Question

Find the volume of the parallelepiped with edges u = i + k, v = 2i + j + 4k, and w = j + k.

Answer 1

To find the volume of the parallelepiped defined by the vectors \( \mathbf{u} = \mathbf{i} + \mathbf{k} \), \( \mathbf{v} = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k} \), and \( \mathbf{w} = \mathbf{j} + \mathbf{k} \), we can use the scalar triple product. The volume \( V \) can be calculated using the formula:

\[
V = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})|
\]

1. **Calculate \( \mathbf{v} \times \mathbf{w} \)**: We shall compute the cross product \( \mathbf{v} \times \mathbf{w} \).

\[
\mathbf{v} = \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix}, \quad \mathbf{w} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}
\]

The cross product \( \mathbf{v} \times \mathbf{w} \) can be computed using the determinant of the following matrix:

\[
\mathbf{v} \times \mathbf{w} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 1 & 4 \\
0 & 1 & 1
\end{vmatrix}
\]

Calculating the determinant:

\[
\mathbf{v} \times \mathbf{w} = \mathbf{i} \begin{vmatrix} 1 & 4 \\ 1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 4 \\ 0 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix}
\]

Calculating each of these 2x2 determinants:

\[
\begin{vmatrix} 1 & 4 \\ 1 & 1 \end{vmatrix} = (1)(1) - (4)(1) = 1 - 4 = -3
\]
\[
\begin{vmatrix} 2 & 4 \\ 0 & 1 \end{vmatrix} = (2)(1) - (4)(0) = 2 - 0 = 2
\]
\[
\begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} = (2)(1) - (1)(0) = 2 - 0 = 2
\]

Now plug these values back into the expression:

\[
\mathbf{v} \times \mathbf{w} = -3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}
\]
Thus,

\[
\mathbf{v} \times \mathbf{w} = \begin{pmatrix} -3 \\ -2 \\ 2 \end{pmatrix}
\]

2. **Calculate \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \)**: Now we compute the dot product \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \).

\[
\mathbf{u} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}
\]

Then,

\[
\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ -2 \\ 2 \end{pmatrix} = 1 \cdot (-3) + 0 \cdot (-2) + 1 \cdot 2 = -3 + 0 + 2 = -1
\]

3. **Calculate the volume**: The volume \( V \) is the absolute value of the dot product computed:

\[
V = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})| = |-1| = 1
\]

Thus, the volume of the parallelepiped is

\[
\boxed{1}
\]