Let $\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c}=\vec{c} \times \vec{b}$ and $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=168$. Then the maximum value of $|\overrightarrow{\mathrm{c}}|^{2}$ is :

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Vector Algebra

Let

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

and

c

be a vector such that

a \times c = c \times b

and

(a + c) \cdot (b + c) = 168

. Then the maximum value of

∣ c ∣^{2}

is :

Answer: (C)

From

a \times c = c \times b

, i.e.

a \times c + b \times c = 0

, we get

(a + b) \times c = 0

, so

c

is parallel to

a + b = (5, - 6, 4)

; write

c = λ (5, - 6, 4)

, so

∣ c ∣^{2} = 77 λ^{2}

.

Expanding the given dot product:

(a + c) \cdot (b + c) = a \cdot b + c \cdot (a + b) + ∣ c ∣^{2} .

With

a \cdot b = 2 (3) + (- 1) (- 5) + 3 (1) = 14

and

c \cdot (a + b) = 77 λ

14 + 77 λ + 77 λ^{2} = 168 ⟹ λ^{2} + λ - 2 = 0 ⟹ λ = 1 or - 2.

The larger

∣ c ∣^{2} = 77 λ^{2}

occurs at

λ = - 2

∣ c ∣^{2} = 77 (4) = 308.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option