Let $\vec{a}=6\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}$. If $\vec{c}$ is a vector such that $|\vec{c}| \geq 6, \vec{a} \cdot \vec{c}=6|\vec{c}|, |\vec{c}-\vec{a}|=2\sqrt{2}$ and the angle between $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ is $60^{\circ}$ then $|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}|$ is equal to

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Vector Algebra

Let

a = 6 \hat{i} + \hat{j} - \hat{k}

and

b = \hat{i} + \hat{j}

. If

c

is a vector such that

∣ c ∣ \geq 6, a \cdot c = 6∣ c ∣, ∣ c - a ∣ = 22

and the angle between

a \times b

and

c

6 0^{\circ}

then

∣ (a \times b) \times c ∣

is equal to

Answer: (C)

a \times b = (6, 1, - 1) \times (1, 1, 0) = (1, - 1, 5)

∣ a \times b ∣ = 33

;

∣ a ∣^{2} = 38

. From

∣ c - a ∣^{2} = ∣ c ∣^{2} - 2 a \cdot c + ∣ a ∣^{2} = 8

and

a \cdot c = 6∣ c ∣

∣ c ∣^{2} - 12∣ c ∣ + 30 = 0 \Rightarrow ∣ c ∣ = 6 \pm 6

. Since

∣ c ∣ \geq 6

, take

∣ c ∣ = 6 + 6

. Then

∣ (a \times b) \times c ∣ = ∣ a \times b ∣ ∣ c ∣ sin 6 0^{\circ} = 33 (6 + 6) \cdot \frac{3}{2} = \frac{9}{2} (6 + 6)

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option