Let $\hat{a}, \hat{b}$ be unit vectors. If $\vec{c}$ be a vector such that the angle between $\hat{\mathrm{a}}$ and $\overrightarrow{\mathrm{c}}$ is $\frac{\pi}{12}$, and $\hat{\mathrm{b}}=\overrightarrow{\mathrm{c}}+2(\overrightarrow{\mathrm{c}} \times \hat{\mathrm{a}})$, then $|6 \overrightarrow{\mathrm{c}}|^{2}$ is equal to

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Vector Algebra

Let

\overset{a}{^}, \hat{b}

be unit vectors. If

c

be a vector such that the angle between

\overset{a}{^}

and

c

\frac{π}{12}

, and

\hat{b} = c + 2 (c \times \overset{a}{^})

, then

∣6 c ∣^{2}

is equal to

Answer: (C)

Taking magnitude squared of

\hat{b} = c + 2 (c \times \overset{a}{^})

∣ \hat{b} ∣^{2} = ∣ c ∣^{2} + 4∣ c \times \overset{a}{^} ∣^{2} + 4 c \cdot (c \times \overset{a}{^})

Since

c ⊥ (c \times \overset{a}{^})

, the last term is zero. Also

∣ c \times \overset{a}{^} ∣ = ∣ c ∣ sin \frac{π}{12}

1 = ∣ c ∣^{2} + 4∣ c ∣^{2} sin^{2} \frac{π}{12}

With

sin \frac{π}{12} = \frac{3 - 1}{2 2}

, we get

sin^{2} \frac{π}{12} = \frac{4 - 2 3}{8} = \frac{2 - 3}{4}

1 = ∣ c ∣^{2} (1 + 2 - 3) = ∣ c ∣^{2} (3 - 3)

Thus

∣ c ∣^{2} = \frac{1}{3 - 3} = \frac{3 + 3}{6}

, so

36∣ c ∣^{2} = 6 (3 + 3)

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option