Consider three vectors \(\vec{a}, \vec{b}, \vec{c}\). Let \(|\vec{a}|=2, |\vec{b}|=3\) and \(\vec{a}=\vec{b} \times \vec{c}\). If \(\alpha \in\left[0, \frac{\pi}{3}\right]\) is the angle between the vectors \(\vec{b}\) and \(\vec{c}\), then the minimum value of \(27|\vec{c}-\vec{a}|^{2}\) is equal to

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Vector Algebra

Consider three vectors

a, b, c

. Let

∣ a ∣ = 2, ∣ b ∣ = 3

and

a = b \times c

. If

α \in [0, \frac{π}{3}]

is the angle between the vectors

b

and

c

, then the minimum value of

27∣ c - a ∣^{2}

is equal to

Answer: (B)

c - a = c - (b \times c)

. Since

a = b \times c ⊥ c

∣ c - a ∣^{2} = ∣ c ∣^{2} + ∣ a ∣^{2} - 2 a \cdot c = ∣ c ∣^{2} + 4

With

∣ a ∣ = ∣ b ∣∣ c ∣ sin α ⟹ 2 = 3∣ c ∣ sin α ⟹ ∣ c ∣ = \frac{2}{3 sin α}

27∣ c - a ∣^{2} = 27 (\frac{4}{9 sin ^{2} α} + 4) = \frac{12}{sin ^{2} α} + 108 = 12 csc^{2} α + 108

For

α \in [0, \frac{π}{3}]

csc^{2} α

is minimised at

α = \frac{π}{3}

, where

csc^{2} α = \frac{4}{3}

27∣ c - a ∣_{min}^{2} = 12 (\frac{4}{3}) + 108 = 16 + 108 = 124

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Single correct option