The points `A(4, 5, 10), B(2, 3, 4) and C(1, 2, -1)` are three vertices of a parallelogram ABCD. Find the vector equations of side AB and BC and also find the coordinates of point D .

Options

A. Vector equation of AB is `2i+3j+4k+lambda(i+j+3k)`
B. Cartesian equation of BC is `(x-2)/(1)=(y-3)/(1)=(z-4)/(-5)`
C. Coordinate of D are `(3, 4,5)`
D. ABCD is a rectangle.

Answer

To solve the problem, we need to find the vector equations of sides AB and BC of the parallelogram ABCD, and also determine the coordinates of point D.

### Step 1: Find the direction vector of AB

The coordinates of points A and B are given as:
- A(4, 5, 10)
- B(2, 3, 4)

The direction vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = \vec{B} - \vec{A} = (2 - 4) \hat{i} + (3 - 5) \hat{j} + (4 - 10) \hat{k} \]
Calculating this gives:
\[ \vec{AB} = -2 \hat{i} - 2 \hat{j} - 6 \hat{k} \]

### Step 2: Write the vector equation of line AB

The vector equation of line AB can be expressed as:
\[ \vec{r} = \vec{A} + \lambda \vec{AB} \]
Substituting the values:
\[ \vec{r} = (4 \hat{i} + 5 \hat{j} + 10 \hat{k}) + \lambda (-2 \hat{i} - 2 \hat{j} - 6 \hat{k}) \]
This simplifies to:
\[ \vec{r} = (4 - 2\lambda) \hat{i} + (5 - 2\lambda) \hat{j} + (10 - 6\lambda) \hat{k} \]

### Step 3: Find the direction vector of BC

The coordinates of points B and C are given as:
- B(2, 3, 4)
- C(1, 2, -1)

The direction vector \( \vec{BC} \) can be calculated as:
\[ \vec{BC} = \vec{C} - \vec{B} = (1 - 2) \hat{i} + (2 - 3) \hat{j} + (-1 - 4) \hat{k} \]
Calculating this gives:
\[ \vec{BC} = -1 \hat{i} - 1 \hat{j} - 5 \hat{k} \]

### Step 4: Write the vector equation of line BC

The vector equation of line BC can be expressed as:
\[ \vec{r} = \vec{B} + \mu \vec{BC} \]
Substituting the values:
\[ \vec{r} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + \mu (-1 \hat{i} - 1 \hat{j} - 5 \hat{k}) \]
This simplifies to:
\[ \vec{r} = (2 - \mu) \hat{i} + (3 - \mu) \hat{j} + (4 - 5\mu) \hat{k} \]

### Step 5: Find the coordinates of point D

In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of AC should be equal to the midpoint of BD.

Calculating the midpoint of AC:
\[ \text{Midpoint of AC} = \left( \frac{4 + 1}{2}, \frac{5 + 2}{2}, \frac{10 - 1}{2} \right) = \left( \frac{5}{2}, \frac{7}{2}, \frac{9}{2} \right) \]

Let the coordinates of D be (x, y, z). The midpoint of BD is:
\[ \text{Midpoint of BD} = \left( \frac{2 + x}{2}, \frac{3 + y}{2}, \frac{4 + z}{2} \right) \]

Setting the midpoints equal:
\[ \frac{2 + x}{2} = \frac{5}{2}, \quad \frac{3 + y}{2} = \frac{7}{2}, \quad \frac{4 + z}{2} = \frac{9}{2} \]

From these equations, we can solve for x, y, and z:
1. \( 2 + x = 5 \) ⇒ \( x = 3 \)
2. \( 3 + y = 7 \) ⇒ \( y = 4 \)
3. \( 4 + z = 9 \) ⇒ \( z = 5 \)

Thus, the coordinates of point D are (3, 4, 5).

### Summary of Results:
- The vector equation of line AB is:
\[ \vec{r} = (4 - 2\lambda) \hat{i} + (5 - 2\lambda) \hat{j} + (10 - 6\lambda) \hat{k} \]
- The vector equation of line BC is:
\[ \vec{r} = (2 - \mu) \hat{i} + (3 - \mu) \hat{j} + (4 - 5\mu) \hat{k} \]
- The coordinates of point D are \( D(3, 4, 5) \).