- JEE Main 2026
- 08 April Evening
- Q14
- MCQ
- 4 marks
Vector Triple Product Q14
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Quick Summary
The equation 2(a×b)+3(b×c)=0 rearranges to b×c = −(2/3)(a×b), a fully known vector V. Any vector equation b×c=V has general solution c = (V×b)/|b|² + t·b, for a free scalar t (valid since b·V=0 automatically here). Plugging in a=4i−j+3k, b=10i+2j−k and computing a×b=(−5,34,18) gives V=(10/3,−68/3,−12), and then (V×b)/|b|²=(4/9,−10/9,20/9). The extra condition a·c=15 pins down t=7/45 exactly. With c fully known, c·(i+j−3k) works out to −5.
Question Replay
Let a = 4î − ĵ + 3k̂, b = 10î + 2ĵ − k̂ and a vector c be such that 2(a × b) + 3(b × c) = 0. If a · c = 15, then c · (î + ĵ − 3k̂) is equal to:
Options
Correct Answer
Correct Answer: Option (B) — −5
Why This Answer
A vector cross-product equation b×c=V does NOT determine c uniquely — c can slide freely along the direction of b without changing b×c (since b×b=0). That is exactly why the question supplies a SECOND, independent piece of information: a·c=15. Together, the two conditions pin c down completely: the cross-product equation fixes the component of c perpendicular to b, and the dot-product condition fixes how far along b the vector sits.
The most common slip here is in the vector-triple-product direction: writing the particular solution as (b×V)/|b|² instead of (V×b)/|b|² — these differ by a sign, and the wrong choice flips the final answer’s sign pattern entirely. Always verify with the identity b×(V×b) = V|b|² − b(b·V), which equals V|b|² exactly when b·V=0 (always true here, since a cross product is perpendicular to both its factors).
Core Concept
General solution of b × x = V. Given b×x=V with b·V=0 (a necessary consistency condition), every solution has the form x = (V×b)/|b|² + t·b, where t is any real number. The first term is one particular solution (verify: b×[(V×b)/|b|²] = [V|b|²−b(b·V)]/|b|² = V when b·V=0); the second term t·b adds nothing to the cross product since b×b=0.
Pinning the free parameter. A single extra scalar equation involving x (here, a·c=15) becomes a linear equation in t as long as a·b≠0, giving exactly one value of t and hence one fully determined c.
2(a×b)+3(b×c)=0 ⇒ b×c = −(2/3)(a×b) = (10/3, −68/3, −12) = V
Check: b·V = 10(10/3)+2(−68/3)+(−1)(−12) = −12+12 = 0 ✓ (consistent)
c = (V×b)/|b|² + t·b, |b|²=105
(V×b) = (−140/3, 350/3, −700/3) ⇒ (V×b)/|b|² = (4/9, −10/9, 20/9)
Step-by-Step Solution
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Compute a×b. a=(4,−1,3), b=(10,2,−1):
a×b = ((−1)(−1)−(3)(2), −[(4)(−1)−(3)(10)], (4)(2)−(−1)(10)) = (−5, 34, 18) -
Find V = b×c from the given equation: 3(b×c) = −2(a×b) ⇒ V = (10/3, −68/3, −12). Check b·V=0 ✓.
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Find the particular solution c₀ = (V×b)/|b|², with |b|²=100+4+1=105:
V×b = (−140/3, 350/3, −700/3)
c0 = V×b / 105 = (4/9, −10/9, 20/9) -
Write c = c₀ + t·b and impose a·c=15 (using a·b=4·10+(−1)·2+3·(−1)=35):
a·c0 = 4(4/9)+(−1)(−10/9)+3(20/9) = 86/9
a·c = 86/9 + 35t = 15 ⇒ 35t = 49/9 ⇒ t = 7/45 -
Compute c·(i+j−3k) using w=(1,1,−3), c0·w=−22/3, b·w=15:
c·w = c0·w + t(b·w) = −22/3 + 15(7/45) = −22/3 + 7/3 = −15/3 = −5
Final Answer
Final Answer: Option (B) — −5
c·(î+ĵ−3k̂) = −5. In the exam, the moment you see a cross-product equation for an unknown vector plus one extra scalar condition, reach for the “general solution + pin the free parameter” template — it turns an under-determined-looking vector problem into two clean linear-algebra steps.
Key Facts to Remember
- b×x=V alone never determines x uniquely — x is free to shift along b.
- General solution: x = (V×b)/|b|² + t·b, valid when b·V=0.
- Vector triple product identity: b×(V×b) = V|b|² − b(b·V); use it to verify the particular solution.
- One extra scalar (dot-product) equation is exactly enough to fix the free parameter t, provided its coefficient (a·b here) is nonzero.
Common Mistakes
Mistake 1: Using (b×V)/|b|² instead of (V×b)/|b|² as the particular solution — these are negatives of each other.
Fix: Always sanity-check by expanding b×[(V×b)/|b|²] via the triple-product identity and confirming it equals V, not −V, before proceeding.
Mistake 2: Forgetting the free parameter t·b entirely and treating the particular solution as the full answer for c.
Fix: A cross-product equation has one dimension of freedom (along b) that only a genuinely independent scalar condition like a·c=15 can remove — always add + t·b and solve for t using the extra given data.
FAQs
Q1. How do you solve a vector equation of the form b × c = V for c?
The general solution is c = (V × b)/|b|² + t·b for any scalar t, provided b·V=0 (a necessary consistency check, since a cross product is always perpendicular to both operands). The free parameter t is then pinned down by any additional scalar condition, such as a dot-product constraint.
Q2. What is the correct answer to this question?
c · (î+ĵ−3k̂) = −5, option (2). The vector c is fully determined by combining the cross-product equation with a·c=15.
Q3. Why is the vector-triple-product order (V × b) and not (b × V)?
Because b×(V×b) expands via the vector triple product identity to V|b|² − b(b·V) = V|b|² when b·V=0. Using (b×V) instead would produce −V, the wrong sign, and is a common slip.
Q4. Why does a·c=15 uniquely fix the free parameter t?
Because a·b (=35 here) is nonzero, so a·c = a·(particular solution) + t(a·b) is a linear equation in t with a nonzero coefficient, giving exactly one solution for t.
Q5. How many marks is this question worth, and how long should it take?
It is a 4-mark MCQ with -1 negative marking. With the general cross-product-equation formula memorised, it is a 3-4 minute question given the amount of arithmetic.
Prerequisites
Before practising this question type, make sure the following are in place:
- Vector triple product identity A×(B×A) = B|A|² − A(A·B).
- Solving b×x=V for x, including the free-parameter structure and consistency check b·V=0.
- Cross and dot product component computations in 3D.
Revise these from the Vector Algebra chapter, then attempt the related questions below.
Solved by Nishant Kumar Gupta for padholikhojee.in · JEE Main 2026 08 April Evening · Last updated: 2026-09-02 · Verified against the official question paper and answer key.