IOE entrance Mathematics · Chapter 6
Vectors and their Products
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26 questions in 2 syllabus topics · 11 are 2-mark questions.
6.1 Vectors, linear combination and dependence
10 questions
1. The magnitude of the vector a⃗ = 2i − 3j + 6k is
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Magnitude = √(x² + y² + z²).
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Answer: B. 7
|a⃗| = √(2² + (−3)² + 6²) = √(4 + 9 + 36) = √49 = 7.
2. The unit vector in the direction of 3i − 4j is
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Divide the vector by its own magnitude.
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Answer: B. (3i − 4j)/5
|3i − 4j| = √(9 + 16) = 5, so the unit vector is (3i − 4j)/5.
3. The vectors 2i + λj + 3k and 4i − 6j + 6k are parallel (collinear). The value of λ is
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Set the ratios of corresponding components equal.
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Answer: C. −3
Parallel vectors have proportional components: 2/4 = λ/(−6) = 3/6 = ½, so λ = −3.
4. The position vectors of points A and B are i + 2j − k and 3i − 2j + 5k respectively. The vector AB⃗ is
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AB⃗ = b⃗ − a⃗, not a⃗ − b⃗.
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Answer: C. 2i − 4j + 6k
AB⃗ = (position vector of B) − (position vector of A) = (3 − 1)i + (−2 − 2)j + (5 + 1)k = 2i − 4j + 6k.
5. A = (1, 2, 3) and B = (4, −1, 6). The point P dividing AB internally in the ratio 2 : 1 is
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Section formula: (m b⃗ + n a⃗)/(m + n) for ratio m : n.
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Answer: B. (3, 0, 5)
P = (2B + 1A)/(2 + 1) = (8 + 1, −2 + 2, 12 + 3)/3 = (9, 0, 15)/3 = (3, 0, 5).
6. Two non-zero vectors a⃗ and b⃗ are linearly dependent if and only if they are
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Linear dependence of two vectors means a⃗ = λb⃗.
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Answer: A. collinear (parallel)
a⃗ and b⃗ are linearly dependent when one is a scalar multiple of the other, i.e. they are collinear.
7. The vectors i + j + k, 2i + 3j − k and 3i + 4j + mk are linearly dependent. The value of m is
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Three vectors are dependent when the determinant of their components is zero.
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Answer: C. 0
Dependence requires |1 1 1; 2 3 −1; 3 4 m| = 0. Expanding: 1(3m + 4) − 1(2m + 3) + 1(8 − 9) = m = 0. Indeed (i + j + k) + (2i + 3j − k) = 3i + 4j.
8. Any four vectors in three-dimensional space are always
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Think about the maximum number of independent vectors in 3-D.
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Answer: C. linearly dependent
Space has dimension 3, so at most three vectors can be linearly independent; any four vectors must be linearly dependent.
9. Let a⃗ = i + 2j and b⃗ = 3i − j. If 7i = x a⃗ + y b⃗, then (x, y) is
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Equate the i and j components and solve the two equations.
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Answer: A. (1, 2)
Comparing components: x + 3y = 7 and 2x − y = 0 ⇒ y = 2x ⇒ 7x = 7 ⇒ x = 1, y = 2. Check: a⃗ + 2b⃗ = 7i + 0j.
10. A vector makes angles α, β, γ with the positive x, y and z axes. Then sin²α + sin²β + sin²γ equals
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Use cos²α + cos²β + cos²γ = 1 and sin² = 1 − cos².
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Answer: C. 2
cos²α + cos²β + cos²γ = 1, so sin² sum = 3 − 1 = 2.
6.2 Scalar, vector and scalar triple products
16 questions
11. If a⃗ = i + 2j + 3k and b⃗ = 2i + 3j − 2k, then the projection of b⃗ on a⃗ is
IOE model question 2080Show hintHide hint
Use (a·b)/|a|.
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Answer: A. 2/√14 or C. √(2/7)
2/√14 and √(2/7) are the same number, so options A and C are both correct as printed in the official model set.
Projection of b on a = (a·b)/|a| = (2 + 6 − 6)/√(1 + 4 + 9) = 2/√14.
12. If a⃗ = 2i − j + 3k and b⃗ = i + 4j + 2k, then a⃗·b⃗ equals
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Multiply corresponding components and add, keeping signs.
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Answer: D. 4
a⃗·b⃗ = (2)(1) + (−1)(4) + (3)(2) = 2 − 4 + 6 = 4.
13. The vectors 2i + λj + k and i − 2j + 3k are perpendicular. Then λ equals
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Perpendicular vectors have zero dot product.
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Answer: A. 5/2
Perpendicular ⇒ dot product = 0: 2 − 2λ + 3 = 0 ⇒ λ = 5/2.
14. The angle between the vectors i + j and j + k is
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cos θ = a⃗·b⃗ /(|a⃗||b⃗|).
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Answer: A. 60°
Dot product = 0 + 1 + 0 = 1; |each| = √2, so cos θ = 1/(√2·√2) = ½, θ = 60°.
15. The value of i·(j × k) + j·(i × k) is
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Use the cyclic order i → j → k for cross products.
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Answer: B. 0
j × k = i so the first term is 1; i × k = −j so the second term is j·(−j) = −1. Sum = 0.
16. The area of the parallelogram whose adjacent sides are a⃗ = i + 2j + 3k and b⃗ = 3i − 2j + k is
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Area of parallelogram = |a⃗ × b⃗|.
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Answer: C. 8√3 sq. units
a⃗ × b⃗ = (2·1 − 3·(−2))i − (1·1 − 3·3)j + (1·(−2) − 2·3)k = 8i + 8j − 8k, with magnitude √192 = 8√3.
17. The area of the triangle with vertices A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) is
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Area of triangle = ½|AB⃗ × AC⃗|.
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Answer: D. √21/2 sq. units
AB⃗ = (0, 1, 2), AC⃗ = (1, 2, 0). AB⃗ × AC⃗ = (1·0 − 2·2, 2·1 − 0·0, 0·2 − 1·1) = (−4, 2, −1), magnitude √21. Area = ½√21.
18. The scalar projection of a⃗ = 2i + 3j + 2k on b⃗ = i + 2j + 2k is
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Divide the dot product by the magnitude of the vector you project onto.
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Answer: D. 4
Projection of a⃗ on b⃗ = a⃗·b⃗/|b⃗| = (2 + 6 + 4)/3 = 12/3 = 4.
19. For any vectors a⃗ and b⃗, the scalar triple product [a⃗ a⃗ b⃗] equals
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What happens to a determinant with two identical rows?
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Answer: D. 0
a⃗ × a⃗ = 0 (or: a determinant with two equal rows is zero), so [a⃗ a⃗ b⃗] = 0.
20. The volume of the parallelepiped with coterminous edges a⃗ = i + 2j + 3k, b⃗ = j + 2k and c⃗ = 2i + k is
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Volume = |[a⃗ b⃗ c⃗]|, the 3×3 determinant of components.
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Answer: C. 3 cubic units
[a⃗ b⃗ c⃗] = |1 2 3; 0 1 2; 2 0 1| = 1(1 − 0) − 2(0 − 4) + 3(0 − 2) = 1 + 8 − 6 = 3. Volume = 3.
21. The vectors 2i − j + k, i + 2j − 3k and 3i + λj + 5k are coplanar. The value of λ is
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Three vectors are coplanar when their scalar triple product is zero.
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Answer: D. −4
Coplanar ⇒ |2 −1 1; 1 2 −3; 3 λ 5| = 0: 2(10 + 3λ) + 1(5 + 9) + 1(λ − 6) = 7λ + 28 = 0 ⇒ λ = −4.
22. If |a⃗| = 2, |b⃗| = 5 and a⃗·b⃗ = 6, then |a⃗ × b⃗| equals
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Use |a⃗ × b⃗|² + (a⃗·b⃗)² = |a⃗|²|b⃗|².
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Answer: D. 8
|a⃗ × b⃗|² = |a⃗|²|b⃗|² − (a⃗·b⃗)² = 100 − 36 = 64, so |a⃗ × b⃗| = 8.
23. If a⃗ and b⃗ are non-zero vectors with a⃗ × b⃗ = 0⃗, then a⃗ and b⃗ are
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Cross product magnitude involves sin θ.
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Answer: A. parallel
|a⃗ × b⃗| = |a⃗||b⃗| sin θ = 0 with non-zero vectors gives sin θ = 0, i.e. θ = 0° or 180°.
24. Which of the following is NOT true for all vectors a⃗ and b⃗?
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Is the cross product commutative?
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Answer: A. a⃗ × b⃗ = b⃗ × a⃗
The cross product is anti-commutative: a⃗ × b⃗ = −(b⃗ × a⃗). The other three always hold.
25. If |a⃗| = 3, |b⃗| = 4 and |a⃗ + b⃗| = 5, then |a⃗ − b⃗| equals
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Square both magnitudes and find a⃗·b⃗ first.
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Answer: A. 5
|a⃗ + b⃗|² = 9 + 16 + 2a⃗·b⃗ = 25 ⇒ a⃗·b⃗ = 0. Then |a⃗ − b⃗|² = 9 + 16 − 0 = 25, so |a⃗ − b⃗| = 5.
26. A force F⃗ = 2i + 3j + k newton moves a particle from the point (1, 1, 1) to (3, 2, 4) (in metres). The work done is
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Work = F⃗·d⃗, where d⃗ is final minus initial position.
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Answer: C. 10 J
Displacement d⃗ = 2i + j + 3k. W = F⃗·d⃗ = 4 + 3 + 3 = 10 J.