If `veca = 2 hati - 3hatj, vecb = hati + hatj -hatk, vecc = 3hati - hatk,` find `[a vecb vecc]`

Answer

To find the scalar triple product of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), we can use the determinant method. Here’s a step-by-step solution:

### Step 1: Write the vectors in component form
Given:
- \(\vec{a} = 2\hat{i} - 3\hat{j} + 0\hat{k}\)
- \(\vec{b} = 1\hat{i} + 1\hat{j} - 1\hat{k}\)
- \(\vec{c} = 3\hat{i} + 0\hat{j} - 1\hat{k}\)

### Step 2: Set up the determinant
The scalar triple product \([\vec{a} \, \vec{b} \, \vec{c}]\) can be represented as the determinant of a 3x3 matrix formed by the components of the vectors:

\[ [\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix} 2 & -3 & 0 \\ 1 & 1 & -1 \\ 3 & 0 & -1 \end{vmatrix} \]

### Step 3: Calculate the determinant
We will expand the determinant along the first row:

\[ [\vec{a} \, \vec{b} \, \vec{c}] = 2 \begin{vmatrix} 1 & -1 \\ 0 & -1 \end{vmatrix} - (-3) \begin{vmatrix} 1 & -1 \\ 3 & -1 \end{vmatrix} + 0 \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} \]

Calculating the first determinant:
\[ \begin{vmatrix} 1 & -1 \\ 0 & -1 \end{vmatrix} = (1)(-1) - (0)(-1) = -1 \]

Calculating the second determinant:
\[ \begin{vmatrix} 1 & -1 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (3)(-1) = -1 + 3 = 2 \]

Now substituting these values back into the determinant calculation:
\[ [\vec{a} \, \vec{b} \, \vec{c}] = 2(-1) + 3(2) + 0 = -2 + 6 + 0 = 4 \]

### Final Answer
Thus, the scalar triple product \([\vec{a} \, \vec{b} \, \vec{c}] = 4\).

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