Given that a = i+j+k, b = i-j+k and c = i+j–k, find (a x b).c:

To find \((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\), where \(\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}\), \(\mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}\), and \(\mathbf{c} = \mathbf{i} + \mathbf{j} - \mathbf{k}\), we will proceed step-by-step:

### Step 1: Compute \(\mathbf{a} \times \mathbf{b}\)

Given:
\[
\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}
\]
\[
\mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}
\]

The cross product \(\mathbf{a} \times \mathbf{b}\) is determined by the following determinant:
\[
\mathbf{a} \times \mathbf{b} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 1 & 1 \\
1 & -1 & 1 
\end{vmatrix}
\]

Expand this determinant:

\[
\mathbf{a} \times \mathbf{b} = \mathbf{i} \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix}
\]

Calculating each 2x2 determinant:
\[
\begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = (1)(1) - (1)(-1) = 1 + 1 = 2
\]
\[
\begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(1) = 1 - 1 = 0
\]
\[
\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = (1)(-1) - (1)(1) = -1 - 1 = -2
\]

Thus, the cross product is:
\[
\mathbf{a} \times \mathbf{b} = 2\mathbf{i} - 0\mathbf{j} - 2\mathbf{k} = 2\mathbf{i} - 2\mathbf{k}
\]

### Step 2: Compute \((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\)

Given:
\[
\mathbf{c} = \mathbf{i} + \mathbf{j} - \mathbf{k}
\]

Now, calculate the dot product:
\[
(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = (2\mathbf{i} - 2\mathbf{k}) \cdot (\mathbf{i} + \mathbf{j} - \mathbf{k})
\]

Compute the dot product by multiplying corresponding components:
\[
(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = (2)(1) + (0)(1) + (-2)(-1) = 2 + 0 + 2 = 4
\]

So, the correct answer is:
\[
\boxed{4}
\]

However, none of the provided options matches this answer, so it appears the choices may have a typo. Given our solution steps, the correct value for \((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\) is indeed \(4\).

Learn how to calculate (a x b) dot c using vector algebra. This problem involves finding the dot product after computing the cross product of vectors a, b, and c. Suitable for advanced high school students.

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