Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$, where $|\vec{a}|=4,|\vec{b}|=3$ $\theta \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right)$. Then $|(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})|^{2}+4(\vec{a} \cdot \vec{b})^{2}$ is equal to ______

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Vector Algebra

Let

θ

be the angle between the vectors

a

and

b

,
where

∣ a ∣ = 4, ∣ b ∣ = 3

θ \in (\frac{π}{4}, \frac{π}{3})

. Then

∣ (a - b) \times (a + b) ∣^{2} + 4 (a \cdot b)^{2}

is equal to ______

Answer: 576

Expand the cross product:

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

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

∣ (a - b) \times (a + b) ∣^{2} = 4∣ a \times b ∣^{2} = 4∣ a ∣^{2} ∣ b ∣^{2} sin^{2} θ

Also

(a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2} cos^{2} θ

Therefore:

∣ (a - b) \times (a + b) ∣^{2} + 4 (a \cdot b)^{2}

= 4∣ a ∣^{2} ∣ b ∣^{2} sin^{2} θ + 4∣ a ∣^{2} ∣ b ∣^{2} cos^{2} θ

= 4∣ a ∣^{2} ∣ b ∣^{2} (sin^{2} θ + cos^{2} θ)

= 4∣ a ∣^{2} ∣ b ∣^{2} = 4 \times 16 \times 9 = 576

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer