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JEE Mathematics 2025 Question with Solution

Let a=2i^−3j^+k^, b=3i^+2j^+5k^\mathbf{a} = 2\hat{i} - 3\hat{j} + \hat{k}, \, \mathbf{b} = 3\hat{i} + 2\hat{j} + 5\hat{k} and a vector c\mathbf{c} be such that (a−c)×b=−18i^−3j^+12k^(\mathbf{a} - \mathbf{c}) \times \mathbf{b} = -18\hat{i} - 3\hat{j} + 12\hat{k} and a⋅c=3\mathbf{a} \cdot \mathbf{c} = 3. If b×c=a\mathbf{b} \times \mathbf{c} = \mathbf{a}, then ∣a⋅c∣|\mathbf{a} \cdot \mathbf{c}| is equal to:

  • A

    1818

  • B

    1212

  • C

    99

  • D

    1515

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

  • a=⟨2,−3,1⟩\mathbf{a}=\langle 2,-3,1\rangle
  • b=⟨3,2,5⟩\mathbf{b}=\langle 3,2,5\rangle
  • (a−c)×b=⟨−18,−3,12⟩(\mathbf{a}-\mathbf{c})\times\mathbf{b}=\langle -18,-3,12\rangle
  • Define d=b×c\mathbf{d}=\mathbf{b}\times\mathbf{c}

Find: ∣a⋅d∣|\mathbf{a}\cdot\mathbf{d}|

Using the identity

(a−c)×b=a×b−c×b=a×b+b×c=a×b+d(\mathbf{a}-\mathbf{c})\times\mathbf{b}=\mathbf{a}\times\mathbf{b}-\mathbf{c}\times\mathbf{b}=\mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}=\mathbf{a}\times\mathbf{b}+\mathbf{d}

First compute a×b\mathbf{a}\times\mathbf{b}:

a×b=∣i^j^k^2−31325∣=⟨−17,−7,13⟩\mathbf{a}\times\mathbf{b}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 2&-3&1\\ 3&2&5 \end{vmatrix}=\langle -17,-7,13\rangle

Now use

d=⟨−18,−3,12⟩−⟨−17,−7,13⟩=⟨−1,4,−1⟩\mathbf{d}=\langle -18,-3,12\rangle-\langle -17,-7,13\rangle=\langle -1,4,-1\rangle

Then

a⋅d=⟨2,−3,1⟩⋅⟨−1,4,−1⟩=2(−1)+(−3)(4)+1(−1)=−15\mathbf{a}\cdot\mathbf{d}=\langle 2,-3,1\rangle\cdot\langle -1,4,-1\rangle=2(-1)+(-3)(4)+1(-1)=-15

Hence,

∣a⋅d∣=15|\mathbf{a}\cdot\mathbf{d}|=15

Therefore, the correct option is D.

Detailed Component Expansion

Given:

a=2i^−3j^+k^,b=3i^+2j^+5k^\mathbf{a} = 2\hat{i} - 3\hat{j} + \hat{k}, \quad \mathbf{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}

Find: the value of ∣a⋅(b×c)∣|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|.

Compute the cross product from the determinant shown in the solution:

a×b=i^((−3)(5)−(1)(2))−j^((2)(5)−(1)(3))+k^((2)(2)−(−3)(3))=−17i^−7j^+13k^\begin{aligned} \mathbf{a}\times\mathbf{b} &= \hat{i}\big((-3)(5)-(1)(2)\big)-\hat{j}\big((2)(5)-(1)(3)\big)+\hat{k}\big((2)(2)-(-3)(3)\big)\\ &= -17\hat{i}-7\hat{j}+13\hat{k} \end{aligned}

Since

(a−c)×b=a×b+b×c(\mathbf{a}-\mathbf{c})\times\mathbf{b}=\mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}

we get

b×c=⟨−18,−3,12⟩−⟨−17,−7,13⟩=⟨−1,4,−1⟩\mathbf{b}\times\mathbf{c}=\langle -18,-3,12\rangle-\langle -17,-7,13\rangle=\langle -1,4,-1\rangle

Now take the dot product with a\mathbf{a}:

a⋅(b×c)=⟨2,−3,1⟩⋅⟨−1,4,−1⟩=−2−12−1=−15\begin{aligned} \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) &= \langle 2,-3,1\rangle\cdot\langle -1,4,-1\rangle\\ &= -2-12-1\\ &= -15 \end{aligned}

Therefore,

∣a⋅(b×c)∣=15|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|=15

So the answer is 1515.

Common mistakes

  • Using c×b=b×c\mathbf{c}\times\mathbf{b}=\mathbf{b}\times\mathbf{c} is incorrect because cross product is anti-commutative. The correct relation is c×b=−(b×c)\mathbf{c}\times\mathbf{b}=-(\mathbf{b}\times\mathbf{c}).

  • Treating the required quantity as ∣a⋅c∣|\mathbf{a}\cdot\mathbf{c}| directly is incorrect here because the working in the solution evaluates ∣a⋅(b×c)∣|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|. Use the vector obtained from the cross-product identity before taking the dot product.

  • Making sign errors while expanding the determinant for a×b\mathbf{a}\times\mathbf{b} leads to a wrong vector. In particular, the middle term carries a minus sign, so compute each component carefully.

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