Find the volume of the parallelopiped with segments AB, AC and AD as concurrent edges, where : (1) `A-=(3,7,4),B-=(5,-2,3),C-=(-4,5,6)andD-=(1,2,3)` (2) the position vectors of A,B,C,D are `hat(i)+hat(j)+hat(k),2hat(i)-hat(j),3hat(i)-2hat(j)-2hat(k)and3hat(i)+3hat(j)+4hat(k)`.

Answer

To find the volume of the parallelepiped formed by the vectors AB, AC, and AD, we will follow these steps:

### Step 1: Find the vectors AB, AC, and AD

Given the points:
- A = (3, 7, 4)
- B = (5, -2, 3)
- C = (-4, 5, 6)
- D = (1, 2, 3)

1. **Calculate vector AB:**
\[ \vec{AB} = \vec{B} - \vec{A} = (5 - 3, -2 - 7, 3 - 4) = (2, -9, -1) \]

2. **Calculate vector AC:**
\[ \vec{AC} = \vec{C} - \vec{A} = (-4 - 3, 5 - 7, 6 - 4) = (-7, -2, 2) \]

3. **Calculate vector AD:**
\[ \vec{AD} = \vec{D} - \vec{A} = (1 - 3, 2 - 7, 3 - 4) = (-2, -5, -1) \]

### Step 2: Set up the scalar triple product

The volume \( V \) of the parallelepiped can be calculated using the scalar triple product:
\[ V = |\vec{AB} \cdot (\vec{AC} \times \vec{AD})| \]

### Step 3: Calculate the cross product \( \vec{AC} \times \vec{AD} \)

Using the determinant to find the cross product:
\[ \vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -7 & -2 & 2 \\ -2 & -5 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{AC} \times \vec{AD} = \hat{i} \begin{vmatrix} -2 & 2 \\ -5 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} -7 & 2 \\ -2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} -7 & -2 \\ -2 & -5 \end{vmatrix} \]

Calculating each of the determinants:
1. For \( \hat{i} \):
\[ (-2)(-1) - (2)(-5) = 2 + 10 = 12 \]

2. For \( \hat{j} \):
\[ (-7)(-1) - (2)(-2) = 7 + 4 = 11 \]
(Note the sign change for \( \hat{j} \) gives us -11)

3. For \( \hat{k} \):
\[ (-7)(-5) - (-2)(-2) = 35 - 4 = 31 \]

Thus, the cross product is:
\[ \vec{AC} \times \vec{AD} = 12\hat{i} + 11\hat{j} + 31\hat{k} \]

### Step 4: Calculate the dot product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \)

Now, we compute:
\[ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (2, -9, -1) \cdot (12, -11, 31) \]
Calculating the dot product:
\[ = 2 \cdot 12 + (-9) \cdot (-11) + (-1) \cdot 31 = 24 + 99 - 31 = 92 \]

### Step 5: Calculate the volume

The volume \( V \) is:
\[ V = |\vec{AB} \cdot (\vec{AC} \times \vec{AD})| = |92| = 92 \text{ cubic units} \]

### Final Answer:
The volume of the parallelepiped is **92 cubic units**.

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