Let a = i + αj + 3k and b = 3i - αj + k. If the area of the parallelogram whose adjacent sides are r

Q: Why do we use the cross product for area? Can't we just use base times height?

You absolutely can use base times height, but in 3D space, finding the height of a parallelogram defined by vectors is often more complex than calculating the cross product. The cross product is a powerful tool because it encodes both the orientation and the magnitude of the area in one operation. The magnitude of the cross product $|\vec{a} \times \vec{b}|$ is mathematically equivalent to the area of the parallelogram, and it saves us from having to perform complex trigonometry to find the height.

Q: Do I need to find $\alpha$ or just $\alpha^2$? I'm worried about the sign.

This is the mark of a seasoned JEE aspirant! You noticed that the dot product $\vec{a} \cdot \vec{b}$ only depends on $\alpha^2$. Solving for $\alpha$ (which would be $\pm 2$) is unnecessary work. Always look at the target expression before diving into calculations. If the target expression only contains $\alpha^2$, you can stop as soon as you find that value. It saves time and reduces the chance of a sign error.

Q: What if the area was zero? Does that change the approach?

If the area is zero, it implies that the magnitude of the cross product is zero, which means $\vec{a} \times \vec{b} = \vec{0}$. Geometrically, this tells us that the vectors $\vec{a}$ and $\vec{b}$ are collinear (parallel). While our problem gives a non-zero area, keeping this in mind is a great sanity check. If you ever calculate an area of zero, you know your vectors are linearly dependent!

Question

Let $\vec{a} = \hat{i} + \alpha\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - \alpha\hat{j} + \hat{k}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $8\sqrt{3}$ square units, then $\vec{a} \cdot \vec{b}$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. You have two vectors, $\vec{a} = \hat{i} + \alpha\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - \alpha\hat{j} + \hat{k}$, originating from the origin.

These vectors are not just lines; they are the adjacent sides of a parallelogram floating in space. The problem asks us to find the dot product $\vec{a} \cdot \vec{b}$, given that the area of this parallelogram is $8\sqrt{3}$.

### The Geometric Bridge

The first step is to connect the physical area to our algebraic vectors. In vector algebra, the area of a parallelogram formed by two vectors $\vec{a}$ and $\vec{b}$ is defined by the magnitude of their cross product:

$$Area = |\vec{a} 	imes \vec{b}|$$

We are given that this area is $8\sqrt{3}$. Therefore, our mission is to compute the cross product, find its magnitude, and equate it to $8\sqrt{3}$.

### The Determinant Dance

To find the cross product $\vec{a} 	imes \vec{b}$, we use the determinant method. We set up the matrix as follows:

$$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & \alpha & 3 \ 3 & -\alpha & 1 \end{vmatrix}$$

Expanding along the first row:

For the $\hat{i}$ component: $(\alpha)(1) - (3)(-\alpha) = \alpha + 3\alpha = 4\alpha$.

For the $\hat{j}$ component: $-[(1)(1) - (3)(3)] = -[1 - 9] = 8$.

For the $\hat{k}$ component: $(1)(-\alpha) - (\alpha)(3) = -\alpha - 3\alpha = -4\alpha$.

Thus, our cross product vector is:

$$\vec{a} 	imes \vec{b} = 4\alpha\hat{i} + 8\hat{j} - 4\alpha\hat{k}$$

### The Algebraic Climax

Now that we have the vector, we calculate its magnitude squared:

$$|\vec{a} 	imes \vec{b}|^2 = (4\alpha)^2 + (8)^2 + (-4\alpha)^2 = 16\alpha^2 + 64 + 16\alpha^2 = 32\alpha^2 + 64$$

We know the area is $8\sqrt{3}$, so the magnitude squared is $(8\sqrt{3})^2 = 64 	imes 3 = 192$. Equating the two:

$$32\alpha^2 + 64 = 192$$

Subtracting $64$ from both sides gives $32\alpha^2 = 128$. Dividing by $32$, we find:

$$\alpha^2 = 4$$

### Final Calculation

The question asks for the dot product $\vec{a} \cdot \vec{b}$. Calculating this:

$$\vec{a} \cdot \vec{b} = (1)(3) + (\alpha)(-\alpha) + (3)(1) = 3 - \alpha^2 + 3 = 6 - \alpha^2$$

We do not need to find $\alpha$ itself, as we only need $\alpha^2$. Substituting $\alpha^2 = 4$ into the expression:

$$\vec{a} \cdot \vec{b} = 6 - 4 = 2$$

The final answer is **2**.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + α \hat{j} + 3 \hat{k}$ and $b = 3 \hat{i} - α \hat{j} + \hat{k}$ . If the area of the parallelogram whose adjacent sides are represented by the vectors $a$ and $b$ is $83$ square units, then $a \cdot b$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors

The Area Formula

Setting up the Cross Product

Expanding the Determinant: $\hat{i}$ component

Expanding the Determinant: $\hat{j}$ component

Expanding the Determinant: $\hat{k}$ component

The Cross Product Vector

Calculating Magnitude Squared

Equating to Given Area

Solving for $α^{2}$

Finding the Dot Product

Final Answer

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Geometric Bridge

The Determinant Dance

The Algebraic Climax

Final Calculation

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