Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\vec{b}=\hat{i}+\hat{j}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c}=|\vec{c}|,|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $(\vec{a} \times \vec{b})$ and $\vec{c}$ is $\frac{\pi}{6}$, then the value of $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is :

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Vector Algebra

Let

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

and

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

. If

c

is a vector such that

a \cdot c = ∣ c ∣, ∣ c - a ∣ = 22

and the angle between

(a \times b)

and

c

\frac{π}{6}

, then the value of

∣ (a \times b) \times c ∣

is :

Answer: (C)

Expanding

∣ c - a ∣^{2} = 8

with

∣ a ∣ = 3

and

a \cdot c = ∣ c ∣

∣ c ∣^{2} - 2 a \cdot c + ∣ a ∣^{2} = 8 \Rightarrow ∣ c ∣^{2} - 2∣ c ∣ + 9 = 8 \Rightarrow (∣ c ∣ - 1)^{2} = 0

∣ c ∣ = 1

. The cross product is

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

, whose magnitude is

3

. With the angle between

a \times b

and

c

equal to

\frac{π}{6}

(a \times b) \times c = a \times b ∣ c ∣ sin \frac{π}{6} = 3 (1) (\frac{1}{2}) = \frac{3}{2}

so the magnitude of the triple cross product is

\frac{3}{2}

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option