Let `vec a = - 5 hat i + hat j - 3 hat k`, `vec b = hat i + 2 hat j - 4 hat k` and `vec c = (((vec a times vec b) times hat i ) times hat i) times hat i`. Then `vec c.(- hat i + hat j + hat k)` is equal to

Options

A. `-12`
B. `-10`
C. `-13`
D. `-15`

Answer

To solve the problem, we need to find the vector \( \vec{c} \) defined as \( \vec{c} = (((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i} \) and then compute \( \vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k}) \).

### Step 1: Calculate \( \vec{a} \times \vec{b} \)

Given:
\[ \vec{a} = -5 \hat{i} + \hat{j} - 3 \hat{k} \]
\[ \vec{b} = \hat{i} + 2 \hat{j} - 4 \hat{k} \]

We can calculate the cross product \( \vec{a} \times \vec{b} \) using the determinant of a matrix:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -3 \\ 1 & 2 & -4 \end{vmatrix} \]

Calculating this determinant:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 1 & -3 \\ 2 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} -5 & -3 \\ 1 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} -5 & 1 \\ 1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & -3 \\ 2 & -4 \end{vmatrix} = (1)(-4) - (-3)(2) = -4 + 6 = 2 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} -5 & -3 \\ 1 & -4 \end{vmatrix} = (-5)(-4) - (-3)(1) = 20 + 3 = 23 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} -5 & 1 \\ 1 & 2 \end{vmatrix} = (-5)(2) - (1)(1) = -10 - 1 = -11 \]

Putting it all together:
\[ \vec{a} \times \vec{b} = 2 \hat{i} - 23 \hat{j} - 11 \hat{k} \]

### Step 2: Calculate \( \vec{c} = ((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i} \)

Now we need to calculate \( \vec{c} = ((2 \hat{i} - 23 \hat{j} - 11 \hat{k}) \times \hat{i}) \times \hat{i} \).

First, calculate \( \vec{d} = \vec{a} \times \hat{i} \):
\[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -23 & -11 \\ 1 & 0 & 0 \end{vmatrix} \]

Calculating this determinant:
\[ \vec{d} = \hat{i} \begin{vmatrix} -23 & -11 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -11 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -23 \\ 1 & 0 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. For \( \hat{i} \): \( 0 \)
2. For \( \hat{j} \): \( -2 \)
3. For \( \hat{k} \): \( 23 \)

Thus:
\[ \vec{d} = 0 \hat{i} + 2 \hat{j} + 23 \hat{k} \]

Now calculate \( \vec{c} = \vec{d} \times \hat{i} \):
\[ \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & 23 \\ 1 & 0 & 0 \end{vmatrix} \]

Calculating this determinant:
\[ \vec{c} = \hat{i} \begin{vmatrix} 2 & 23 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 23 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 2 \\ 1 & 0 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. For \( \hat{i} \): \( 0 \)
2. For \( \hat{j} \): \( -23 \)
3. For \( \hat{k} \): \( -2 \)

Thus:
\[ \vec{c} = 0 \hat{i} + 23 \hat{j} - 2 \hat{k} \]

### Step 3: Calculate \( \vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k}) \)

Now we compute:
\[ \vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k}) = (0 \hat{i} + 23 \hat{j} - 2 \hat{k}) \cdot (-\hat{i} + \hat{j} + \hat{k}) \]

Calculating the dot product:
\[ = 0 \cdot (-1) + 23 \cdot 1 + (-2) \cdot 1 = 0 + 23 - 2 = 21 \]

### Final Answer
Thus, \( \vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 21 \).

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