Let A(2, 3, 5) and C(-3, 4, -2) be opposite vertices of a parallelogram ABCD if the diagonal BD = i

Q: Wait, why are we using the cross product of diagonals? Isn't the area just base times height?

That is a great question! While base times height is the standard definition, in 3D vector geometry, the cross product of the diagonals $\vec{d_1} \times \vec{d_2}$ gives a vector whose magnitude is twice the area of the parallelogram. It is a powerful shortcut that saves us from finding the side lengths and the angle between them!

Q: Does the order of subtraction in $\vec{AC} = C - A$ matter?

It matters for the direction of the vector, but since we are taking the magnitude of the cross product, the sign will be squared away. However, it is best practice to be consistent: always define your vector as (Terminal Point - Initial Point).

Q: Why is there a factor of 1/2 in the formula?

Think of the parallelogram as being composed of four smaller triangles formed by the intersection of the diagonals. The cross product $\vec{d_1} \times \vec{d_2}$ actually calculates the area of a parallelogram formed by the diagonals themselves, which is twice the area of the original shape. The 1/2 is the scaling factor that brings us back to the area of the original parallelogram.

Q: What if I accidentally swapped the order of the cross product, like $\vec{BD} \times \vec{AC}$?

Don't worry! The cross product is anti-commutative, meaning $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$. Since you are taking the magnitude (the absolute value), the negative sign disappears. The result will be identical.

Q: Is it possible to solve this using the side vectors instead?

Yes, but it is much harder! You would need to find the coordinates of $B$ or $D$ first, which requires more steps. Using the diagonals directly is the most elegant and efficient path in JEE Advanced.

Question

Let $A(2, 3, 5)$ and $C(-3, 4, -2)$ be opposite vertices of a parallelogram $ABCD$ if the diagonal $\vec{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$ then the area of the parallelogram is equal to

Options:

(A) $\frac{1}{2}\sqrt{410}$

(B) $\frac{1}{2}\sqrt{474}$

(C) $\frac{1}{2}\sqrt{586}$

(D) $\frac{1}{2}\sqrt{306}$

Sigma Percentile · Accepted Answer

### The Geometry of Diagonals: A Masterclass in Vector Elegance

Welcome, future engineer. Today, we are not just solving a problem; we are uncovering the hidden symmetry of a parallelogram floating in 3D space.

Often, when we see a parallelogram, our minds jump to the classic formula: $Area = base 	imes height$. But in the realm of JEE Advanced, we must learn to see beyond the obvious.

When you are given the diagonals of a parallelogram, you hold the keys to its entire geometry.

### Phase 1: The Vision

Imagine you are standing in a 3D coordinate system. You have two points, $A(2, 3, 5)$ and $C(-3, 4, -2)$, which are opposite vertices of a parallelogram $ABCD$. You are also given the diagonal $\vec{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$.

Most students immediately try to find the coordinates of $B$ and $D$. Stop! That is a trap.

It is a long, winding road that leads to unnecessary calculations. Instead, let us embrace the beauty of vector algebra. The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the elegant relation:

$$Area = \frac{1}{2} |\vec{d_1} 	imes \vec{d_2}|$$

This formula is a gift. It allows us to bypass the side lengths entirely and work directly with the diagonals.

### Phase 2: Constructing the First Diagonal

We already have $\vec{d_2} = \vec{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$. Now, we need $\vec{d_1} = \vec{AC}$.

Since we know the coordinates of $A$ and $C$, we simply find the displacement vector:

$$\vec{AC} = (x_C - x_A)\hat{i} + (y_C - y_A)\hat{j} + (z_C - z_A)\hat{k}$$

Substituting our values: $\vec{AC} = (-3 - 2)\hat{i} + (4 - 3)\hat{j} + (-2 - 5)\hat{k}$. This simplifies beautifully to $\vec{AC} = -5\hat{i} + \hat{j} - 7\hat{k}$.

We have our two vectors. The heavy lifting is done; now, we just need to execute the cross product.

### Phase 3: The Determinant Dance

To find the cross product $\vec{AC} 	imes \vec{BD}$, we set up our determinant. I want you to be meticulous here—a single sign error can derail the entire process.

$$\vec{AC} 	imes \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -5 & 1 & -7 \ 1 & 2 & 3 \end{vmatrix}$$

Expanding this, we calculate the components:

- For $\hat{i}$: $(1)(3) - (-7)(2) = 3 + 14 = 17\hat{i}$
- For $\hat{j}$: $-[(-5)(3) - (-7)(1)] = -[-15 + 7] = 8\hat{j}$
- For $\hat{k}$: $(-5)(2) - (1)(1) = -10 - 1 = -11\hat{k}$

So, our resulting vector is $17\hat{i} + 8\hat{j} - 11\hat{k}$.

### Phase 4: The Final Magnitude

We are at the finish line. The area is half the magnitude of this vector. Let us calculate the magnitude $|\vec{AC} 	imes \vec{BD}| = \sqrt{17^2 + 8^2 + (-11)^2}$.

Squaring these, we get $289 + 64 + 121$. Summing them up, we arrive at $\sqrt{474}$.

Finally, applying our scaling factor of $1/2$, the area is:

$$Area = \frac{\sqrt{474}}{2}$$

Look at that result. It is precise, it is clean, and it was achieved through the sheer power of vector logic. You didn't need to guess; you followed the geometry.

Keep this confidence, and remember: in physics and math, the most elegant path is almost always the right one.

Animated Solution for Mathematics - Vector Algebra: Let $A (2, 3, 5)$ and $C (- 3, 4, - 2)$ be opposite vertices of a parallelogram $A B C D$ if the diagonal $B D = \hat{i} + 2 \hat{j} + 3 \hat{k}$ then the area of the parallelogram is equal to

Select Answer:

Visualized Solution

Visualize the Parallelogram

The Second Diagonal

Area Formula using Diagonals

Finding Vector $A C$

Substitute Coordinates for $A C$

Finalize Vector $A C$

Set up the Cross Product

Expand Determinant: $\hat{i}$ Component

Expand Determinant: $\hat{j}$ Component

Expand Determinant: $\hat{k}$ Component

Magnitude of the Cross Product

Square the Components

Sum the Squares

Final Area Calculation

The Sigma Insight: Vector (Cross) Product

The Geometry of Diagonals

Phase 1

Phase 2

Phase 3

Phase 4

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