Question

Use the scalar triple product to show that the vectors a = (1, 4, −7), b = (2, −1, 4) and
c = (0, −9, 18) are coplanar

Answer 1

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:

1. For \(\det\begin{bmatrix}
-1 & 4 \\
-9 & 18
\end{bmatrix} = (-1)(18) - (4)(-9) = -18 + 36 = 18\)

2. For \(\det\begin{bmatrix}
2 & 4 \\
0 & 18
\end{bmatrix} = (2)(18) - (4)(0) = 36\)

3. 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.