Let $\vec{a}=\alpha \hat{i}+2 \hat{j}-\hat{k}$ and $\vec{b}=-2 \hat{i}+\alpha \hat{j}+\hat{k}$, where $\alpha \in \mathbf{R}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $\sqrt{15\left(\alpha^{2}+4\right)}$, then the value of $2|\overrightarrow{\mathrm{a}}|^{2}+(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})|\overrightarrow{\mathrm{b}}|^{2}$ is equal to

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Vector Algebra

Let

a = α \hat{i} + 2 \hat{j} - \hat{k}

and

b = - 2 \hat{i} + α \hat{j} + \hat{k}

, where

α \in R

. If the area of the parallelogram whose adjacent sides are represented by the vectors

a

and

b

15 (α^{2} + 4)

, then the value of

2∣ a ∣^{2} + (a \cdot b) ∣ b ∣^{2}

is equal to

Answer: (D)

The area of the parallelogram is

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

. Given this equals

15 (α^{2} + 4)

, we have

2 (α^{2} + 4) + (α^{2} + 4)^{2} = 15 (α^{2} + 4)

. Simplifying:

(α^{2} + 4)^{2} = 13 (α^{2} + 4)

, so

α^{2} + 4 = 13

, giving

α^{2} = 9

. Now

∣ a ∣^{2} = α^{2} + 5 = 14

∣ b ∣^{2} = α^{2} + 5 = 14

, and

a \cdot b = - 2 α + 2 α - 1 = - 1

. Therefore,

2∣ a ∣^{2} + (a \cdot b) ∣ b ∣^{2} = 2 (14) + (- 1) (14) = 28 - 14 = 14

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option