If A(3, 1, -1), B(5/3, 7/3, 1/3), C(2, 2, 1) and D(10/3, 2/3, -1/3) are the vertices of a quadrilate

Q: Wait, why can't we just split the quadrilateral into two triangles and find their areas separately?

You absolutely can! You could calculate the area of $\triangle ABC$ and $\triangle ADC$ and add them. However, in JEE Advanced, time is your most precious resource. The diagonal vector formula $\text{Area} = \frac{1}{2} |\vec{AC} \times \vec{BD}|$ is a direct, one-step method that minimizes the chance of arithmetic errors compared to calculating two separate triangle areas.

Q: Does the order of vertices matter when calculating the cross product?

Great question! The cross product $\vec{AC} \times \vec{BD}$ is the standard formula. If you were to accidentally swap the order to $\vec{BD} \times \vec{AC}$, you would get the negative of the original vector. However, since we are taking the magnitude $|\vec{AC} \times \vec{BD}|$, the negative sign disappears, and the final area remains the same. So, while the vector itself changes, the area is invariant!

Q: What if the quadrilateral is not planar?

In the context of this problem, we are given four vertices of a quadrilateral in 3D space. By definition, a quadrilateral is a planar figure. If the four points were not coplanar, they would form a tetrahedron, not a quadrilateral. In JEE problems, you can safely assume the points are coplanar unless the question asks you to prove it.

Question

If $A(3, 1, -1), B(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}), C(2, 2, 1)$ and $D(\frac{10}{3}, \frac{2}{3}, -\frac{1}{3})$ are the vertices of a quadrilateral $ABCD$, then its area is

Options:

(A) $\frac{2\sqrt{2}}{3}$

(B) $\frac{5\sqrt{2}}{3}$

(C) $2\sqrt{2}$

(D) $\frac{4\sqrt{2}}{3}$

Sigma Percentile · Accepted Answer

### The Geometry of 3D Space: A Vector Odyssey

Imagine you are standing in a vast, three-dimensional coordinate system. You have four points, $A(3, 1, -1)$, $B(\frac{5}{3}, \frac{7}{3}, \frac{1}{3})$, $C(2, 2, 1)$, and $D(\frac{10}{3}, \frac{2}{3}, -\frac{1}{3})$, floating like stars in the void.

These are the vertices of a quadrilateral $ABCD$. While the fractional coordinates might seem intimidating, in the world of JEE Advanced, we embrace them as the building blocks of precision.

### The Elegance of Diagonals

The secret to unlocking this problem lies in the diagonals. If we define the diagonal vectors $\vec{AC}$ and $\vec{BD}$, we can use the beautiful vector area formula:

$$	ext{Area} = \frac{1}{2} |\vec{AC} 	imes \vec{BD}|$$

This formula is a masterpiece of geometry. It tells us that the area of a quadrilateral is simply half the magnitude of the cross product of its diagonals. It is efficient, robust, and turns a complex spatial problem into a straightforward algebraic one.

### Step 1: Defining the Diagonals

Let's begin our journey by finding the diagonal vectors. To find $\vec{AC}$, we subtract the position vector of $A$ from $C$:

$$\vec{AC} = (2-3)\hat{i} + (2-1)\hat{j} + (1-(-1))\hat{k} = -\hat{i} + \hat{j} + 2\hat{k}$$

Now, let's tackle $\vec{BD}$ with the same precision. We subtract $B$ from $D$:

$$\vec{BD} = \left( \frac{10}{3} - \frac{5}{3} ight)\hat{i} + \left( \frac{2}{3} - \frac{7}{3} ight)\hat{j} + \left( -\frac{1}{3} - \frac{1}{3} ight)\hat{k} = \frac{5}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{2}{3}\hat{k}$$

### Step 2: The Power of the Cross Product

Now, we set up the determinant to find the cross product $\vec{AC} 	imes \vec{BD}$:

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

Expanding this determinant along the first row reveals the core components. For the $\hat{i}$ component:

$$\hat{i} \left[ (1)\left(-\frac{2}{3}ight) - (2)\left(-\frac{5}{3}ight) ight] = \hat{i} \left[ -\frac{2}{3} + \frac{10}{3} ight] = \frac{8}{3}\hat{i}$$

For the $\hat{j}$ component, remember the negative sign:

$$-\hat{j} \left[ (-1)\left(-\frac{2}{3}ight) - (2)\left(\frac{5}{3}ight) ight] = -\hat{j} \left[ \frac{2}{3} - \frac{10}{3} ight] = -\hat{j} \left[ -\frac{8}{3} ight] = \frac{8}{3}\hat{j}$$

Finally, for the $\hat{k}$ component:

$$\hat{k} \left[ (-1)\left(-\frac{5}{3}ight) - (1)\left(\frac{5}{3}ight) ight] = \hat{k} \left[ \frac{5}{3} - \frac{5}{3} ight] = 0\hat{k}$$

The $\hat{k}$ component vanishes, meaning the quadrilateral lies in a plane parallel to the $xy$-plane. Our resultant vector is $\vec{AC} 	imes \vec{BD} = \frac{8}{3}\hat{i} + \frac{8}{3}\hat{j}$.

### Final Calculation

We now find the magnitude of this vector:

$$|\vec{AC} 	imes \vec{BD}| = \sqrt{\left(\frac{8}{3}ight)^2 + \left(\frac{8}{3}ight)^2} = \sqrt{2 \cdot \left(\frac{8}{3}ight)^2} = \frac{8\sqrt{2}}{3}$$

Finally, we apply the area formula:

$$	ext{Area} = \frac{1}{2} \cdot \left( \frac{8\sqrt{2}}{3} ight) = \frac{4\sqrt{2}}{3}$$

The final area of the quadrilateral is **$\frac{4\sqrt{2}}{3}$** square units.

Animated Solution for Mathematics - Vector Algebra: If $A (3, 1, - 1), B (\frac{5}{3}, \frac{7}{3}, \frac{1}{3}), C (2, 2, 1)$ and $D (\frac{10}{3}, \frac{2}{3}, - \frac{1}{3})$ are the vertices of a quadrilateral $A B C D$ , then its area is

Select Answer:

Visualized Solution

Visualizing the Quadrilateral $A B C D$

The Vector Area Formula

Finding Diagonal Vector $A C$

Simplifying $A C$

Finding Diagonal Vector $B D$

Simplifying $B D$

Setting up the Cross Product

Expanding Determinant: $\hat{i}$ component

Expanding Determinant: $\hat{j}$ component

Expanding Determinant: $\hat{k}$ component

The Resultant Cross Product Vector

Calculating the Magnitude

Final Area Calculation

Conclusion & Key Takeaway

The Sigma Insight: Vector (Cross) Product

The Geometry of 3D Space

The Elegance of Diagonals

Step 1

Step 2

Final Calculation

Similar Questions

If $A (1, - 1, 2)$ , $B (5, 7, - 6)$ , $C (3, 4, - 10)$ and $D (- 1, - 4, - 2)$ are the vertices of a quadrilateral $A B C D$ , then its area is :

The area of the quadrilateral $A B C D$ with vertices $A (2, 1, 1)$ , $B (1, 2, 5)$ , $C (- 2, - 3, 5)$ and $D (1, - 6, - 7)$ is equal to

$A (2, 6, 2), B (- 4, 0, λ), C (2, 3, - 1)$ and $D (4, 5, 0), ∣ λ ∣ \leq 5$ are the vertices of a quadrilateral ABCD. If its area is 18 square units, then $5 - 6 λ$ is equal to

The area of the triangle whose vertices are $A (1, - 1, 2)$ , $B (2, 1, - 1)$ , $C (3, - 1, 2)$ is .........

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

Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that $\frac{Q A}{A R} = \frac{R B}{B P} = \frac{P C}{C Q} = \frac{1}{2}$ . Then $\frac{Area ( Δ P QR )}{Area ( Δ A B C )}$ is equal to

If $A, B, C, D$ are any four points in space, prove that $∣ A B \times C D + B C \times A D + C A \times B D ∣ = 4 (area of triangle A B C)$ .

Let $O A = a$ , $O B = 12 a + 4 b$ and $O C = b$ , where $O$ is the origin. If $S$ is the parallelogram with adjacent sides $O A$ and $O C$ , then $\frac{area of the quadrilateral OABC}{area of S}$ is equal to

Let $O A = 2 a, O B = 6 a + 5 b$ and $O C = 3 b$ , where $O$ is the origin. If the area of the parallelogram with adjacent sides $O A$ and $O C$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $O A B C$ is equal to :

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

Select Answer:

Visualized Solution

Visualizing the Quadrilateral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABCD

The Vector Area Formula

Finding Diagonal Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Finding Diagonal Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BD

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BD