Key Concepts and Formulas

1. Orthogonality of Cross Product: A vector resulting from a cross product, $\overrightarrow A \times \overrightarrow B$, is orthogonal to both $\overrightarrow A$ and $\overrightarrow B$. This implies $\overrightarrow A \cdot (\overrightarrow A \times \overrightarrow B) = 0$ and $\overrightarrow B \cdot (\overrightarrow A \times \overrightarrow B) = 0$.
2. Vector Triple Product Identity: For any three vectors $\overrightarrow a, \overrightarrow b, \overrightarrow c$, the identity $\overrightarrow a \times (\overrightarrow b \times \overrightarrow c) = (\overrightarrow a \cdot \overrightarrow c)\overrightarrow b - (\overrightarrow a \cdot \overrightarrow b)\overrightarrow c$ holds.
3. Properties of Dot Product: The dot product is distributive and commutative. Also, $\overrightarrow X \cdot \overrightarrow X = |\overrightarrow X|^2$.

Step-by-Step Solution

We are given the vectors $\overrightarrow u = \widehat i - \widehat j - 2\widehat k$ and $\overrightarrow v = 2\widehat i + \widehat j - \widehat k$. We are also given that $\overrightarrow v \cdot \overrightarrow w = 2$ and $\overrightarrow v \times \overrightarrow w = \overrightarrow u + \lambda \overrightarrow v$. We need to find $\overrightarrow u \cdot \overrightarrow w$.

Step 1: Calculate $\overrightarrow u \cdot \overrightarrow v$ and $|\overrightarrow v|^2$

To simplify later calculations, we first compute the dot product of $\overrightarrow u$ and $\overrightarrow v$, and the square of the magnitude of $\overrightarrow v$. $\overrightarrow u \cdot \overrightarrow v = (1)(2) + (-1)(1) + (-2)(-1) = 2 - 1 + 2 = 3$ $|\overrightarrow v|^2 = \overrightarrow v \cdot \overrightarrow v = (2)^2 + (1)^2 + (-1)^2 = 4 + 1 + 1 = 6$

Step 2: Determine the scalar $\lambda$

We are given the equation $\overrightarrow v \times \overrightarrow w = \overrightarrow u + \lambda \overrightarrow v$. Why this step? The cross product $\overrightarrow v \times \overrightarrow w$ is orthogonal to $\overrightarrow v$. Therefore, their dot product must be zero. This property will allow us to solve for $\lambda$. Take the dot product of both sides of the given equation with $\overrightarrow v$: $\overrightarrow v \cdot (\overrightarrow v \times \overrightarrow w) = \overrightarrow v \cdot (\overrightarrow u + \lambda \overrightarrow v)$ The left side, $\overrightarrow v \cdot (\overrightarrow v \times \overrightarrow w)$, is zero because $\overrightarrow v \times \overrightarrow w$ is orthogonal to $\overrightarrow v$. $0 = \overrightarrow v \cdot \overrightarrow u + \lambda (\overrightarrow v \cdot \overrightarrow v)$ Substitute the values calculated in Step 1: $0 = 3 + \lambda (6)$ $6\lambda = -3$ $\lambda = -\frac{1}{2}$

Step 3: Substitute $\lambda$ and rearrange the given equation

Now that we have the value of $\lambda$, we substitute it back into the given equation: $\overrightarrow v \times \overrightarrow w = \overrightarrow u - \frac{1}{2}\overrightarrow v$

Step 4: Use the Vector Triple Product Identity to relate $\overrightarrow u \cdot \overrightarrow w$

We want to find $\overrightarrow u \cdot \overrightarrow w$. We can introduce this term by using the vector triple product identity. Why this step? The vector triple product identity allows us to expand expressions involving nested cross products and can help isolate scalar products like $\overrightarrow u \cdot \overrightarrow w$. Cross-multiply both sides of the equation from Step 3 with $\overrightarrow u$: $\overrightarrow u \times (\overrightarrow v \times \overrightarrow w) = \overrightarrow u \times \left( \overrightarrow u - \frac{1}{2}\overrightarrow v \right)$ Apply the vector triple product identity to the left side: $\overrightarrow a \times (\overrightarrow b \times \overrightarrow c) = (\overrightarrow a \cdot \overrightarrow c)\overrightarrow b - (\overrightarrow a \cdot \overrightarrow b)\overrightarrow c$. Here, $\overrightarrow a = \overrightarrow u$, $\overrightarrow b = \overrightarrow v$, and $\overrightarrow c = \overrightarrow w$. $(\overrightarrow u \cdot \overrightarrow w)\overrightarrow v - (\overrightarrow u \cdot \overrightarrow v)\overrightarrow w = \overrightarrow u \times \overrightarrow u - \frac{1}{2}(\overrightarrow u \times \overrightarrow v)$ Since $\overrightarrow u \times \overrightarrow u = \overrightarrow 0$: $(\overrightarrow u \cdot \overrightarrow w)\overrightarrow v - (\overrightarrow u \cdot \overrightarrow v)\overrightarrow w = -\frac{1}{2}(\overrightarrow u \times \overrightarrow v)$ Substitute $\overrightarrow u \cdot \overrightarrow v = 3$ (from Step 1): $(\overrightarrow u \cdot \overrightarrow w)\overrightarrow v - 3\overrightarrow w = -\frac{1}{2}(\overrightarrow u \times \overrightarrow v)$

Step 5: Calculate $\overrightarrow u \times \overrightarrow v$

We need to compute the cross product $\overrightarrow u \times \overrightarrow v$ to substitute into the equation from Step 4. $\overrightarrow u \times \overrightarrow v = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & -1 & -2 \\ 2 & 1 & -1 \end{vmatrix}$ $= \widehat i ((-1)(-1) - (-2)(1)) - \widehat j ((1)(-1) - (-2)(2)) + \widehat k ((1)(1) - (-1)(2))$ $= \widehat i (1 + 2) - \widehat j (-1 + 4) + \widehat k (1 + 2)$ $= 3\widehat i - 3\widehat j + 3\widehat k$

Step 6: Substitute $\overrightarrow u \times \overrightarrow v$ and solve for $\overrightarrow u \cdot \overrightarrow w$

Substitute the result from Step 5 into the equation from Step 4: $(\overrightarrow u \cdot \overrightarrow w)\overrightarrow v - 3\overrightarrow w = -\frac{1}{2}(3\widehat i - 3\widehat j + 3\widehat k)$ $(\overrightarrow u \cdot \overrightarrow w)\overrightarrow v - 3\overrightarrow w = -\frac{3}{2}\widehat i + \frac{3}{2}\widehat j - \frac{3}{2}\widehat k$ To isolate $\overrightarrow u \cdot \overrightarrow w$, we can dot product this entire equation with $\overrightarrow v$. Why this step? Dotting with $\overrightarrow v$ will allow us to use the given $\overrightarrow v \cdot \overrightarrow w = 2$ and the calculated $|\overrightarrow v|^2 = \overrightarrow v \cdot \overrightarrow v = 6$, simplifying the equation to solve for $\overrightarrow u \cdot \overrightarrow w$. $\overrightarrow v \cdot \left( (\overrightarrow u \cdot \overrightarrow w)\overrightarrow v - 3\overrightarrow w \right) = \overrightarrow v \cdot \left( -\frac{3}{2}\widehat i + \frac{3}{2}\widehat j - \frac{3}{2}\widehat k \right)$ Apply the distributive property on the left side: $(\overrightarrow u \cdot \overrightarrow w)(\overrightarrow v \cdot \overrightarrow v) - 3(\overrightarrow v \cdot \overrightarrow w) = (2\widehat i + \widehat j - \widehat k) \cdot \left( -\frac{3}{2}\widehat i + \frac{3}{2}\widehat j - \frac{3}{2}\widehat k \right)$ Substitute the known values: $\overrightarrow v \cdot \overrightarrow v = 6$, $\overrightarrow v \cdot \overrightarrow w = 2$. $(\overrightarrow u \cdot \overrightarrow w)(6) - 3(2) = (2)\left(-\frac{3}{2}\right) + (1)\left(\frac{3}{2}\right) + (-1)\left(-\frac{3}{2}\right)$ $6(\overrightarrow u \cdot \overrightarrow w) - 6 = -3 + \frac{3}{2} + \frac{3}{2}$ $6(\overrightarrow u \cdot \overrightarrow w) - 6 = -3 + 3$ $6(\overrightarrow u \cdot \overrightarrow w) - 6 = 0$ $6(\overrightarrow u \cdot \overrightarrow w) = 6$ $\overrightarrow u \cdot \overrightarrow w = 1$

The final answer is $\boxed{1}$. This corresponds to option (D).