Mathematics · Vector Algebra

JEE Main 2026 — 22 January, Evening Shift — Question 8

Let a⃗=2i^−j^+k^\vec{a}=2 \hat{i}-\hat{j}+\hat{k} and b⃗=λj^+2k^,λ∈Z\vec{b}=\lambda \hat{j}+2 \hat{k}, \lambda \in Z be two vectors, Let c→=a→×b→\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} and d→\overrightarrow{\mathrm{d}} be a vector of magnitude 2 in yz-plane. If ∣c→∣=53|\overrightarrow{\mathrm{c}}|=\sqrt{53}, then the maximum possible value of (c→⋅d→)2(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}})^{2} is equal to :

  1. Option A:

    2626

  2. Option B:

    104104

  3. Option C:

    208208

    Correct
  4. Option D:

    5252

Answer: C

Step-by-step solution

a⃗=2i^−j^+k^\vec{a}=2 \hat{i}-\hat{j}+\hat{k}

b→=λj^+2k^;λ∈Z\overrightarrow{\mathrm{b}}=\lambda \hat{\mathrm{j}}+2 \hat{\mathrm{k}} ; \lambda \in \mathrm{Z}

c→=a→×b→=(−2−λ)i^−4j^+2λk^\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=(-2-\lambda) \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+2 \lambda \hat{\mathrm{k}}

∣c→∣=53|\overrightarrow{\mathrm{c}}|=\sqrt{53} ⇒5λ2+4λ−33=0\Rightarrow 5 \lambda^{2}+4 \lambda-33=0

λ=2.2\lambda=2.2 or -3 ⇒λ=−3\Rightarrow \lambda=-3

c→=i^−4j^−6k^\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}

let d⃗=yj^+zk^\vec{d}=y \hat{j}+z \hat{k}

∣d→∣=2|\overrightarrow{\mathrm{d}}|=2

⇒y2+z2=4\Rightarrow \mathrm{y}^{2}+\mathrm{z}^{2}=4

(c⃗⋅d⃗)2=(4y+6z)2≤(42+62×y2+z2)2≤208(\vec{c} \cdot \vec{d})^{2}=(4 y+6 z)^{2} \leq\left(\sqrt{4^{2}+6^{2}} \times \sqrt{y^{2}+z^{2}}\right)^{2} \leq 208

Answer key and solution verified before publishing.

Practise Vector Algebra

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2026
Subject
Mathematics
Chapter
Vector Algebra
Topic
Vector or Cross Product of Two Vectors