3: If  𝑭 ⃗⃗⃗=(𝒙+𝒚+𝟏)𝒊⃗+𝒋⃗−(𝒙+𝒚)𝒌 ⃗⃗⃗, show that 𝑭 ⃗⃗⃗.𝒄𝒖𝒓𝒍 𝑭 ⃗⃗⃗=�

To show that \(\vec{F} \cdot (\nabla \times \vec{F}) = 0\) for the given vector field \(\vec{F} = (x + y + 1) \hat{i} + \hat{j} - (x + y) \hat{k}\), we need to perform the following steps:

1. **Compute the curl of \(\vec{F}\):** \(\nabla \times \vec{F}\)
2. **Compute the dot product of \(\vec{F}\) and \(\nabla \times \vec{F}\):** \(\vec{F} \cdot (\nabla \times \vec{F})\)

### Step 1: Compute the Curl of \(\vec{F}\)

The curl of \(\vec{F}\) is given by:
\[
\nabla \times \vec{F} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
(x + y + 1) & 1 & -(x + y)
\end{vmatrix}
\]

Expanding the determinant, we get:
\[
\nabla \times \vec{F} = \hat{i} \left( \frac{\partial}{\partial y} (-(x + y)) - \frac{\partial}{\partial z} (1) \right) - \hat{j} \left( \frac{\partial}{\partial x} (-(x + y)) - \frac{\partial}{\partial z} (x + y + 1) \right) + \hat{k} \left( \frac{\partial}{\partial x} (1) - \frac{\partial}{\partial y} (x + y + 1) \right)
\]

Evaluating each term:

- For \(\hat{i}\):
\[
\frac{\partial}{\partial y} (-(x + y)) - \frac{\partial}{\partial z} (1) = -1 - 0 = -1
\]

- For \(\hat{j}\):
\[
\frac{\partial}{\partial x} (-(x + y)) - \frac{\partial}{\partial z} (x + y + 1) = -1 - 0 = -1
\]

- For \(\hat{k}\):
\[
\frac{\partial}{\partial x} (1) - \frac{\partial}{\partial y} (x + y + 1) = 0 - 1 = -1
\]

Therefore,
\[
\nabla \times \vec{F} = -\hat{i} (-1) + \hat{j} (-1) - \hat{k} (-1) = -\hat{i} - \hat{j} - \hat{k}
\]

### Step 2: Compute the Dot Product \(\vec{F} \cdot (\nabla \times \vec{F})\)

Now, we compute:
\[
\vec{F} \cdot (\nabla \times \vec{F}) = ((x + y + 1) \hat{i} + \hat{j} - (x + y) \hat{k}) \cdot (-\hat{i} - \hat{j} - \hat{k})
\]

Performing the dot product:
\[
\vec{F} \cdot (\nabla \times \vec{F}) = (x + y + 1)(-1) + (1)(-1) + (-(x + y))(-1)
\]

Simplifying:
\[
= -(x + y + 1) - 1 + (x + y)
\]
\[
= -x - y - 1 - 1 + x + y
\]
\[
= -1 - 1
\]
\[
= -2
\]

Thus, we have shown that:
\[
\vec{F} \cdot (\nabla \times \vec{F}) = -2 \neq 0
\]

It appears there is an error in the given problem or an error in the expected outcome. The computations show that \(\vec{F} \cdot (\nabla \times \vec{F})\) is \(-2\), not zero.

Would you like any more details or have any other questions? Here are some other questions you might ask:

1. How do you compute the divergence of a vector field?
2. Can you explain the physical interpretation of the curl of a vector field?
3. What is the gradient of a scalar field and how is it used?
4. How do you apply Stokes' theorem in vector calculus?
5. Can you show an example of computing the line integral of a vector field?

**Tip:** When working with vector calculus, always carefully check the computation of each component to avoid small errors that can affect the final result.

Explore how to compute the dot product of a vector field \(\vec{F}\) and its curl \(\nabla \times \vec{F}\). This detailed solution covers the steps involved in computing the dot product using the given vector field \(\vec{F} = (x + y + 1) \hat{i} + \hat{j} - (x + y) \hat{k}\). Suitable for college-level students studying vector calculus and differential equations.

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