Question

Calculate the volume of a parallelepiped whose sides are given by the vectors
a = 3ˆi + 2ˆj + kˆ, b = −ˆi + 3ˆj and c = 2ˆi + 2ˆj + 5kˆ

Answer 1

Question #74267, Physics / Mechanics | Relativity

Calculate the volume of a parallelepiped whose sides are given by the vectors

$$a = 3 ^ {\wedge} i + 2 ^ {\wedge} j + k ^ {\wedge}, b = - ^ {\wedge} i + 3 ^ {\wedge} j \text{ and } c = 2 ^ {\wedge} i + 2 ^ {\wedge} j + 5 k ^ {\wedge}$$

Need to find the volume of the parallelepiped.

Coordinates of vectors – $a = (3, 2, 1)$, $b = (-1, 3, 0)$ and $c = (2, 2, 5)$.

Solution:

Volume of parallelepiped $V = |a \cdot (b \times c)|$

We find a triple product $V = \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix} = \begin{vmatrix} 3 & 2 & 1 \\ -1 & 3 & 0 \\ 2 & 2 & 5 \end{vmatrix} = 3 \cdot 3 \cdot 5 + 2 \cdot 0 \cdot 2 + 1 \cdot (-1) \cdot 2 - 1 \cdot 3 \cdot 2 - 2 \cdot (-1) \cdot 5 - 3 \cdot 0 \cdot 2 = 47$

Answer: $V = 47$.

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