Vector Algebra: 1 mark Questions (CBSE Class 12)
13 different 1 mark questions on Vector Algebra from CBSE Class 12 Maths board exams 2026, newest first.
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
Show answer & solution
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
Show answer & solution
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
Show answer & solution
∣ 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
Show answer & solution
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 ^
Show answer & solution
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.
Show answer & solution
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
Show answer & solution
( 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.
Show answer & solution
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
Show answer & solution
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 .
Three points A(0, 1, 1), B(2, 0, − 1 ) and C(1, 0, 3) form △ A B C . The ar (△ A B C ) is :
(A) 2 53 sq. units(B) 53 sq. units(C) 2 11 sq. units(D) 11 sq. units
Show answer & solution
Answer: (A)
2 53 sq. units
A B = 2 i ^ − j ^ − 2 k ^ , A C = i ^ − j ^ + 2 k ^ .A B × A C = − 4 i ^ − 6 j ^ − k ^ , with magnitude 16 + 36 + 1 = 53 .Area = 2 1 ∣ A B × A C ∣ = 2 53 sq. units.
If position vector p of a point (24, n) is such that ∣ p ∣ = 25 , then the value of n is :
(A) ± 49 (B) ± 5 (C) ± 1 (D) ± 7
Show answer & solution
Answer: (D) ± 7
2 4 2 + n 2 = 2 5 2 ⇒ n 2 = 625 − 576 = 49 ⇒ n = ± 7 .
If vectors a = 3 i ^ + 2 j ^ + λ k ^ and b = 2 i ^ − 4 j ^ + 5 k ^ , represent the two strips of the Red Cross sign placed outside a doctor’s clinic, then the value of λ is :
(A) 1(B) 2 5 (C) 5 2 (D) 0
Show answer & solution
Answer: (C) 5 2
The two strips of a Red Cross sign are perpendicular, so a ⋅ b = 0 . 6 − 8 + 5 λ = 0 ⇒ λ = 5 2 .
Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on
MonoMath CBSE →