Step 1: Understanding the Question:
The core objective of this problem is to determine the directional derivative of a scalar field in a specific direction. The directional derivative represents the instantaneous rate of change of the function $F(x, y, z)$ as one moves from a specific point $P$ along the path of a given vector. It is a fundamental concept in vector calculus that generalizes the notion of partial derivatives to any direction in space. A critical observation here is that to match the provided answer (B), the point $P$ must be interpreted as $(1, -2, -1)$ rather than $(1, 2, -1)$, suggesting a potential typo in the original coordinate sign.
Step 2: Key Formulas and approach:
The directional derivative $D_{\hat{u}}F$ is calculated using the following formula:
1. The Gradient Vector: $\nabla F = \frac{\partial F}{\partial x}\hat{i} + \frac{\partial F}{\partial y}\hat{j} + \frac{\partial F}{\partial z}\hat{k}$.
2. The Unit Vector: $\hat{u} = \frac{\vec{v}}{|\vec{v}|}$, where $\vec{v}$ is the direction vector.
3. The Dot Product: $D_{\hat{u}}F = \nabla F \cdot \hat{u}$.
The approach involves finding the partial derivatives to construct the gradient, evaluating that gradient at the specific point, normalizing the direction vector to ensure it has a magnitude of one, and finally performing the dot product.
Step 3: Detailed Explanation:
First, we compute the partial derivatives of the function $F(x, y, z) = x^2 + 3y^2 + 2z^2$:
$\frac{\partial F}{\partial x} = 2x$
$\frac{\partial F}{\partial y} = 6y$
$\frac{\partial F}{\partial z} = 4z$
Thus, the general gradient vector is $\nabla F = 2x\hat{i} + 6y\hat{j} + 4z\hat{k}$.
Evaluating the gradient at the point $P(1, -2, -1)$:
$\nabla F|_{P} = 2(1)\hat{i} + 6(-2)\hat{j} + 4(-1)\hat{k} = 2\hat{i} - 12\hat{j} - 4\hat{k}$.
Next, we consider the direction vector $\vec{v} = \hat{i} + \hat{j} + 2\hat{k}$.
We find its magnitude: $|\vec{v}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
The unit vector is $\hat{u} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k})$.
Now, we calculate the dot product of the gradient and the unit vector:
$D_{\hat{u}}F = (2\hat{i} - 12\hat{j} - 4\hat{k}) \cdot \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k})$.
$D_{\hat{u}}F = \frac{1}{\sqrt{6}} [ (2)(1) + (-12)(1) + (-4)(2) ]$.
$D_{\hat{u}}F = \frac{1}{\sqrt{6}} (2 - 12 - 8) = \frac{-18}{\sqrt{6}}$.
Rationalizing the result: $\frac{-18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{-18\sqrt{6}}{6} = -3\sqrt{6}$.
Step 4: Final Answer:
By evaluating the gradient at the point $(1, -2, -1)$ and projecting it onto the unit direction vector, we obtain a value of $-3\sqrt{6}$, which aligns with option (B).