A triangle defined by A(2, -5, 1), B(0, 2, 4) and C(0, 3, 1). What is area of the triangle?

Triangle Area Calculation in 3D Space

To determine the area of a triangle in three-dimensional space, given its vertices, we can utilize vector operations, specifically the cross product. The area of a triangle formed by two vectors is half the magnitude of their cross product. This method is effective for finding the geometric area regardless of the triangle's orientation in space.

Understanding the Given Triangle Vertices

We are provided with the coordinates of the three vertices of the triangle. These vertices define the specific triangle whose area we need to find.Forming Position Vectors from Vertices

First, we need to form two vectors from a common vertex. Let's choose vertex A as the common point and create vectors AB and AC. These vectors will represent two sides of the triangle originating from the same point.

- Vector AB is calculated by subtracting the coordinates of vertex A from vertex B: $$\overrightarrow{AB}=B-A=(0-2,2-(-5),4-1)$$ $$\overrightarrow{AB}=(-2,7,3)$$
- Vector AC is calculated by subtracting the coordinates of vertex A from vertex C: $$\overrightarrow{AC}=C-A=(0-2,3-(-5),1-1)$$ $$\overrightarrow{AC}=(-2,8,0)$$

Calculating the Cross Product of Vectors

Next, we calculate the cross product of the two vectors, $\overrightarrow{AB}$ and $\overrightarrow{AC}$. The cross product of two vectors $\vec{u}=⟨u_1,u_2,u_3⟩$ and $\vec{v}=⟨v_1,v_2,v_3⟩$ is a new vector perpendicular to both original vectors, and its magnitude is equal to the area of the parallelogram formed by $\vec{u}$ and $\vec{v}$. The formula for the cross product is:

$$\vec{u}\times \vec{v}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ u_1& u_2& u_3\\ v_1& v_2& v_3\end{array}\right|=(u_2{v}_3-u_3{v}_2)\mathbf{i}-(u_1{v}_3-u_3{v}_1)\mathbf{j}+(u_1{v}_2-u_2{v}_1)\mathbf{k}$$

Applying this formula for $\overrightarrow{AB}=⟨-2,7,3⟩$ and $\overrightarrow{AC}=⟨-2,8,0⟩$:

$$\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -2& 7& 3\\ -2& 8& 0\end{array}\right|$$

$$=\mathbf{i}((7)(0)-(3)(8))-\mathbf{j}((-2)(0)-(3)(-2))+\mathbf{k}((-2)(8)-(7)(-2))$$

$$=\mathbf{i}(0-24)-\mathbf{j}(0-(-6))+\mathbf{k}(-16-(-14))$$

$$=-24\mathbf{i}-6\mathbf{j}-2\mathbf{k}$$

So, the resultant cross product vector is $⟨-24,-6,-2⟩$.

Magnitude of the Cross Product Vector

The magnitude of a vector $\vec{v}=⟨x,y,z⟩$ is given by $|\vec{v}|=\sqrt{x^2+y^2+z^2}$. This magnitude represents the area of the parallelogram formed by the two vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. We need to find the magnitude of the cross product vector $⟨-24,-6,-2⟩$:

$$|\overrightarrow{AB}\times \overrightarrow{AC}|=\sqrt{(-24{)}^2+(-6{)}^2+(-2{)}^2}$$

$$=\sqrt{576+36+4}$$

$$=\sqrt{616}$$

Calculating the Final Triangle Area

The area of the triangle is half the magnitude of the cross product of the two vectors that form two sides of the triangle from a common vertex. This is because a triangle is exactly half of the parallelogram formed by the two vectors. The area of the triangle with vertices A, B, and C is given by:

$$\text{Area}=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$$

$$\text{Area}=\frac{1}{2}\sqrt{616}$$

Now, let's calculate the numerical value:

$$\sqrt{616}\approx 24.81934$$

$$\text{Area}\approx \frac{1}{2}\times 24.81934$$

$$\text{Area}\approx 12.40967$$

Rounding to two decimal places, the area of the triangle is approximately 12.41 square units.