https://file.aiquickdraw.com/mathbot/file/17300479397434rgogsu0.jpg

To solve number 6 (a), we need to determine the area of the triangle \( ABC \) and find the length of one side of this triangle.

The points provided in option (a) are:
\[
A(1, -1, 0), \quad B(3, -3, 2), \quad C(0, 2, 1).
\]

### Step 1: Calculate Vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)
First, find the vectors from \( A \) to \( B \) and from \( A \) to \( C \).

1. **Vector \( \overrightarrow{AB} \):**
   \[
   \overrightarrow{AB} = B - A = (3 - 1, -3 - (-1), 2 - 0) = (2, -2, 2)
   \]

2. **Vector \( \overrightarrow{AC} \):**
   \[
   \overrightarrow{AC} = C - A = (0 - 1, 2 - (-1), 1 - 0) = (-1, 3, 1)
   \]

### Step 2: Calculate the Cross Product \( \overrightarrow{AB} \times \overrightarrow{AC} \)
To find the area of the triangle, we need the magnitude of the cross product of \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).

\[
\overrightarrow{AB} \times \overrightarrow{AC} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & -2 & 2 \\
-1 & 3 & 1 \\
\end{vmatrix}
\]

Expanding this determinant:
\[
= \mathbf{i}((-2)(1) - (2)(3)) - \mathbf{j}((2)(1) - (2)(-1)) + \mathbf{k}((2)(3) - (-2)(-1))
\]
\[
= \mathbf{i}(-2 - 6) - \mathbf{j}(2 + 2) + \mathbf{k}(6 - 2)
\]
\[
= \mathbf{i}(-8) - \mathbf{j}(4) + \mathbf{k}(4)
\]
\[
= (-8, -4, 4)
\]

### Step 3: Find the Magnitude of the Cross Product
The magnitude of \( \overrightarrow{AB} \times \overrightarrow{AC} \) is:
\[
\sqrt{(-8)^2 + (-4)^2 + (4)^2} = \sqrt{64 + 16 + 16} = \sqrt{96} = 4\sqrt{6}
\]

### Step 4: Calculate the Area of Triangle \( ABC \)
The area \( A \) of the triangle is half the magnitude of the cross product:
\[
\text{Area} = \frac{1}{2} \times 4\sqrt{6} = 2\sqrt{6}
\]

### Step 5: Calculate the Length of One Side (Example: \( AB \))
To find the length of \( AB \), use the distance formula:
\[
AB = \sqrt{(3 - 1)^2 + (-3 + 1)^2 + (2 - 0)^2} = \sqrt{2^2 + (-2)^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}
\]

### Answer for Part (a):
- **Area of Triangle \( ABC \):** \( 2\sqrt{6} \)
- **Length of \( AB \):** \( 2\sqrt{3} \)

Would you like further explanation on any of these steps or help with another part?

---

### Additional Questions to Explore:
1. How would you find the area if the points were in a 2D plane instead of 3D?
2. What is the geometrical meaning of the cross product result in this context?
3. How can we find the perimeter of triangle \( ABC \) using distances between all points?
4. How would the area calculation change if the coordinates of \( B \) and \( C \) were altered?
5. Can you verify the area using a different method, such as Heron's formula?

#### Tip:
The cross product of two vectors in 3D is useful for finding areas of parallelograms and triangles by relating it to the magnitude.

Дараах цэгүүд дээр оройтой ABC гурвалжны талбай болон аль нэг өнцгийн уртыг ол. (a) A(1, -1, 0), B(3, -3, 2), C(0, 2, 1)

This solution details how to find the area of triangle ABC with vertices A(1, -1, 0), B(3, -3, 2), and C(0, 2, 1) using vector calculations and the cross product. Steps include calculating vectors AB and AC, finding their cross product, and applying the distance formula. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation