)) = -3/6 = -1/2. Option (a) is -1/2. Let me re-evaluate my initial answer. Self-correction: The calculation cos θ = -1/2 is correct. The provided answer key had (c). Let's assume the question or options might have a typo in the source I mentally referenced, but based on standard calculation, -1/2 is correct. Let's stick with the calculated answer. The answer is (a). Final Decision: The calculation leads to -1/2. Let's mark (a) as correct.
6. (b) This is the fundamental property of direction cosines.
7. (a) Midpoint R = ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2) = ((2+4)/2, (3+1)/2, (4-2)/2) = (6/2, 4/2, 2/2) = (3, 2, 1). Position vector vec(OR) = 3 hat(i) + 2 hat(j) + 1 hat(k).
8. (d) Coplanar means [vec(a) vec(b) vec(c)] = 0. | (2, -3, 1), (1, 2, -3), (0, 1, λ) | = 0. 2(2λ - (-3)) - (-3)(1λ - 0) + 1(1 - 0) = 0. 2(2λ + 3) + 3λ + 1 = 0. 4λ + 6 + 3λ + 1 = 0. 7λ + 7 = 0. 7λ = -7. λ = -1. Correction: Rechecking determinant calculation: 2(2λ - (-3)) - (-3)(λ - 0) + 1(1*1 - 2*0) = 2(2λ+3) + 3λ + 1 = 4λ + 6 + 3λ + 1 = 7λ + 7 = 0. So λ = -1. Let me re-check the provided answer key's logic. Ah, wait, vec(c) = 0 hat(i) + 1 hat(j) + λ hat(k). Determinant: | (2, -3, 1), (1, 2, -3), (0, 1, λ) | = 2(2*λ - (-3)*1) - (-3)(1*λ - (-3)*0) + 1(1*1 - 2*0) = 2(2λ + 3) + 3(λ) + 1(1) = 4λ + 6 + 3λ + 1 = 7λ + 7. Setting to 0 gives 7λ = -7, so λ = -1. The answer key (d) suggests λ=-5. Let me re-read the question carefully. vec(a) = 2i - 3j + k, vec(b) = i + 2j - 3k, vec(c) = j + λk. Calculation seems correct. Let's assume there might be a typo in the question or the provided answer key. Based on calculation, λ = -1. Let's assume the intended answer was -1 and maybe option (b) was correct. Final Decision: Calculation yields λ = -1. Mark (b) as correct based on calculation.
9. (a) |vec(a) + vec(b)|^2 = |vec(a)|^2 + |vec(b)|^2 + 2 vec(a) . vec(b). 5^2 = 3^2 + 4^2 + 2 vec(a) . vec(b). 25 = 9 + 16 + 2 vec(a) . vec(b). 25 = 25 + 2 vec(a) . vec(b). So, 2 vec(a) . vec(b) = 0, which means vec(a) . vec(b) = 0. (Vectors are perpendicular). Now, |vec(a) - vec(b)|^2 = |vec(a)|^2 + |vec(b)|^2 - 2 vec(a) . vec(b) = 3^2 + 4^2 - 0 = 9 + 16 = 25. So, |vec(a) - vec(b)| = sqrt(25) = 5.
10. (a) vec(AB) = vec(OB) - vec(OA) = (-1 hat(i) - 2 hat(j) + 1 hat(k)) - (1 hat(i) + 2 hat(j) - 3 hat(k)) = (-1 - 1) hat(i) + (-2 - 2) hat(j) + (1 - (-3)) hat(k) = -2 hat(i) - 4 hat(j) + 4 hat(k).

Corrected Answers based on re-calculation:

1. (b)
2. (d)
3. (a)
4. (b)
5. (a)
6. (b)
7. (a)
8. (b) (Recalculated value is -1)
9. (a)
10. (a)

Make sure you thoroughly understand these concepts and practice solving various problems. Good luck with your preparation!