To determine if the vectors \(\mathbf{a} = (1, 4, -7)\), \(\mathbf{b} = (2, -1, 4)\), and \(\mathbf{c} = (0, -9, 18)\) are coplanar, we can use the scalar triple product. The scalar triple product of three vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) is given by the determinant of the matrix formed by these vectors:
\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \det(\begin{bmatrix} x_1 & y_1 & z_1 \ x_2 & y_2 & z_2 \ x_3 & y_3 & z_3 \end{bmatrix}) \]
where \(\mathbf{a} = (x_1, y_1, z_1)\), \(\mathbf{b} = (x_2, y_2, z_2)\), and \(\mathbf{c} = (x_3, y_3, z_3)\).
Plugging in the values from our vectors:
\[ \mathbf{a} = (1, 4, -7), \quad \mathbf{b} = (2, -1, 4), \quad \mathbf{c} = (0, -9, 18) \]
we construct the following determinant:
\[ \det\begin{bmatrix} 1 & 4 & -7 \ 2 & -1 & 4 \ 0 & -9 & 18 \end{bmatrix} \]
Now, we can compute this determinant:
\[ = 1 \cdot \det\begin{bmatrix} -1 & 4 \ -9 & 18 \end{bmatrix} - 4 \cdot \det\begin{bmatrix} 2 & 4 \ 0 & 18 \end{bmatrix} - 7 \cdot \det\begin{bmatrix} 2 & -1 \ 0 & -9 \end{bmatrix} \]
Calculating the smaller determinants:
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For (\det\begin{bmatrix} -1 & 4 \ -9 & 18 \end{bmatrix} = (-1)(18) - (4)(-9) = -18 + 36 = 18)
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For (\det\begin{bmatrix} 2 & 4 \ 0 & 18 \end{bmatrix} = (2)(18) - (4)(0) = 36)
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For (\det\begin{bmatrix} 2 & -1 \ 0 & -9 \end{bmatrix} = (2)(-9) - (-1)(0) = -18)
Putting these values back into the determinant calculation:
\[ = 1(18) - 4(36) - 7(-18) \] \[ = 18 - 144 + 126 \] \[ = 18 - 144 + 126 = 0 \]
Since the scalar triple product is zero:
\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0 \]
The result indicates that the vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are coplanar. Thus, we have shown that these vectors are coplanar.
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