Vector Algebra: CBSE Class 12 Previous Year Questions
28 different questions from Vector Algebra (NCERT Chapter 10) asked in CBSE Class 12 Maths board exams 2026.
Pick a mark group to practise, each with answers and step-by-step solutions.
Most asked Vector Algebra questions
The value of p for which vectors i ^ + 2 j ^ + 3 k ^ and 2 i ^ − p j ^ + k ^ are perpendicular to each other is
(A) 0(B) 1(C) 2 5 (D) − 2 5
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Answer: (C) 2 5
Perpendicular vectors have zero dot product. ( 1 ) ( 2 ) + ( 2 ) ( − p ) + ( 3 ) ( 1 ) = 0 ⇒ 5 − 2 p = 0 ⇒ p = 2 5 .
The value of m for which the points with position vectors − i ^ − j ^ + 2 k ^ , 2 i ^ + m j ^ + 5 k ^ and 3 i ^ + 11 j ^ + 6 k ^ are collinear, is
(A) 8(B) − 8 (C) 2(D) 2 5
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Answer: (A) 8
Let the points be A, B, C. A B = 3 i ^ + ( m + 1 ) j ^ + 3 k ^ , A C = 4 i ^ + 12 j ^ + 4 k ^ . For collinearity, A B ∥ A C : 4 3 = 12 m + 1 = 4 3 . So m + 1 = 9 , i.e. m = 8 .
If ∣ a ∣ = 8 , ∣ b ∣ = 3 and ∣ a × b ∣ = 12 , then the value of ∣ a ⋅ b ∣
(A) 6 3 (B) 8 3 (C) 12 3 (D) 3 12
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∣ a × b ∣ = ∣ a ∣∣ b ∣ sin θ ⇒ 12 = 24 sin θ ⇒ sin θ = 2 1 .So ∣ cos θ ∣ = 2 3 . ∣ a ⋅ b ∣ = ∣ a ∣∣ b ∣∣ cos θ ∣ = 24 × 2 3 = 12 3 .
If ∣ a ∣ = 5 and − 2 ≤ λ ≤ 1 , then the sum of greatest and the smallest value of ∣ λ a ∣ is
(A) − 5 (B) 5(C) 10(D) 15
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Answer: (C) 10
∣ λ a ∣ = ∣ λ ∣ ⋅ 5 .For − 2 ≤ λ ≤ 1 , 0 ≤ ∣ λ ∣ ≤ 2 . Greatest value = 10 (at λ = − 2 ), smallest = 0 (at λ = 0 ); sum = 10 .
Vector of magnitude 3 making equal angles with x and y axes and perpendicular to z axis is
(A) i ^ + 2 2 j ^ (B) 3 k ^ (C) 2 3 2 i ^ + 2 3 2 j ^ (D) 3 i ^ + 3 j ^ + 3 k ^
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Answer: (C)
2 3 2 i ^ + 2 3 2 j ^
Perpendicular to the z-axis: n = 0 . Equal angles with x and y axes: l = m . l 2 + m 2 = 1 ⇒ l = m = 2 1 (taking positive values).Vector = 3 ( 2 1 i ^ + 2 1 j ^ ) = 2 3 2 i ^ + 2 3 2 j ^ .
For two vectors a and b Assertion (A) : ∣ a × b ∣ 2 + ( a ⋅ b ) 2 = ∣ a ∣ 2 ∣ b ∣ 2 Reason (R) : ∣ a × b ∣ = ( a ⋅ b ) tan θ , ( θ = 2 π )
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (B) Both A and R are true, but R is not the correct explanation of A.
∣ a × b ∣ 2 + ( a ⋅ b ) 2 = ∣ a ∣ 2 ∣ b ∣ 2 sin 2 θ + ∣ a ∣ 2 ∣ b ∣ 2 cos 2 θ = ∣ a ∣ 2 ∣ b ∣ 2 for all vectors, so A is true.( a ⋅ b ) tan θ = ∣ a ∣∣ b ∣ cos θ ⋅ c o s θ s i n θ = ∣ a ∣∣ b ∣ sin θ = ∣ a × b ∣ for θ = 2 π , so R is true.A holds for all angles (including θ = 2 π ) and follows directly from sin 2 θ + cos 2 θ = 1 , not from R; so R is not the correct explanation of A.
For any two vectors a and b , which of the following statements is always true ?
(A) a ⋅ b ≤ ∣ a ∣ ∣ b ∣ (B) ∣ a + b ∣ ≥ ∣ a ∣ + ∣ b ∣ (C) ∣ a − b ∣ = ∣ a ∣ − ∣ b ∣ (D) ∣ a × b ∣ ≥ ∣ a ∣ ∣ b ∣
Show answer & solution
a ⋅ b = ∣ a ∣ ∣ b ∣ cos θ and cos θ ≤ 1 , so (A) always holds.(B) and (C) fail in general (triangle inequality gives ≤ ), and ∣ a × b ∣ = ∣ a ∣ ∣ b ∣ sin θ ≤ ∣ a ∣ ∣ b ∣ , so (D) fails.
If ( a + b ) ⋅ ( a − b ) = 198 and ∣ a ∣ = 10∣ b ∣ , then :
(A) ∣ a ∣ = 2 (B) ∣ b ∣ = 2 (C) ∣ b ∣ = 10 2 (D) ∣ a ∣ = 2 10
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( a + b ) ⋅ ( a − b ) = ∣ a ∣ 2 − ∣ b ∣ 2 = 198 .100∣ b ∣ 2 − ∣ b ∣ 2 = 99∣ b ∣ 2 = 198 , so ∣ b ∣ 2 = 2 .∣ b ∣ = 2 .
Assertion (A) : The vectors a and ( − 2 a ) , where a = 0 are collinear vectors. Reason (R) : a ⋅ ( − 2 a ) = 0 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (C) Assertion (A) is true, but Reason (R) is false.
− 2 a is a scalar multiple of a , so they are collinear: A is true.a ⋅ ( − 2 a ) = − 2∣ a ∣ 2 = 0 as a = 0 : R is false.
If ( 3 i ^ − 2 j ^ + 5 k ^ ) × ( 4 i ^ + p j ^ + q k ^ ) = 0 , then the values of p and q are :
(A) p = − 3 2 , q = 3 5 (B) p = − 3 8 , q = 3 20 (C) p = 3 20 , q = − 3 8 (D) p = 0 , q = 0
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Answer: (B) p = − 3 8 , q = 3 20
Cross product is zero, so the vectors are parallel: 3 4 = − 2 p = 5 q . p = − 3 8 , q = 3 20 .
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