Calculate the vector triple product (a x b) x c given vectors a = i + j + k, b = i + j, and c = i.

We are given the vectors:

$$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$ $$\vec{b} = \hat{i} + \hat{j}$$ $$\vec{c} = \hat{i}$$

The vector triple product identity states that for any three vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$:

$$(\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{B} \cdot \vec{C})\vec{A}$$

In this problem, $\vec{A} = \vec{a}$, $\vec{B} = \vec{b}$, and $\vec{C} = \vec{c}$.
So, we can write:

$$(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$$

First, calculate $\vec{a} \cdot \vec{c}$:

$$\vec{a} \cdot \vec{c} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i})$$ $$\vec{a} \cdot \vec{c} = (1)(1) + (1)(0) + (1)(0)$$ $$\vec{a} \cdot \vec{c} = 1$$

Next, calculate $\vec{b} \cdot \vec{c}$:

$$\vec{b} \cdot \vec{c} = (\hat{i} + \hat{j}) \cdot (\hat{i})$$ $$\vec{b} \cdot \vec{c} = (1)(1) + (1)(0)$$ $$\vec{b} \cdot \vec{c} = 1$$

$$(\vec{a} \times \vec{b}) \times \vec{c} = (1)\vec{b} - (1)\vec{a}$$ $$(\vec{a} \times \vec{b}) \times \vec{c} = \vec{b} - \vec{a}$$

We are given that $(\vec{a} \times \vec{b}) \times \vec{c} = \lambda \vec{a} + \mu \vec{b}$.
Equating the two expressions for $(\vec{a} \times \vec{b}) \times \vec{c}$:

$$\vec{b} - \vec{a} = \lambda \vec{a} + \mu \vec{b}$$

Rearranging the terms to match the form $\lambda \vec{a} + \mu \vec{b}$:

$$(-1)\vec{a} + (1)\vec{b} = \lambda \vec{a} + \mu \vec{b}$$

By comparing the coefficients of $\vec{a}$ and $\vec{b}$ on both sides, we find:

$$\lambda = -1$$ $$\mu = 1$$

$$\lambda + \mu = -1 + 1$$ $$\lambda + \mu = 0$$

The value of $\lambda + \mu$ is 0.