https://file.aiquickdraw.com/mathbot/file/1729571595756lej77lq9.jpg

To solve the problem of finding the volume of the parallelepiped with adjacent edges \( PQ \), \( PR \), and \( PS \), we can use the formula for the volume of a parallelepiped, which is the scalar triple product of the vectors formed by these edges.

The volume is given by:

\[
V = |\mathbf{PQ} \cdot (\mathbf{PR} \times \mathbf{PS})|
\]

### Problem 35:
The points are:
- \( P(-2, 1, 0) \)
- \( Q(2, 3, 2) \)
- \( R(1, 4, -1) \)
- \( S(3, 6, 1) \)

1. First, find the vectors \( PQ \), \( PR \), and \( PS \):

\[
\mathbf{PQ} = Q - P = (2 - (-2), 3 - 1, 2 - 0) = (4, 2, 2)
\]
\[
\mathbf{PR} = R - P = (1 - (-2), 4 - 1, -1 - 0) = (3, 3, -1)
\]
\[
\mathbf{PS} = S - P = (3 - (-2), 6 - 1, 1 - 0) = (5, 5, 1)
\]

2. Compute the cross product \( \mathbf{PR} \times \mathbf{PS} \):

\[
\mathbf{PR} \times \mathbf{PS} = \begin{vmatrix} 
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 3 & -1 \\
5 & 5 & 1
\end{vmatrix}
\]

Expanding this determinant:

\[
\mathbf{PR} \times \mathbf{PS} = \mathbf{i} [(3)(1) - (-1)(5)] - \mathbf{j} [(3)(1) - (-1)(5)] + \mathbf{k} [(3)(5) - (3)(5)]
\]
\[
= \mathbf{i} (3 + 5) - \mathbf{j} (3 + 5) + \mathbf{k} (15 - 15)
\]
\[
= 8\mathbf{i} - 8\mathbf{j} + 0\mathbf{k}
\]
\[
\mathbf{PR} \times \mathbf{PS} = (8, -8, 0)
\]

3. Now, compute the dot product \( \mathbf{PQ} \cdot (\mathbf{PR} \times \mathbf{PS}) \):

\[
\mathbf{PQ} \cdot (8, -8, 0) = (4)(8) + (2)(-8) + (2)(0)
\]
\[
= 32 - 16 + 0 = 16
\]

4. The volume of the parallelepiped is the absolute value of the scalar triple product:

\[
V = |16| = 16
\]

---

### Problem 36:
The points are:
- \( P(3, 0, 1) \)
- \( Q(-1, 2, 5) \)
- \( R(5, 1, -1) \)
- \( S(0, 4, 2) \)

1. First, find the vectors \( PQ \), \( PR \), and \( PS \):

\[
\mathbf{PQ} = Q - P = (-1 - 3, 2 - 0, 5 - 1) = (-4, 2, 4)
\]
\[
\mathbf{PR} = R - P = (5 - 3, 1 - 0, -1 - 1) = (2, 1, -2)
\]
\[
\mathbf{PS} = S - P = (0 - 3, 4 - 0, 2 - 1) = (-3, 4, 1)
\]

2. Compute the cross product \( \mathbf{PR} \times \mathbf{PS} \):

\[
\mathbf{PR} \times \mathbf{PS} = \begin{vmatrix} 
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 1 & -2 \\
-3 & 4 & 1
\end{vmatrix}
\]

Expanding this determinant:

\[
\mathbf{PR} \times \mathbf{PS} = \mathbf{i} [(1)(1) - (-2)(4)] - \mathbf{j} [(2)(1) - (-2)(-3)] + \mathbf{k} [(2)(4) - (1)(-3)]
\]
\[
= \mathbf{i} (1 + 8) - \mathbf{j} (2 - 6) + \mathbf{k} (8 + 3)
\]
\[
= 9\mathbf{i} + 4\mathbf{j} + 11\mathbf{k}
\]
\[
\mathbf{PR} \times \mathbf{PS} = (9, 4, 11)
\]

3. Now, compute the dot product \( \mathbf{PQ} \cdot (\mathbf{PR} \times \mathbf{PS}) \):

\[
\mathbf{PQ} \cdot (9, 4, 11) = (-4)(9) + (2)(4) + (4)(11)
\]
\[
= -36 + 8 + 44 = 16
\]

4. The volume of the parallelepiped is the absolute value of the scalar triple product:

\[
V = |16| = 16
\]

---

Both problems have a parallelepiped volume of \( \boxed{16} \).

Would you like any further details or have questions about the solution?

---

Here are 5 related questions to expand on this concept:
1. How would you compute the volume if the points were in 2D instead of 3D?
2. Can the scalar triple product be negative, and what does that signify?
3. How do you geometrically interpret the cross product of two vectors?
4. What happens to the volume if two of the vectors are parallel?
5. How can the concept of the scalar triple product be used in physics?

**Tip:** The cross product of two vectors results in a vector perpendicular to the plane formed by the original vectors.

Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS for the given points in 3D space.

This problem involves finding the volume of a parallelepiped formed by three vectors using the scalar triple product. It includes calculating vector differences, cross products, and dot products to obtain the volume in 3D space.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation