If `a=3i+2j+k, b=5i-j+2k" and "c=i+j+k`, find [a b c]

Answer

To find the scalar triple product \([a, b, c]\) for the vectors \(a = 3i + 2j + k\), \(b = 5i - j + 2k\), and \(c = i + j + k\), we will use the determinant method. The scalar triple product can be calculated using the determinant of a matrix formed by the components of the vectors.

### Step-by-Step Solution:

1. **Write the vectors in matrix form**:
\[ [a, b, c] = \begin{vmatrix} 3 & 2 & 1 \\ 5 & -1 & 2 \\ 1 & 1 & 1 \end{vmatrix} \]

2. **Calculate the determinant**:
The determinant of a 3x3 matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]
For our matrix:
\[ \begin{vmatrix} 3 & 2 & 1 \\ 5 & -1 & 2 \\ 1 & 1 & 1 \end{vmatrix} \]

3. **Expand the determinant**:
\[ = 3 \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} - 2 \begin{vmatrix} 5 & 2 \\ 1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 5 & -1 \\ 1 & 1 \end{vmatrix} \]

4. **Calculate the 2x2 determinants**:
- For the first determinant:
\[ \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} = (-1)(1) - (2)(1) = -1 - 2 = -3 \]
- For the second determinant:
\[ \begin{vmatrix} 5 & 2 \\ 1 & 1 \end{vmatrix} = (5)(1) - (2)(1) = 5 - 2 = 3 \]
- For the third determinant:
\[ \begin{vmatrix} 5 & -1 \\ 1 & 1 \end{vmatrix} = (5)(1) - (-1)(1) = 5 + 1 = 6 \]

5. **Substitute back into the determinant expansion**:
\[ = 3(-3) - 2(3) + 1(6) \]
\[ = -9 - 6 + 6 \]
\[ = -9 \]

6. **Conclusion**:
The value of the scalar triple product \([a, b, c]\) is \(-9\).