Let a = αi + 2j - k and b = -2i + αj + k, where α ∈ R. If the area of the parallelogram whose adjace

Q: Wait, why do we square the area? Can't we just work with the square root?

Working with square roots is a recipe for algebraic errors and unnecessary complexity. Squaring both sides of the equation $|\vec{a} \times \vec{b}| = \sqrt{15(\alpha^2 + 4)}$ transforms the problem into a polynomial equation, which is much easier to handle. It is a standard, professional strategy in JEE problems to eliminate radicals early to keep your workspace clean and your logic sharp.

Q: Is it safe to divide by $(\alpha^2 + 4)$? What if it equals zero?

That is a brilliant question! In mathematics, we must always check for division by zero. However, since $\alpha$ is a real number, $\alpha^2$ is always non-negative. Therefore, $\alpha^2 + 4$ is at least $4$, which is strictly greater than zero. We are perfectly safe to divide, and doing so reveals the elegant structure of the problem.

Q: Why did the $\alpha$ terms cancel out in the dot product calculation?

It is not a coincidence! The problem is designed with symmetry. The vectors $\vec{a}$ and $\vec{b}$ are constructed such that their dot product simplifies to a constant. This is a common feature in well-crafted JEE problems—the complexity often collapses into elegance at the final step. Trust the process, and the math will reward you.

Question

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

Options:

(A) 10

(B) 7

(C) 9

(D) 14

Sigma Percentile · Accepted Answer

### The Symphony of Vectors: A Journey into Geometric Elegance

My dear student, welcome to a problem that is not merely a calculation, but a masterclass in the elegance of vector algebra. When you first look at vectors $\vec{a} = \alpha\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -2\hat{i} + \alpha\hat{j} + \hat{k}$, you might feel a slight hesitation.

There is an unknown parameter, $\alpha$, lurking in the components. But I want you to take a deep breath. In the world of JEE Advanced, variables are not obstacles; they are invitations to find hidden symmetries.

### Phase 1: The Geometric Foundation

The problem asks us to consider the area of a parallelogram formed by these two vectors. The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product: $	ext{Area} = |\vec{a} 	imes \vec{b}|$.

We are given that this area equals $\sqrt{15(\alpha^2 + 4)}$. Immediately, our intuition tells us that working with square roots is cumbersome.

Let us square both sides to work with the square of the area:
$$	ext{Area}^2 = |\vec{a} 	imes \vec{b}|^2 = 15(\alpha^2 + 4)$$
This is our North Star. Everything we do from here is to reach this equality.

### Phase 2: The Determinant Dance

Now, we must compute the cross product $\vec{a} 	imes \vec{b}$. We set up our determinant:
$$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \alpha & 2 & -1 \ -2 & \alpha & 1 \end{vmatrix}$$

As we expand this, we must be meticulous. For the $\hat{i}$ component, we have $(2 - (-\alpha)) = \alpha + 2$. For the $\hat{j}$ component, we have $-(\alpha - 2) = -(\alpha - 2)$.

For the $\hat{k}$ component, we have $(\alpha^2 - (-4)) = \alpha^2 + 4$. Thus, our cross product vector is $(\alpha + 2)\hat{i} - (\alpha - 2)\hat{j} + (\alpha^2 + 4)\hat{k}$.

### Phase 3: The Algebraic Shortcut

Here is where the magic happens. We need the square of the magnitude:
$$|\vec{a} 	imes \vec{b}|^2 = (\alpha + 2)^2 + (\alpha - 2)^2 + (\alpha^2 + 4)^2$$

A novice would expand every single term, risking a sign error. But you are a JEE aspirant; you look for patterns. Notice the first two terms: $(\alpha + 2)^2 + (\alpha - 2)^2$.

This is the classic identity $(x+y)^2 + (x-y)^2 = 2(x^2 + y^2)$. Applying this, the expression simplifies instantly to $2(\alpha^2 + 4)$. Our total area squared is now $2(\alpha^2 + 4) + (\alpha^2 + 4)^2$.

### Phase 4: The Revelation

We equate our result to the given area squared:
$$2(\alpha^2 + 4) + (\alpha^2 + 4)^2 = 15(\alpha^2 + 4)$$

Since $\alpha^2 + 4$ is strictly positive, we divide both sides by it. We are left with $2 + (\alpha^2 + 4) = 15$.

This simplifies to $\alpha^2 + 6 = 15$, which means $\alpha^2 = 9$. The fog has lifted! We have found the value of $\alpha^2$.

### Phase 5: The Final Evaluation

With $\alpha^2 = 9$, the rest is a victory lap. We need to calculate $2|\vec{a}|^2 + (\vec{a} \cdot \vec{b})|\vec{b}|^2$.

First, $|\vec{a}|^2 = \alpha^2 + 2^2 + (-1)^2 = \alpha^2 + 5 = 9 + 5 = 14$. Similarly, $|\vec{b}|^2 = (-2)^2 + \alpha^2 + 1^2 = 14$.

Now, the dot product:
$$\vec{a} \cdot \vec{b} = (\alpha)(-2) + (2)(\alpha) + (-1)(1) = -2\alpha + 2\alpha - 1 = -1$$

The $\alpha$ terms vanish, leaving us with a constant. Finally, we compute $2(14) + (-1)(14) = 28 - 14 = 14$.

The answer is **14**. By staying calm and looking for the structure within the algebra, we turned a daunting problem into a beautiful, logical sequence.

Select Answer:

Visualized Solution

Defining the Vectors $a$ and $b$

The Area of a Parallelogram Formula

Setting up the Cross Product $a \times b$

Expanding the Determinant

Finding the Magnitude Squared $∣ a \times b ∣^{2}$

Simplifying the Magnitude Squared

Equating to the Given Area

Solving for $α^{2}$

Calculating $∣ a ∣^{2}$ and $∣ b ∣^{2}$

Calculating the Dot Product $a \cdot b$

Final Evaluation of the Expression

The Sigma Insight: Vector (Cross) Product

The Symphony of Vectors

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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