Question

Find the area of triangle if a. (1,-1,0), B.( 2,1,-1) And c.(-1,1,2)

Answer 1

To find the area of a triangle given its vertices A, B, and C in three-dimensional space, we can use the formula for the area of a triangle determined by the vectors given by its vertices.

Let \( A = (1, -1, 0) \), \( B = (2, 1, -1) \), and \( C = (-1, 1, 2) \).

1. First, we'll find the vectors \( \vec{AB} \) and \( \vec{AC} \):
\[
\vec{AB} = B - A = (2 - 1, 1 - (-1), -1 - 0) = (1, 2, -1)
\]
\[
\vec{AC} = C - A = (-1 - 1, 1 - (-1), 2 - 0) = (-2, 2, 2)
\]

2. Next, we find the cross product \( \vec{AB} \times \vec{AC} \):
\[
\vec{AB} \times \vec{AC} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & -1 \\
-2 & 2 & 2
\end{vmatrix}
\]

Calculating the determinant:

\[
\vec{AB} \times \vec{AC} = \mathbf{i}(2 \cdot 2 - (-1) \cdot 2) - \mathbf{j}(1 \cdot 2 - (-1) \cdot -2) + \mathbf{k}(1 \cdot 2 - 2 \cdot -2)
\]
\[
= \mathbf{i}(4 + 2) - \mathbf{j}(2 - 2) + \mathbf{k}(2 + 4)
\]
\[
= \mathbf{i}(6) - \mathbf{j}(0) + \mathbf{k}(6)
\]
\[
= (6, 0, 6)
\]

3. Now we find the magnitude of the cross product:
\[
|\vec{AB} \times \vec{AC}| = \sqrt{6^2 + 0^2 + 6^2} = \sqrt{36 + 0 + 36} = \sqrt{72} = 6\sqrt{2}
\]

4. The area \( A \) of triangle \( ABC \) is given by:
\[
A = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} (6\sqrt{2}) = 3\sqrt{2}
\]

Thus, the area of triangle ABC is \( \boxed{3\sqrt{2}} \).