AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Mathematics

Vector Algebra

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The magnitude of 3î + 2ĵ − 6k̂ is:
  1. 7
  2. 49
  3. 11
  4. 8
2.
If a = −2î − 5ĵ + 4k̂ and b = −4î − 4ĵ − k̂, then a · b is:
  1. 32
  2. −12
  3. −24
  4. 24
3.
The unit vector in the direction of −12î − 12ĵ + k̂ is:
  1. (1/17)(12î + 12ĵ − k̂)
  2. (1/17)(−12î − 12ĵ + k̂)
  3. (1/289)(−12î − 12ĵ + k̂)
  4. (1/25)(−12î − 12ĵ + k̂)
4.
If a = î − 3ĵ + 4k̂ and b = 4î, then a × b is:
  1. 16î + 12k̂
  2. −16ĵ − 12k̂
  3. 16ĵ + 12k̂
  4. 4î
5.
A quantity with both magnitude and direction is:
  1. an angle
  2. a scalar
  3. a ratio
  4. a vector
6.
The projection of a = −4î + ĵ + 3k̂ on b = −6î + 7ĵ − 6k̂ is:
  1. 13/11
  2. 13/121
  3. 13
  4. −13/11
7.
The vectors λî + 6ĵ − 4k̂ and 4î + 4ĵ + k̂ are perpendicular when λ is:
  1. 5
  2. 20
  3. −4
  4. −5
8.
Assertion (A): The vectors 6î − 4ĵ + 2k̂ and −32î − 34ĵ + 28k̂ are perpendicular.
Reason (R): Two non-zero vectors are perpendicular when their vector product is zero.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
If a = 3î + ĵ + 5k̂ and b = −î + 3ĵ + 5k̂, find a + b, a − b and 4a − 4b.
10.
Find the area of the parallelogram whose adjacent sides are 3î + 2ĵ + 2k̂ and −î.
11.
Find the area of the triangle with vertices A(−5, −5, −3), B(5, 4, 2) and C(3, 2, 0).
12.
Find the position vector of the point R dividing the join of P(−2î − ĵ + 2k̂) and Q(2î + 3ĵ + 2k̂) internally in the ratio 4 : 1.
13.
Find the direction cosines of the vector −6î + 9ĵ + 2k̂.
14.
Find the projection of −î + 2ĵ on 6î − 2ĵ − 9k̂.
15.
If a = −4î − 2ĵ + k̂, b = −3î + 2ĵ − 5k̂ and c = −4î − 5ĵ + 2k̂, find λ such that a + λb is perpendicular to c.
16.
Find a unit vector perpendicular to both 4î − 3ĵ − k̂ and −î − 3ĵ − 4k̂.
Section C — Word problems4 × 4 = 16 marks
17.
Show that the points A(3, −1, 5), B(6, 1, 4) and C(1, 0, 1) form a right-angled triangle, and find its area.
18.
Find the angle between a = 12î − 12ĵ − k̂ and b = −12î − 9ĵ − 8k̂.
19.
The adjacent sides of a parallelogram are a = 4ĵ + 2k̂ and b = 2î − 2ĵ + 4k̂. Find its diagonals and its area.
20.
Vectors a, b and c satisfy a + b + c = 0, with |a| = 10, |b| = 9 and |c| = 7. Find a · b + b · c + c · a.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Mathematics

Answer key

Vector Algebra · AW-12M-VEC-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(a) 7√(32 + 22 + (−6)2) = √49
2(d) 24(−2)(−4) + (−5)(−4) + 4(−1)
3(b) (1/17)(−12î − 12ĵ + k̂)divide by |a| = 17
4(c) 16ĵ + 12k̂expand the determinant îĵk̂a1a2a3b1b2b3
5(d) a vector—
6(a) 13/11(a · b)/|b| = 13/11
7(d) −54λ + 6(4) + (−4)(1) = 0
8(c) A is true, but R is false.A is true: their scalar product is 0. R is false.
9a + b = 2î + 4ĵ + 10k̂; a − b = 4î − 2ĵ; 4a − 4b = 16î − 8ĵadd or subtract matching components
102√2 square unitsa × b = −2ĵ + 2k̂; |a × b| = √8
11√42 square unitsAB × AC = −8î + 10ĵ − 2k̂; area = ½√168
12(6/5)î + (11/5)ĵ + 2k̂(p + 4q)/5
13−6/11, 9/11, 2/11|a| = 11
14−10/11a · b = −10, |b| = 11
15λ = 7/2(a + λb) · c = 0 ⇒ 28 + (−8)λ = 0
16±(1/√595)(9î + 17ĵ − 15k̂)a × b = 9î + 17ĵ − 15k̂; divide by the length
17AB = 3î + 2ĵ − k̂, AC = −2î + ĵ − 4k̂; AB · AC = 0, so the right angle is at A; area = (7/2)√6 square units|AB × AC| = √294
18θ = cos−1(−28/289)a · b = −28, |a| = 17, |b| = 17; cos θ = −28/289
19diagonals 2î + 2ĵ + 6k̂ and −2î + 6ĵ − 2k̂; area 4√30 square unitsarea = |a × b|, a × b = 20î + 4ĵ − 8k̂
20−115|a + b + c|2 = 0 ⇒ 100 + 81 + 49 + 2(a · b + b · c + c · a) = 0