Math Problem Statement

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

Solution

To solve the problem of finding the volume of the parallelepiped with adjacent edges PQPQ, PRPR, and PSPS, 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=PQ(PR×PS)V = |\mathbf{PQ} \cdot (\mathbf{PR} \times \mathbf{PS})|

Problem 35:

The points are:

  • P(2,1,0)P(-2, 1, 0)
  • Q(2,3,2)Q(2, 3, 2)
  • R(1,4,1)R(1, 4, -1)
  • S(3,6,1)S(3, 6, 1)
  1. First, find the vectors PQPQ, PRPR, and PSPS:

PQ=QP=(2(2),31,20)=(4,2,2)\mathbf{PQ} = Q - P = (2 - (-2), 3 - 1, 2 - 0) = (4, 2, 2) PR=RP=(1(2),41,10)=(3,3,1)\mathbf{PR} = R - P = (1 - (-2), 4 - 1, -1 - 0) = (3, 3, -1) PS=SP=(3(2),61,10)=(5,5,1)\mathbf{PS} = S - P = (3 - (-2), 6 - 1, 1 - 0) = (5, 5, 1)

  1. Compute the cross product PR×PS\mathbf{PR} \times \mathbf{PS}:
\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.

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Math Problem Analysis

Mathematical Concepts

Vector Operations
Cross Product
Dot Product
Scalar Triple Product

Formulas

Volume of parallelepiped: V = |PQ ⋅ (PR × PS)|
Vector subtraction to find PQ, PR, PS

Theorems

Scalar Triple Product

Suitable Grade Level

Undergraduate or Advanced High School