Vector Algebra — JEE Main Questions

32 JEE Main practice questions on Vector Algebra, part of Mathematics. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. If r × i = j, find the general solution for r.

  1. r = k + t j
  2. r = j + t i
  3. r = i + t j
  4. r = k + t i

Answer: r = k + t i

Given r × i = j. Since cross product is perpendicular to i, j must be perpendicular to i, which holds. Take cross product with i: i × (r × i) = i × j. Using BAC-CAB: (i·i)r - (i·r)i = k. So |i|² r - (i·r)i = k, giving r = k + (i·r)i. Let t = i·r, then r = k + t i. Verify: (k + t i) × i = k × i + 0 = j. Hence general solution is r = k + t i.

2. Find r such that r·i = 2 and r × i = j.

  1. r = 2 i + k
  2. r = 2 i + j
  3. r = i + 2 k
  4. r = 2 j + k

Answer: r = 2 i + k

Given r·i = 2 and r × i = j. The general solution of r × i = j is r = k + t i. Substitute into dot condition: (k + t i)·i = 0 + t = 2, so t = 2. Hence r = k + 2 i = 2 i + k. Verify: r·i = 2, r × i = (2 i + k) × i = 2(i × i) + k × i = 0 + j = j.

3. Given u = (1,1,1), v = (1,2,3), w = (2,3,4), which of the following is true?

  1. u, v, w are coplanar and w = u + v
  2. u, v, w are coplanar but linearly independent
  3. u, v, w are non-coplanar and [u v w] = 1
  4. u, v, w are non-coplanar and [u v w] = -1

Answer: u, v, w are coplanar and w = u + v

Compute scalar triple product [u v w] = det|1 1 1; 1 2 3; 2 3 4| = 1(8-9) - 1(4-6) + 1(3-4) = -1 + 2 - 1 = 0, so vectors are coplanar. By inspection, w = (2,3,4) = (1,1,1)+(1,2,3) = u+v, confirming linear dependence.

4. Which of the following statements about the dot product is true?

  1. Both (b) and (c) are true
  2. a·b = b·a
  3. a·(b+c) = a·b + a·c
  4. a·(b·c) = (a·b)·c

Answer: Both (b) and (c) are true

The dot product is commutative (a·b = b·a) and distributive over addition (a·(b+c) = a·b + a·c). It is not associative because a·b is a scalar, so (a·b)·c is meaningless. Hence both (b) and (c) are true.

5. For vectors a = (1,2,1), b = (2,1,0), c = (0,1,2), what is a × (b × c)?

  1. (8, -8, 0)
  2. (8, 0, 8)
  3. (0, 8, -8)
  4. (8, 0, -8)

Answer: (8, 0, -8)

Using the vector triple product identity: a × (b × c) = (a·c)b − (a·b)c. Compute a·c = 1·0+2·1+1·2 = 4, a·b = 1·2+2·1+1·0 = 4. Then (a·c)b − (a·b)c = 4(2,1,0) − 4(0,1,2) = (8,4,0) − (0,4,8) = (8,0,−8).

6. If vectors a = (1,2,1), b = (2,1,3), c = (4,5,5) are linearly dependent, find the value of k such that c = k a + b.

  1. 3
  2. 2
  3. 1
  4. 4

Answer: 2

The scalar triple product [a b c] = a·(b×c) = 0 confirms dependence. Solving c = k a + b gives (4,5,5) = k(1,2,1)+(2,1,3). Equating components: 4 = k+2 → k=2; 5 = 2k+1 → k=2; 5 = k+3 → k=2. So k=2.

7. If a = i, b = j, c = i, d = k, then (a × b)·(c × d) equals

  1. 1
  2. -1
  3. 0
  4. 2

Answer: 0

Compute directly: a×b = i×j = k, c×d = i×k = -j. Dot product: k·(-j) = 0. Using identity (a×b)·(c×d) = (a·c)(b·d) - (a·d)(b·c) = (i·i)(j·k) - (i·k)(j·i) = (1)(0) - (0)(0) = 0.

8. For vectors a = (1,2,2) and b = (2,2,1), what is the value of sin²θ + cos²θ?

  1. 1
  2. 64/81
  3. 17/81
  4. 81/64

Answer: 1

Using dot product, cosθ = (a·b)/(|a||b|) = 8/9. Using cross product, |a×b| = √17, so sinθ = √17/9. Then sin²θ + cos²θ = 17/81 + 64/81 = 1, verifying Lagrange's identity.

9. If r·i = 3 and r × i = j, then r is

  1. 3i + j
  2. 3i - k
  3. 3i + k
  4. 3i - j

Answer: 3i + k

Using formula r = (p a + (a×b))/|a|² with a=i, b=j, p=3, |a|=1. a×b = i×j = k. So r = (3i + k)/1 = 3i + k. Verification: r·i = 3, r×i = (3i+k)×i = 3(i×i)+k×i = 0+j = j.

10. If a = (1,1,0) and b = (1,0,1), what is the angle between a and b?

  1. 30°
  2. 90°
  3. 60°
  4. 45°

Answer: 60°

Using dot product formula: cosθ = (a·b)/(|a||b|). a·b = 1·1 + 1·0 + 0·1 = 1. |a| = √(1²+1²+0²) = √2, |b| = √(1²+0²+1²) = √2. So cosθ = 1/(√2·√2) = 1/2, hence θ = 60°.

11. Are the points (1,2,3), (2,3,4), (3,4,5), (4,5,7) coplanar?

  1. Only if the points are collinear
  2. No
  3. Cannot be determined
  4. Yes

Answer: Yes

Take A(1,2,3), B(2,3,4), C(3,4,5), D(4,5,7). Vectors AB=(1,1,1), AC=(2,2,2), AD=(3,3,4). The scalar triple product [AB AC AD] = 0, so points are coplanar.

12. For vectors a = i, b = i, c = j, which of the following is true?

  1. a × (b × c) = (a × b) × c
  2. a × (b × c) = -j and (a × b) × c = 0
  3. a × (b × c) = 0 and (a × b) × c = -j
  4. a × (b × c) = (a × b) × c = 0

Answer: a × (b × c) = -j and (a × b) × c = 0

Compute: a × (b × c) = i × (i × j) = i × k = -j. (a × b) × c = (i × i) × j = 0 × j = 0. So they are not equal, showing cross product is non-associative.

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