The vertices of a `triangleABC` are `A(2,3,1),B(-2,2,0) and C(0,1,-1)`
What is the magnitude of the line joining mid-points of the side AC and BC?

Options

A. `1/sqrt2` unit
B. 1 unit
C. `3/sqrt2` unit
D. 2 unit

Answer

To find the magnitude of the line joining the mid-points of the sides AC and BC of triangle ABC with vertices A(2, 3, 1), B(-2, 2, 0), and C(0, 1, -1), we will follow these steps:

### Step 1: Find the mid-point D of side AC
The mid-point D of line segment AC can be calculated using the formula:
\[ D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]
where \( A(2, 3, 1) \) and \( C(0, 1, -1) \).

Calculating the coordinates:
- \( x \)-coordinate: \( \frac{2 + 0}{2} = 1 \)
- \( y \)-coordinate: \( \frac{3 + 1}{2} = 2 \)
- \( z \)-coordinate: \( \frac{1 + (-1)}{2} = 0 \)

Thus, the mid-point D is:
\[ D(1, 2, 0) \]

### Step 2: Find the mid-point E of side BC
The mid-point E of line segment BC can be calculated similarly:
\[ E = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]
where \( B(-2, 2, 0) \) and \( C(0, 1, -1) \).

Calculating the coordinates:
- \( x \)-coordinate: \( \frac{-2 + 0}{2} = -1 \)
- \( y \)-coordinate: \( \frac{2 + 1}{2} = \frac{3}{2} \)
- \( z \)-coordinate: \( \frac{0 + (-1)}{2} = -\frac{1}{2} \)

Thus, the mid-point E is:
\[ E\left(-1, \frac{3}{2}, -\frac{1}{2}\right) \]

### Step 3: Calculate the distance DE between points D and E
The distance DE can be calculated using the distance formula in 3D:
\[ DE = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates of D(1, 2, 0) and E\((-1, \frac{3}{2}, -\frac{1}{2})\):
- \( x_1 = 1, y_1 = 2, z_1 = 0 \)
- \( x_2 = -1, y_2 = \frac{3}{2}, z_2 = -\frac{1}{2} \)

Calculating the differences:
- \( x_2 - x_1 = -1 - 1 = -2 \)
- \( y_2 - y_1 = \frac{3}{2} - 2 = \frac{3}{2} - \frac{4}{2} = -\frac{1}{2} \)
- \( z_2 - z_1 = -\frac{1}{2} - 0 = -\frac{1}{2} \)

Now substituting these into the distance formula:
\[ DE = \sqrt{(-2)^2 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} \]
Calculating each term:
- \( (-2)^2 = 4 \)
- \( \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \)
- \( \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \)

Thus:
\[ DE = \sqrt{4 + \frac{1}{4} + \frac{1}{4}} = \sqrt{4 + \frac{2}{4}} = \sqrt{4 + \frac{1}{2}} = \sqrt{4.5} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} \]

### Final Answer
The magnitude of the line joining the mid-points of sides AC and BC is:
\[ \frac{3}{\sqrt{2}} \]