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

Q: Why can't I just use the 2D area formula $\frac{1}{2} \times \text{base} \times \text{height}$?

You absolutely can, but in 3D space, finding the 'height' (the perpendicular distance from a vertex to the opposite side) involves complex projection calculations. The vector method is a shortcut that bypasses the need for explicit height calculation by using the properties of the cross product.

Q: Does the order of vectors in the cross product matter?

The order changes the direction of the resulting vector (it becomes the negative of the original), but the magnitude $|\vec{BA} \times \vec{BC}|$ remains identical. Since area is a scalar magnitude, the order does not affect your final answer.

Q: Can I pick any vertex as the common initial point?

Yes! You can use $A$, $B$, or $C$ as the common vertex. The area of the triangle will be the same regardless of which vertex you choose as the origin for your two vectors. Consistency is the only requirement.

Question

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

Sigma Percentile · Accepted Answer

### The Elegance of 3D Geometry

Imagine you are standing in a vast, empty room. You have three points, $A(1, -1, 2)$, $B(2, 1, -1)$, and $C(3, -1, 2)$, floating in the air.

Your task is to find the area of the triangle they form. In a 2D world, this would be a simple exercise, but here, in three-dimensional space, the challenge is to avoid the trap of trying to project these points onto a flat plane.

Instead, we embrace the power of vectors.

### The Vector Tool: Our Secret Weapon

When we deal with 3D geometry, standard formulas often fail us. Instead, we turn to the vector area formula:

$$	ext{Area} = \frac{1}{2} |\vec{BA} 	imes \vec{BC}|$$

The magnitude of the cross product of two vectors, $|\vec{BA} 	imes \vec{BC}|$, geometrically represents the area of the parallelogram formed by those two vectors. Since a triangle is exactly half of that parallelogram, we simply divide by two.

It is a beautiful, efficient, and robust method that works regardless of how the triangle is tilted in space.

### Step 1: Defining the Vectors

First, we must define our vectors. We choose point $B$ as our common starting point.

We calculate $\vec{BA}$ by subtracting the position vector of $B$ from $A$:

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

Similarly, for $\vec{BC}$, we subtract $B$ from $C$:

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

### Step 2: The Cross Product

Now, we set up our determinant to find the cross product $\vec{BA} 	imes \vec{BC}$. This is where we must be precise:

$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & -2 & 3 \ 1 & -2 & 3 \end{vmatrix}$$

Expanding this, the $\hat{i}$ component is $((-2)(3) - (3)(-2)) = -6 - (-6) = 0$.

The $\hat{j}$ component, remembering the negative sign, is $-((-1)(3) - (3)(1)) = -(-3 - 3) = 6$.

Finally, the $\hat{k}$ component is $((-1)(-2) - (-2)(1)) = 2 - (-2) = 4$. Our resulting vector is $0\hat{i} + 6\hat{j} + 4\hat{k}$.

### Step 3: The Final Calculation

We are almost there. We need the magnitude of this vector:

$$\sqrt{0^2 + 6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}$$

Simplifying $\sqrt{52}$, we get $2\sqrt{13}$. This is the area of the parallelogram.

Finally, we apply our triangle formula:

$$	ext{Area} = \frac{1}{2} (2\sqrt{13}) = \sqrt{13}$$

The twos cancel out, leaving us with the elegant result of **$\sqrt{13}$ square units**. You have successfully navigated the 3D space and arrived at the solution with the precision of a master.

Animated Solution for Mathematics - Vector Algebra: The area of the triangle whose vertices are $A (1, - 1, 2)$ , $B (2, 1, - 1)$ , $C (3, - 1, 2)$ is .........

Visualized Solution

Visualizing the Vertices in 3D Space

Forming the Triangle

The Vector Area Formula

Geometric Meaning of Cross Product

Calculating Vector $B A$

Calculating Vector $B C$

Setting up the Cross Product

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

Expanding the Determinant: $\hat{j}$ and $\hat{k}$ components

The Cross Product Vector

Magnitude of the Cross Product

Computing the Magnitude

Final Area of the Triangle

The Sigma Insight: Vector (Cross) Product

The Elegance of 3D Geometry

The Vector Tool

Step 1

Step 2

Step 3

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the triangle whose vertices are A(1,−1,2), B(2,1,−1), C(3,−1,2) is .........