The value of scalar triple product `i-2j+3k, 2i+j-k" and "j+k` is

Options

A. 12
B. 10
C. 14
D. 16

Answer

To find the value of the scalar triple product of the vectors \( \mathbf{a} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \), \( \mathbf{b} = 2\mathbf{i} + \mathbf{j} - \mathbf{k} \), and \( \mathbf{c} = \mathbf{j} + \mathbf{k} \), we can use the determinant of a matrix formed by these vectors.

### Step-by-Step Solution:

1. **Write the vectors in component form:**
- \( \mathbf{a} = (1, -2, 3) \)
- \( \mathbf{b} = (2, 1, -1) \)
- \( \mathbf{c} = (0, 1, 1) \)

2. **Set up the determinant:**
The scalar triple product can be calculated using the determinant of the following matrix:
\[ \begin{vmatrix} 1 & -2 & 3 \\ 2 & 1 & -1 \\ 0 & 1 & 1 \end{vmatrix} \]

3. **Calculate the determinant:**
We can calculate the determinant using the formula:
\[ \text{det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]
For our matrix:
- \( a = 1, b = -2, c = 3 \)
- \( d = 2, e = 1, f = -1 \)
- \( g = 0, h = 1, i = 1 \)

Plugging in the values:
\[ \text{det} = 1 \cdot (1 \cdot 1 - (-1) \cdot 1) - (-2) \cdot (2 \cdot 1 - (-1) \cdot 0) + 3 \cdot (2 \cdot 1 - 1 \cdot 0) \]

Simplifying each term:
- First term: \( 1 \cdot (1 + 1) = 1 \cdot 2 = 2 \)
- Second term: \( -(-2) \cdot (2 + 0) = 2 \cdot 2 = 4 \)
- Third term: \( 3 \cdot (2 - 0) = 3 \cdot 2 = 6 \)

Now, combine these results:
\[ \text{det} = 2 + 4 + 6 = 12 \]

4. **Conclusion:**
The value of the scalar triple product is \( 12 \).