If `a = 1,2,4, b =2,-3,-1, c=1,4-4`, then the vector `a xx(b xxc)` is orthogonal to

Options

A. a
B. b
C. c
D. `a+b+c`

Answer

To solve the problem, we need to find the vector \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \) and determine which vector it is orthogonal to.

### Step 1: Define the vectors
Given:
- \( \mathbf{a} = (1, 2, 4) \)
- \( \mathbf{b} = (2, -3, -1) \)
- \( \mathbf{c} = (1, 4, -4) \)

### Step 2: Calculate the cross product \( \mathbf{b} \times \mathbf{c} \)
Using the formula for the cross product:
\[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & -1 \\ 1 & 4 & -4 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} -3 & -1 \\ 4 & -4 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & -1 \\ 1 & -4 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -3 & -1 \\ 4 & -4 \end{vmatrix} = (-3)(-4) - (-1)(4) = 12 + 4 = 16 \)
2. \( \begin{vmatrix} 2 & -1 \\ 1 & -4 \end{vmatrix} = (2)(-4) - (-1)(1) = -8 + 1 = -7 \)
3. \( \begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} = (2)(4) - (-3)(1) = 8 + 3 = 11 \)

Putting it all together:
\[ \mathbf{b} \times \mathbf{c} = 16\mathbf{i} + 7\mathbf{j} + 11\mathbf{k} = (16, 7, 11) \]

### Step 3: Calculate \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \)
Now we calculate \( \mathbf{a} \times (16, 7, 11) \):
\[ \mathbf{a} \times (16, 7, 11) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 4 \\ 16 & 7 & 11 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 2 & 4 \\ 7 & 11 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 4 \\ 16 & 11 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 16 & 7 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 4 \\ 7 & 11 \end{vmatrix} = (2)(11) - (4)(7) = 22 - 28 = -6 \)
2. \( \begin{vmatrix} 1 & 4 \\ 16 & 11 \end{vmatrix} = (1)(11) - (4)(16) = 11 - 64 = -53 \)
3. \( \begin{vmatrix} 1 & 2 \\ 16 & 7 \end{vmatrix} = (1)(7) - (2)(16) = 7 - 32 = -25 \)

Putting it all together:
\[ \mathbf{a} \times (16, 7, 11) = -6\mathbf{i} + 53\mathbf{j} - 25\mathbf{k} = (-6, 53, -25) \]

### Step 4: Determine orthogonality
A vector \( \mathbf{v} \) is orthogonal to another vector \( \mathbf{u} \) if their dot product is zero:
\[ \mathbf{u} \cdot \mathbf{v} = 0 \]

To find the vectors that are orthogonal to \( (-6, 53, -25) \), we can check the dot product with the original vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \).

### Step 5: Check orthogonality with \( \mathbf{a} \)
\[ \mathbf{a} \cdot (-6, 53, -25) = 1(-6) + 2(53) + 4(-25) = -6 + 106 - 100 = 0 \]

Thus, \( \mathbf{a} \) is orthogonal to \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \).

### Conclusion
The vector \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \) is orthogonal to \( \mathbf{a} \).

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