Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonal of a parallelogram having area $2\sqrt{2}$. Let the angle between $\vec{a}$ and $\vec{b}$ be acute. $|\vec{a}|=1$ and $|\vec{a}\cdot\vec{b}|=|\vec{a}\times\vec{b}|$. If $\vec{c}=2\sqrt{2}(\vec{a}\times\vec{b})-2\vec{b}$, then an angle between $\vec{b}$ and $\vec{c}$ is :

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Vector Algebra

Let

a

and

b

be the vectors along the diagonal of a parallelogram having area

22

. Let the angle between

a

and

b

be acute.

∣ a ∣ = 1

and

∣ a \cdot b ∣ = ∣ a \times b ∣

. If

c = 22 (a \times b) - 2 b

, then an angle between

b

and

c

is :

Answer: (D)

The area of the parallelogram formed by diagonals is

\frac{1}{2} ∣ a \times b ∣ = 22

, so

∣ a \times b ∣ = 42

.

Given

∣ a \cdot b ∣ = ∣ a \times b ∣

, we have

∣ a ∣∣ b ∣∣ cos θ ∣ = ∣ a ∣∣ b ∣∣ sin θ ∣

, which gives

∣ cos θ ∣ = ∣ sin θ ∣

. Since the angle is acute,

θ = \frac{π}{4}

.

From

∣ a ∣∣ b ∣ sin \frac{π}{4} = 42

and

∣ a ∣ = 1

, we get

∣ b ∣ = 8

.

Now

c = 22 (a \times b) - 2 b

. Since

a \times b

is perpendicular to

b

∣ c ∣ = (22)^{2} ∣ a \times b ∣^{2} + 4∣ b ∣^{2} = 8 \cdot 32 + 4 \cdot 64 = 162

Also,

b \cdot c = - 2∣ b ∣^{2} = - 128

.

Therefore,

cos α = \frac{b \cdot c}{∣ b ∣∣ c ∣} = \frac{- 128}{8 \cdot 16 2} = - \frac{1}{2}

, giving

α = \frac{3 π}{4}

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option