Skip to the question JEE Main › Mathematics › Vector Algebra +4 marks−1 if incorrectSingle correct pyq
Vector Algebra: JEE Main Mathematics Question with Solution Let
a ⃗ \vec{\mathrm{a}} a and
b ⃗ \vec{\mathrm{b}} b be two vectors, Let
∣ a ⃗ ∣ = 1 , ∣ b ⃗ ∣ = 4 |\vec{\mathrm{a}}| = 1, |\vec{\mathrm{b}}| = 4 ∣ a ∣ = 1 , ∣ b ∣ = 4 and
a ⃗ ⋅ b ⃗ = 2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}} = 2 a ⋅ b = 2 . If
c ⃗ = ( 2 a ⃗ × b ⃗ ) − 3 b ⃗ \vec{\mathrm{c}} = (2\vec{\mathrm{a}} \times \vec{\mathrm{b}}) - 3\vec{\mathrm{b}} c = ( 2 a × b ) − 3 b , then the value of
b ⃗ ⋅ c ⃗ \vec{\mathrm{b}} \cdot \vec{\mathrm{c}} b ⋅ c is
Step-by-step solution View Correct answer
Taking the dot product of b ⃗ \vec{\mathrm{b}} b c ⃗ \vec{\mathrm{c}} c b ⃗ ⋅ ( a ⃗ × b ⃗ ) = 0 \vec{\mathrm{b}} \cdot (\vec{\mathrm{a}} \times \vec{\mathrm{b}}) = 0 b ⋅ ( a × b ) = 0 − 3 ∣ b ⃗ ∣ 2 = − 3 ( 16 ) = − 48 -3|\vec{\mathrm{b}}|^2 = -3(16) = -48 − 3∣ b ∣ 2 = − 3 ( 16 ) = − 48 Option analysis
Why each option works or fails A · − 24 -24 − 24 Believing that ∣ b ⃗ ∣ 2 = 8 |\vec{\mathrm{b}}|^2 = 8 ∣ b ∣ 2 = 8 4 4 4 4 × 2 4 \times 2 4 × 2 − 3 × 8 = − 24 -3 \times 8 = -24 − 3 × 8 = − 24 ∣ b ⃗ ∣ 2 |\vec{\mathrm{b}}|^2 ∣ b ∣ 2 4 2 = 16 4^2 = 16 4 2 = 16 − 3 ( 16 ) = − 48 -3(16) = -48 − 3 ( 16 ) = − 48
B · − 84 -84 − 84 Attempting to compute b ⃗ ⋅ ( 2 a ⃗ × b ⃗ ) \vec{\mathrm{b}} \cdot (2\vec{\mathrm{a}} \times \vec{\mathrm{b}}) b ⋅ ( 2 a × b ) sin θ \sin\theta sin θ b ⃗ ⋅ ( a ⃗ × b ⃗ ) \vec{\mathrm{b}} \cdot (\vec{\mathrm{a}} \times \vec{\mathrm{b}}) b ⋅ ( a × b )
C · − 48 -48 − 48 None. The cross product a ⃗ × b ⃗ \vec{\mathrm{a}} \times \vec{\mathrm{b}} a × b b ⃗ \vec{\mathrm{b}} b b ⃗ ⋅ ( 2 a ⃗ × b ⃗ ) = 0 \vec{\mathrm{b}} \cdot (2\vec{\mathrm{a}} \times \vec{\mathrm{b}}) = 0 b ⋅ ( 2 a × b ) = 0 − 3 ( b ⃗ ⋅ b ⃗ ) = − 3 ( 16 ) = − 48 -3(\vec{\mathrm{b}} \cdot \vec{\mathrm{b}}) = -3(16) = -48 − 3 ( b ⋅ b ) = − 3 ( 16 ) = − 48
D · − 60 -60 − 60 Mistakenly adding or subtracting extraneous multiples of a ⃗ ⋅ b ⃗ = 2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}} = 2 a ⋅ b = 2 b ⃗ ⋅ ( 2 a ⃗ × b ⃗ ) − 3 b ⃗ ⋅ b ⃗ = 0 − 3 ∣ b ⃗ ∣ 2 \vec{\mathrm{b}} \cdot (2\vec{\mathrm{a}} \times \vec{\mathrm{b}}) - 3\vec{\mathrm{b}} \cdot \vec{\mathrm{b}} = 0 - 3|\vec{\mathrm{b}}|^2 b ⋅ ( 2 a × b ) − 3 b ⋅ b = 0 − 3∣ b ∣ 2
Step Working
01 given ∣ a ⃗ ∣ = 1 |\vec{\mathrm{a}}| = 1 ∣ a ∣ = 1 ∣ b ⃗ ∣ = 4 |\vec{\mathrm{b}}| = 4 ∣ b ∣ = 4 a ⃗ ⋅ b ⃗ = 2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}} = 2 a ⋅ b = 2 c ⃗ = 2 ( a ⃗ × b ⃗ ) − 3 b ⃗ \vec{\mathrm{c}} = 2(\vec{\mathrm{a}} \times \vec{\mathrm{b}}) - 3\vec{\mathrm{b}} c = 2 ( a × b ) − 3 b
02 goal Find the value of the scalar product b ⃗ ⋅ c ⃗ \vec{\mathrm{b}} \cdot \vec{\mathrm{c}} b ⋅ c
03 approach Take the dot product of b ⃗ \vec{\mathrm{b}} b c ⃗ \vec{\mathrm{c}} c b ⃗ ⋅ c ⃗ = 2 ( a ⃗ × b ⃗ ) ⋅ b ⃗ − 3 ( b ⃗ ⋅ b ⃗ ) \vec{\mathrm{b}} \cdot \vec{\mathrm{c}} = 2(\vec{\mathrm{a}} \times \vec{\mathrm{b}}) \cdot \vec{\mathrm{b}} - 3(\vec{\mathrm{b}} \cdot \vec{\mathrm{b}}) b ⋅ c = 2 ( a × b ) ⋅ b − 3 ( b ⋅ b ) ( a ⃗ × b ⃗ ) ⋅ b ⃗ = 0 (\vec{\mathrm{a}} \times \vec{\mathrm{b}}) \cdot \vec{\mathrm{b}} = 0 ( a × b ) ⋅ b = 0
04 execute Compute b ⃗ ⋅ c ⃗ = 0 − 3 ∣ b ⃗ ∣ 2 = − 3 ( 4 2 ) = − 48 \vec{\mathrm{b}} \cdot \vec{\mathrm{c}} = 0 - 3|\vec{\mathrm{b}}|^2 = -3(4^2) = -48 b ⋅ c = 0 − 3∣ b ∣ 2 = − 3 ( 4 2 ) = − 48
✓ verify Check that the extra information ∣ a ⃗ ∣ = 1 |\vec{\mathrm{a}}| = 1 ∣ a ∣ = 1 a ⃗ ⋅ b ⃗ = 2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}} = 2 a ⋅ b = 2 ∣ b ⃗ ∣ |\vec{\mathrm{b}}| ∣ b ∣
Your next move We think you should solve this next Same idea, new question If the four points, whose position vectors are… Solve → Same idea, new question Let : a ⃗ = i ^ + 2 j ^ + 3 k ^ , b ⃗ = i ^ − j ^ + 2 k ^ \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \vec{b} = \hat{i} - \hat{j} + 2\hat{k} a = i ^ + 2 j ^ + 3 k ^ , b = i ^ − j ^ + 2 k ^ Solve → Same idea, new question If a ⃗ , b ⃗ , c ⃗ \vec{a}, \vec{b}, \vec{c} a , b , c n ^ \hat{n} n ^ Solve → ✓ Source and academic review↓
Question type Single correct
Exam relevance JEE Main · Mathematics
Concepts assessed Mathematics
Academic status Reviewed by official_key
Source pyq
Editorial review 9 September 2026 Quick checks
Students also ask Why was the value of a ⃗ ⋅ b ⃗ = 2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}} = 2 a ⋅ b = 2 ∣ a ⃗ ∣ = 1 |\vec{\mathrm{a}}| = 1 ∣ a ∣ = 1 The examiner provided extraneous parameters as deliberate distractors to make candidates spend time trying to compute the cross product or angle between vectors.
Answer Taking the dot product of b ⃗ \vec{\mathrm{b}} b c ⃗ \vec{\mathrm{c}} c b ⃗ ⋅ ( a ⃗ × b ⃗ ) = 0 \vec{\mathrm{b}} \cdot (\vec{\mathrm{a}} \times \vec{\mathrm{b}}) = 0 b ⋅ ( a × b ) = 0 − 3 ∣ b ⃗ ∣ 2 = − 3 ( 16 ) = − 48 -3|\vec{\mathrm{b}}|^2 = -3(16) = -48 − 3∣ b ∣ 2 = − 3 ( 16 ) = − 48
Why each option works or fails A: − 24 -24 − 24 - Believing that ∣ b ⃗ ∣ 2 = 8 |\vec{\mathrm{b}}|^2 = 8 ∣ b ∣ 2 = 8 4 4 4 4 × 2 4 \times 2 4 × 2 − 3 × 8 = − 24 -3 \times 8 = -24 − 3 × 8 = − 24 ∣ b ⃗ ∣ 2 |\vec{\mathrm{b}}|^2 ∣ b ∣ 2 4 2 = 16 4^2 = 16 4 2 = 16 − 3 ( 16 ) = − 48 -3(16) = -48 − 3 ( 16 ) = − 48 B: − 84 -84 − 84 - Attempting to compute b ⃗ ⋅ ( 2 a ⃗ × b ⃗ ) \vec{\mathrm{b}} \cdot (2\vec{\mathrm{a}} \times \vec{\mathrm{b}}) b ⋅ ( 2 a × b ) sin θ \sin\theta sin θ b ⃗ ⋅ ( a ⃗ × b ⃗ ) \vec{\mathrm{b}} \cdot (\vec{\mathrm{a}} \times \vec{\mathrm{b}}) b ⋅ ( a × b ) C · correct: − 48 -48 − 48 - None. The cross product a ⃗ × b ⃗ \vec{\mathrm{a}} \times \vec{\mathrm{b}} a × b b ⃗ \vec{\mathrm{b}} b b ⃗ ⋅ ( 2 a ⃗ × b ⃗ ) = 0 \vec{\mathrm{b}} \cdot (2\vec{\mathrm{a}} \times \vec{\mathrm{b}}) = 0 b ⋅ ( 2 a × b ) = 0 − 3 ( b ⃗ ⋅ b ⃗ ) = − 3 ( 16 ) = − 48 -3(\vec{\mathrm{b}} \cdot \vec{\mathrm{b}}) = -3(16) = -48 − 3 ( b ⋅ b ) = − 3 ( 16 ) = − 48 D: − 60 -60 − 60 - Mistakenly adding or subtracting extraneous multiples of a ⃗ ⋅ b ⃗ = 2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}} = 2 a ⋅ b = 2 b ⃗ ⋅ ( 2 a ⃗ × b ⃗ ) − 3 b ⃗ ⋅ b ⃗ = 0 − 3 ∣ b ⃗ ∣ 2 \vec{\mathrm{b}} \cdot (2\vec{\mathrm{a}} \times \vec{\mathrm{b}}) - 3\vec{\mathrm{b}} \cdot \vec{\mathrm{b}} = 0 - 3|\vec{\mathrm{b}}|^2 b ⋅ ( 2 a × b ) − 3 b ⋅ b = 0 − 3∣ b ∣ 2 Step-by-step solution given: ∣ a ⃗ ∣ = 1 |\vec{\mathrm{a}}| = 1 ∣ a ∣ = 1 ∣ b ⃗ ∣ = 4 |\vec{\mathrm{b}}| = 4 ∣ b ∣ = 4 a ⃗ ⋅ b ⃗ = 2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}} = 2 a ⋅ b = 2 c ⃗ = 2 ( a ⃗ × b ⃗ ) − 3 b ⃗ \vec{\mathrm{c}} = 2(\vec{\mathrm{a}} \times \vec{\mathrm{b}}) - 3\vec{\mathrm{b}} c = 2 ( a × b ) − 3 b goal: Find the value of the scalar product b ⃗ ⋅ c ⃗ \vec{\mathrm{b}} \cdot \vec{\mathrm{c}} b ⋅ c approach: Take the dot product of b ⃗ \vec{\mathrm{b}} b c ⃗ \vec{\mathrm{c}} c b ⃗ ⋅ c ⃗ = 2 ( a ⃗ × b ⃗ ) ⋅ b ⃗ − 3 ( b ⃗ ⋅ b ⃗ ) \vec{\mathrm{b}} \cdot \vec{\mathrm{c}} = 2(\vec{\mathrm{a}} \times \vec{\mathrm{b}}) \cdot \vec{\mathrm{b}} - 3(\vec{\mathrm{b}} \cdot \vec{\mathrm{b}}) b ⋅ c = 2 ( a × b ) ⋅ b − 3 ( b ⋅ b ) ( a ⃗ × b ⃗ ) ⋅ b ⃗ = 0 (\vec{\mathrm{a}} \times \vec{\mathrm{b}}) \cdot \vec{\mathrm{b}} = 0 ( a × b ) ⋅ b = 0 execute: Compute b ⃗ ⋅ c ⃗ = 0 − 3 ∣ b ⃗ ∣ 2 = − 3 ( 4 2 ) = − 48 \vec{\mathrm{b}} \cdot \vec{\mathrm{c}} = 0 - 3|\vec{\mathrm{b}}|^2 = -3(4^2) = -48 b ⋅ c = 0 − 3∣ b ∣ 2 = − 3 ( 4 2 ) = − 48 verify: Check that the extra information ∣ a ⃗ ∣ = 1 |\vec{\mathrm{a}}| = 1 ∣ a ∣ = 1 a ⃗ ⋅ b ⃗ = 2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}} = 2 a ⋅ b = 2 ∣ b ⃗ ∣ |\vec{\mathrm{b}}| ∣ b ∣ Shortcut: When to use it: Recognizing immediately that b ⃗ \vec{\mathrm{b}} b b ⃗ \vec{\mathrm{b}} b
quick_kill: Since ( a ⃗ × b ⃗ ) ⊥ b ⃗ (\vec{\mathrm{a}} \times \vec{\mathrm{b}}) \perp \vec{\mathrm{b}} ( a × b ) ⊥ b c ⃗ \vec{\mathrm{c}} c b ⃗ \vec{\mathrm{b}} b − 3 ∣ b ⃗ ∣ 2 = − 3 ( 16 ) = − 48 -3|\vec{\mathrm{b}}|^2 = -3(16) = -48 − 3∣ b ∣ 2 = − 3 ( 16 ) = − 48