Let \(\vec{a} = 2\hat{i} - \hat{j} + \hat{k}\) and \(\vec{b} = \lambda\hat{j} + 2\hat{k}\), \(\lambda \in \mathbb{Z}\) be two vectors. Let \(\vec{c} = \vec{a} \times \vec{b}\) and \(\vec{d}\) be a vector of magnitude 2 in yz-plane. If \(|\vec{c}| = \sqrt{53}\), then the maximum possible value of \((\vec{c} \cdot \vec{d})^2\) is equal to :

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Vector Algebra

Let

a = 2 \hat{i} - \hat{j} + \hat{k}

and

b = λ \hat{j} + 2 \hat{k}

λ \in Z

be two vectors. Let

c = a \times b

and

d

be a vector of magnitude 2 in yz-plane. If

∣ c ∣ = 53

, then the maximum possible value of

(c \cdot d)^{2}

is equal to :

Answer: (C)

The cross product is

c = a \times b = (- 2 - λ) \hat{i} - 4 \hat{j} + 2 λ \hat{k}

. From

∣ c ∣^{2} = 53

(2 + λ)^{2} + 16 + 4 λ^{2} = 53 \Rightarrow 5 λ^{2} + 4 λ - 33 = 0,

whose integer root is

λ = - 3

, giving

c = \hat{i} - 4 \hat{j} - 6 \hat{k}

. Writing

d = y \hat{j} + z \hat{k}

with

y^{2} + z^{2} = 4

, the

\hat{i}

-component of

c

does not contribute, so

c \cdot d = - 4 y - 6 z

. By the Cauchy\textendash Schwarz inequality,

(c \cdot d)^{2} \leq (4^{2} + 6^{2}) (y^{2} + z^{2}) = 52 \times 4 = 208,

and the bound is attained. The maximum possible value of

(c \cdot d)^{2}

208

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option