Vectors & 3D GeometrymediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Vector Combination Dot Products: |c|^2 = 12 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let the vectors a⃗=−i^+j^+3k^\vec a=-\hat i+\hat j+3\hat k and b⃗=i^+3j^+k^\vec b=\hat i+3\hat j+\hat k. For some λ,μ∈R\lambda,\mu\in\mathbb{R}, let c⃗=λa⃗+μb⃗\vec c=\lambda\vec a+\mu\vec b. If c⃗⋅(3i^−6j^+2k^)=10\vec c\cdot(3\hat i-6\hat j+2\hat k)=10 and c⃗⋅(i^+j^+k^)=−2\vec c\cdot(\hat i+\hat j+\hat k)=-2, then ∣c⃗∣2|\vec c|^2 is equal to
A88
B1212correct
C1414
D1515
Solution
Step 1: Write c⃗\vec c in components:
c⃗=λa⃗+μb⃗=(μ−λ)i^+(λ+3μ)j^+(3λ+μ)k^.\vec c=\lambda\vec a+\mu\vec b=(\mu-\lambda)\hat i+(\lambda+3\mu)\hat j+(3\lambda+\mu)\hat k.
 Step 2: Apply c⃗⋅(3i^−6j^+2k^)=10\vec c\cdot(3\hat i-6\hat j+2\hat k)=10:
3(μ−λ)−6(λ+3μ)+2(3λ+μ)=10 ⇒ −3λ−13μ=10.(1)3(\mu-\lambda)-6(\lambda+3\mu)+2(3\lambda+\mu)=10\ \Rightarrow\ -3\lambda-13\mu=10.\quad(1)
 Step 3: Apply c⃗⋅(i^+j^+k^)=−2\vec c\cdot(\hat i+\hat j+\hat k)=-2:
(μ−λ)+(λ+3μ)+(3λ+μ)=−2 ⇒ 3λ+5μ=−2.(2)(\mu-\lambda)+(\lambda+3\mu)+(3\lambda+\mu)=-2\ \Rightarrow\ 3\lambda+5\mu=-2.\quad(2)
 Step 4: Solve (1) and (2): adding, −8μ=8⇒μ=−1-8\mu=8\Rightarrow\mu=-1; then 3λ−5=−2⇒λ=13\lambda-5=-2\Rightarrow\lambda=1. Step 5: So c⃗=(−1−1)i^+(1−3)j^+(3−1)k^=−2i^−2j^+2k^\vec c=(-1-1)\hat i+(1-3)\hat j+(3-1)\hat k=-2\hat i-2\hat j+2\hat k, and
∣c⃗∣2=(−2)2+(−2)2+22=12.|\vec c|^2=(-2)^2+(-2)^2+2^2=12.
 Correct answer: (2)
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