If the three successive vertices of a parallelogram have the position vectors as, A(–3, –2, 0), B(3, –3, 1) and C(5, 0, 2). Then find
The angle between `vec(AC)` and `vec(BD)`.

Answer

To find the angle between the vectors \( \vec{AC} \) and \( \vec{BD} \) in the given parallelogram, we will follow these steps:

### Step 1: Find the coordinates of point D
Given the position vectors of points A, B, and C:
- \( A(-3, -2, 0) \)
- \( B(3, -3, 1) \)
- \( C(5, 0, 2) \)

Since the diagonals of a parallelogram bisect each other, we can find the coordinates of point D using the midpoint formula. Let the coordinates of point D be \( D(x, y, z) \).

The midpoint O of diagonal AC can be calculated as:
\[ O = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}, \frac{z_A + z_C}{2} \right) \]
Substituting the coordinates of A and C:
\[ O = \left( \frac{-3 + 5}{2}, \frac{-2 + 0}{2}, \frac{0 + 2}{2} \right) = \left( \frac{2}{2}, \frac{-2}{2}, \frac{2}{2} \right) = (1, -1, 1) \]

Now, the midpoint O of diagonal BD can also be expressed as:
\[ O = \left( \frac{x_B + x_D}{2}, \frac{y_B + y_D}{2}, \frac{z_B + z_D}{2} \right) \]
Substituting the coordinates of B:
\[ O = \left( \frac{3 + x}{2}, \frac{-3 + y}{2}, \frac{1 + z}{2} \right) \]

Setting the two expressions for O equal to each other:
1. \( \frac{3 + x}{2} = 1 \)
2. \( \frac{-3 + y}{2} = -1 \)
3. \( \frac{1 + z}{2} = 1 \)

Solving these equations:
1. From \( \frac{3 + x}{2} = 1 \):
\[ 3 + x = 2 \implies x = -1 \]
2. From \( \frac{-3 + y}{2} = -1 \):
\[ -3 + y = -2 \implies y = 1 \]
3. From \( \frac{1 + z}{2} = 1 \):
\[ 1 + z = 2 \implies z = 1 \]

Thus, the coordinates of point D are \( D(-1, 1, 1) \).

### Step 2: Find the vectors \( \vec{AC} \) and \( \vec{BD} \)
Now, we can find the vectors \( \vec{AC} \) and \( \vec{BD} \).

**For \( \vec{AC} \)**:
\[ \vec{AC} = \vec{C} - \vec{A} = (5, 0, 2) - (-3, -2, 0) = (5 + 3, 0 + 2, 2 - 0) = (8, 2, 2) \]

**For \( \vec{BD} \)**:
\[ \vec{BD} = \vec{D} - \vec{B} = (-1, 1, 1) - (3, -3, 1) = (-1 - 3, 1 + 3, 1 - 1) = (-4, 4, 0) \]

### Step 3: Calculate the dot product \( \vec{AC} \cdot \vec{BD} \)
The dot product is given by:
\[ \vec{AC} \cdot \vec{BD} = (8, 2, 2) \cdot (-4, 4, 0) = 8 \cdot (-4) + 2 \cdot 4 + 2 \cdot 0 = -32 + 8 + 0 = -24 \]

### Step 4: Find the magnitudes of \( \vec{AC} \) and \( \vec{BD} \)
**Magnitude of \( \vec{AC} \)**:
\[ |\vec{AC}| = \sqrt{8^2 + 2^2 + 2^2} = \sqrt{64 + 4 + 4} = \sqrt{72} = 6\sqrt{2} \]

**Magnitude of \( \vec{BD} \)**:
\[ |\vec{BD}| = \sqrt{(-4)^2 + 4^2 + 0^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \]

### Step 5: Use the dot product to find the angle \( \theta \)
Using the formula for the dot product:
\[ \vec{AC} \cdot \vec{BD} = |\vec{AC}| |\vec{BD}| \cos \theta \]
Substituting the values we have:
\[ -24 = (6\sqrt{2})(4\sqrt{2}) \cos \theta \]
\[ -24 = 48 \cos \theta \implies \cos \theta = \frac{-24}{48} = -\frac{1}{2} \]

### Step 6: Find the angle \( \theta \)
The angle \( \theta \) whose cosine is \( -\frac{1}{2} \) is:
\[ \theta = \frac{2\pi}{3} \text{ radians} \text{ or } 120^\circ \]

Thus, the angle between \( \vec{AC} \) and \( \vec{BD} \) is \( \frac{2\pi}{3} \) radians.