K2KARAKORAM
मराठीEnglish
Open the app

State Board · Class 12 · Mathematics Part 1

Lesson 5: Vectors

5 concepts · 40 practice questions with answers · Free to read

This lesson is also in: मराठी

Vector basics, position vectors and the section formula

From daily life

"3 km north" carries a direction as well as a size, so it is a vector; "3 km" alone is a scalar. Building on class 11 : points and the distance formula came in class 11; now a point is seen as a position vector.

Simple definition

A vector has magnitude and direction. The position vector of P from the origin O is OP = xi + yj + zk with magnitude |OP| = √(x² + y² + z²). Two vectors are collinear if a = λb.

Rule 1: Addition — Triangle law: AB + BC = AC; add component-wise.

Rule 2: Section formula — The point dividing A, B internally in the ratio m : n is (n·a + m·b)/(m + n); the midpoint is (a + b)/2.

Rule 3: Collinearity — If AB = λ·AC then A, B, C are collinear.

aba + b

In the figure : a + b by the triangle law: put the tail of b at the head of a.

How is the formula obtained? : A(1, 2, 3), B(4, 5, 9) divided internally in 1 : 2: (2·1 + 1·4)/3 = 2, (2·2 + 1·5)/3 = 3, (2·3 + 1·9)/3 = 5, so (2, 3, 5).

  • Easy : |3i + 4j + 12k| = √(9 + 16 + 144) = 13.
  • Medium : Midpoint of A(1, 2, 3), B(4, 5, 9) is (5/2, 7/2, 6).
  • Exam-type : A(1, 2, 3), B(2, 4, 6), C(3, 6, 9): AB = (1, 2, 3), AC = (2, 4, 6) = 2AB, so collinear.
Remember

AB = b − a (head minus tail).

Common mistake

Swapping m and n in the section formula.

Shortcut

Apply the formula separately to each component. Memory clue : "Head minus tail"; "the nearer part multiplies the far point".

CET trap

Do not add magnitudes for a vector sum; |a + b| ≤ |a| + |b|.

CET-style example : The point dividing A(1, 2, 3), B(4, 5, 9) internally in 1 : 2 is: (A) (2, 3, 5) (B) (3, 4, 7) (C) (5/2, 7/2, 6) (D) (2, 3, 4)

Answer : (A). Time-saving method : Avoid the midpoint option; the ratio is 1 : 2.

Spot the difference

प्रकारType
सदिशvector
अदिशscalar
सदिशvector

Practice questions

Easy
  1. 1Find |3i + 4j|.
  2. 2Magnitude of (1, 2, 2)?
  3. 3Find AB for A(1, 2, 3), B(4, 5, 9).
Medium
  1. 4Midpoint of A and B?
  2. 5Point dividing A, B internally in 1 : 2?
  3. 6Are A(1, 2, 3), B(2, 4, 6), C(3, 6, 9) collinear?
Challenging
  1. 7Magnitude and unit vector of (3, 4, 12)?
  2. 8Find the vector of magnitude 6 along a = (1, 2, 2).
Show answers (try them yourself first)
  1. 15.
  2. 2√9 = 3.
  3. 3(3, 3, 6).
  4. 4(5/2, 7/2, 6).
  5. 5(2, 3, 5).
  6. 6Yes, AC = 2AB.
  7. 713; (3/13, 4/13, 12/13).
  8. 8|a| = 3, so 6·(1/3, 2/3, 2/3) = (2, 4, 4).

Dot product, angle and projection

From daily life

When dragging a box with a rope, only the part of the force along the ground does work; that "along-the-direction part" is the dot product.

Simple definition

a·b = |a||b| cos θ = a₁b₁ + a₂b₂ + a₃b₃ (the result is a scalar). a ⊥ b ⇔ a·b = 0. The projection of a on b is a·b / |b|.

Rule 1: Angle — cos θ = a·b / (|a||b|).

Rule 2: Properties — a·a = |a|²; a·b = b·a; i·i = j·j = k·k = 1; i·j = j·k = k·i = 0.

How is the formula obtained? : a = (1, 2, 2), b = (2, −1, 2): a·b = 2 − 2 + 4 = 4; |a| = |b| = 3, so cos θ = 4/9; projection of a on b = 4/3.

  • Easy : (1, 2, 3)·(2, −1, 1) = 2 − 2 + 3 = 3.
  • Medium : (1, 1, 1)·(1, −2, 1) = 1 − 2 + 1 = 0, so they are perpendicular.
  • Exam-type : For a = (1, 2, 2), b = (2, −1, 2): cos θ = 4/9, so θ ≈ 63.6°.
Remember

The dot product is a scalar; the cross product is a vector.

Common mistake

Forgetting to divide by |b| in a projection.

Shortcut

To test perpendicularity just check that the dot product is 0. Memory clue : "Dot = cos = scalar".

CET trap

a·b = 0 does not force a = 0 or b = 0; the vectors may be perpendicular.

CET-style example : (1, 2, 3)·(2, −1, 1) = ? (A) 3 (B) 7 (C) 0 (D) −3

Answer : (A). Time-saving method : Multiply component-wise and add.

Spot the difference

उत्तरResult
अदिशscalar
सदिशvector

Practice questions

Easy
  1. 1What is i·j?
  2. 2Find (2, 3, 1)·(1, 0, 4).
  3. 3What does (1, 1, 1)·(1, −2, 1) tell us?
Medium
  1. 4Dot product of (1, 2, 2) and (2, −1, 2)?
  2. 5cos of the angle between them?
  3. 6Projection of a on b?
Challenging
  1. 7For which k are (k, 2, −1) and (1, 3, 2) perpendicular?
  2. 8When is |a + b|² = |a|² + |b|² true?
Show answers (try them yourself first)
  1. 10.
  2. 22 + 0 + 4 = 6.
  3. 30, so the vectors are perpendicular.
  4. 44.
  5. 54/9.
  6. 64/3.
  7. 7k + 6 − 2 = 0 ⇒ k = −4.
  8. 8When a·b = 0 (perpendicular).

Cross product and area

From daily life

When a spanner turns a nut, the nut moves perpendicular to both force and spanner, which is the direction of the cross product (right-hand rule).

Simple definition

a × b = |a||b| sin θ · n̂ (n̂ is the unit vector perpendicular to both). In components: a × b = (a₂b₃ − a₃b₂, a₃b₁ − a₁b₃, a₁b₂ − a₂b₁). a ∥ b ⇔ a × b = 0.

Rule 1: Area — Area of the parallelogram = |a × b|; area of the triangle = ½|a × b|.

Rule 2: Properties — a × b = −(b × a); a × b is perpendicular to a and b; |a × b|² = |a|²|b|² − (a·b)².

aba × bक्षेत्रफळ

In the figure : The area of the parallelogram is |a × b|.

How is the formula obtained? : a = (1, 2, 2), b = (2, −1, 2): a × b = (2·2 − 2·(−1), 2·2 − 1·2, 1·(−1) − 2·2) = (6, 2, −5); |a × b|² = 65 = 81 − 16 (check), so area = √65 ≈ 8.06.

  • Easy : i × j = k; j × k = i; k × i = j.
  • Medium : (a × b)·a = 6 + 4 − 10 = 0 and (a × b)·b = 12 − 2 − 10 = 0, so it is perpendicular to both.
  • Exam-type : Area of the triangle = ½√65 ≈ 4.03.
Remember

Swapping the order changes the sign: a × b = −b × a.

Common mistake

Forgetting to reverse the sign of the middle component.

Shortcut

Set it as a determinant [i j k; a₁ a₂ a₃; b₁ b₂ b₃]. Memory clue : "Cross = sine = vector = area".

CET trap

a × b = 0 means parallel, not perpendicular.

CET-style example : i × j = ? (A) k (B) −k (C) 0 (D) 1

Answer : (A). Time-saving method : The cycle i → j → k.

Spot the difference

dot / crossdot / cross
cos / sincos / sin
अदिश / सदिशscalar / vector
लंब / समांतरperpendicular / parallel

Practice questions

Easy
  1. 1What is j × k?
  2. 2What is i × i?
  3. 3What is a × a?
Medium
  1. 4Find (1, 2, 2) × (2, −1, 2).
  2. 5Area of the parallelogram on them?
  3. 6Area of the triangle on them?
Challenging
  1. 7Compute |a|²|b|² − (a·b)² and compare with |a × b|².
  2. 8Area of the triangle A(1, 0, 0), B(0, 2, 0), C(0, 0, 3)?
Show answers (try them yourself first)
  1. 1i.
  2. 20.
  3. 30.
  4. 4(6, 2, −5).
  5. 5√65 ≈ 8.06.
  6. 6½√65 ≈ 4.03.
  7. 781 − 16 = 65 = |a × b|².
  8. 8AB = (−1, 2, 0), AC = (−1, 0, 3); AB × AC = (6, 3, 2); area = ½√49 = 7/2.

Scalar triple product and coplanarity

From daily life

Three sticks from one corner form a box (parallelepiped); its volume is the scalar triple product.

Simple definition

[a b c] = a·(b × c) = the determinant |a₁ a₂ a₃; b₁ b₂ b₃; c₁ c₂ c₃|. Three vectors are coplanar ⇔ [a b c] = 0. The volume of the parallelepiped is |[a b c]|.

Rule 1: Volume — Parallelepiped = |[a b c]|; tetrahedron = ⅙|[a b c]|.

Rule 2: Cyclic order — [a b c] = [b c a] = [c a b]; reversing the order changes the sign.

How is the formula obtained? : a = (1, 2, 2), b = (2, −1, 2), c = (1, 0, 1): b × c = (−1·1 − 2·0, 2·1 − 2·1, 2·0 − (−1)·1) = (−1, 0, 1); a·(b × c) = −1 + 0 + 2 = 1.

  • Easy : [i j k] = 1.
  • Medium : The volume of the box on (2, 0, 0), (0, 3, 0), (0, 0, 4) is 24.
  • Exam-type : (1, 2, 3), (2, 1, 0), (4, 5, 6): determinant = 1(6) − 2(12) + 3(6) = 0, so coplanar.
Remember

Coplanar means volume 0 (the box is flat).

Common mistake

Getting the order of the rows wrong in [a b c].

Shortcut

Use the determinant directly rather than doing cross and dot separately. Memory clue : "Three vectors, one box, one determinant".

CET trap

A tetrahedron has ⅙ of the triple product, not ½.

CET-style example : [i j k] = ? (A) 1 (B) −1 (C) 0 (D) 3

Answer : (A). Time-saving method : The unit box has volume 1.

Spot the difference

अर्थMeaning
समतलीयcoplanar
घनफळमूल्यvolumevalue

Practice questions

Easy
  1. 1What is [i j k]?
  2. 2What is [j i k]?
  3. 3Volume of the box on (2, 0, 0), (0, 3, 0), (0, 0, 4)?
Medium
  1. 4Find [a b c] for (1, 2, 2), (2, −1, 2), (1, 0, 1).
  2. 5Are (1, 2, 3), (2, 1, 0), (4, 5, 6) coplanar?
  3. 6Volume for (1, 1, 0), (0, 1, 1), (1, 0, 1)?
Challenging
  1. 7Volume of the tetrahedron on those three vectors?
  2. 8If (1, 2, 3), (2, 1, 0), (4, k, 6) are coplanar, find k.
Show answers (try them yourself first)
  1. 11.
  2. 2−1.
  3. 324.
  4. 41.
  5. 5Yes, the determinant is 0.
  6. 62.
  7. 72/6 = 1/3.
  8. 81(6 − 0) − 2(12 − 0) + 3(2k − 4) = 6 − 24 + 6k − 12 = 0 ⇒ k = 5.

Direction ratios and direction cosines

From daily life

A room corner is where three edges meet; direction cosines tell what angles any rod makes with these three axes.

Simple definition

If r = (x, y, z) makes angles α, β, γ with the X, Y, Z axes, its direction cosines are l = cos α = x/|r|, m = cos β = y/|r|, n = cos γ = z/|r|, and l² + m² + n² = 1. Any multiple of (x, y, z) gives direction ratios.

Rule 1: Ratios to cosines — For direction ratios (a, b, c), l = a/√(a² + b² + c²) and so on.

How is the formula obtained? : r = (1, 2, 2): |r| = 3, so l = 1/3, m = 2/3, n = 2/3; l² + m² + n² = 1/9 + 4/9 + 4/9 = 1.

  • Easy : Direction cosines of the X-axis are (1, 0, 0).
  • Medium : Direction ratios (2, 3, 6): √49 = 7, so (2/7, 3/7, 6/7).
  • Exam-type : If l = m = n then 3l² = 1, so l = ±1/√3 (the diagonal of a cube).
Remember

The squares of direction cosines always add up to 1.

Common mistake

Treating direction ratios as cosines.

Shortcut

Multiplying ratios by the same number keeps them valid; divide by the magnitude to get cosines. Memory clue : "Divide by the magnitude to get cosines".

CET trap

The reverse vector has direction cosines of opposite sign.

CET-style example : Direction cosines of (2, 3, 6): (A) 2/7, 3/7, 6/7 (B) 2/11, 3/11, 6/11 (C) 1/3, 1/3, 1/3 (D) 2/√11, 3/√11, 6/√11

Answer : (A). √(4 + 9 + 36) = 7. Time-saving method : Find the magnitude first.

Spot the difference

अटCondition
l² + m² + n² = 1l² + m² + n² = 1
कोणत्याही समान गुणकany common multiple

Practice questions

Easy
  1. 1Direction cosines of the Y-axis?
  2. 2Direction cosines of (1, 2, 2)?
  3. 3Direction cosines of (2, 3, 6)?
Medium
  1. 4If l = 1/2 and m = 1/2, find n.
  2. 5Direction cosines of a cube diagonal?
  3. 6Direction cosines of (−1, −2, −2)?
Challenging
  1. 7Angle of (1, 2, 2) with the X-axis?
  2. 8Use direction cosines of (1, 2, 2) and (2, −1, 2) to find cos θ = ll′ + mm′ + nn′.
Show answers (try them yourself first)
  1. 1(0, 1, 0).
  2. 2(1/3, 2/3, 2/3).
  3. 3(2/7, 3/7, 6/7).
  4. 4n² = 1/2 ⇒ n = ±1/√2.
  5. 5(±1/√3, ±1/√3, ±1/√3).
  6. 6(−1/3, −2/3, −2/3).
  7. 7cos α = 1/3, so α ≈ 70.5°.
  8. 8(1/3)(2/3) + (2/3)(−1/3) + (2/3)(2/3) = 4/9.