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