Answer :
Answer:
To find the directional derivative of \( \phi = \sqrt{x^2 + y^2 + z^2} \) at the point \( (3, 1, 2) \) in the direction of the vector \( \vec{v} = yz \, \mathbf{i} + zx \, \mathbf{j} + xy \, \mathbf{k} \), we proceed as follows:
1. **Compute the gradient of \( \phi \)**:
The gradient of \( \phi \) is given by:
\[
\nabla \phi = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right) = \left( \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \frac{z}{\sqrt{x^2 + y^2 + z^2}} \right)
\]
2. **Evaluate the gradient at \( (3, 1, 2) \)**:
Substitute \( x = 3 \), \( y = 1 \), \( z = 2 \):
\[
\nabla \phi (3, 1, 2) = \left( \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}} \right)
\]
3. **Normalize the vector \( \vec{v} \)**:
Calculate the magnitude of \( \vec{v} \):
\[
|\vec{v}| = \sqrt{(yz)^2 + (zx)^2 + (xy)^2} = \sqrt{y^2 z^2 + z^2 x^2 + x^2 y^2} = |xyz|
\]
Since \( x = 3 \), \( y = 1 \), \( z = 2 \):
\[
|\vec{v}| = |3 \cdot 1 \cdot 2| = 6
\]
Normalize \( \vec{v} \):
\[
\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{yz \, \mathbf{i} + zx \, \mathbf{j} + xy \, \mathbf{k}}{6}
\]
4. **Compute the directional derivative**:
The directional derivative \( D_{\vec{v}} \phi \) is given by:
\[
D_{\vec{v}} \phi = \nabla \phi \cdot \hat{v}
\]
Substitute \( \nabla \phi (3, 1, 2) \) and \( \hat{v} \):
\[
D_{\vec{v}} \phi = \left( \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}} \right) \cdot \frac{yz \, \mathbf{i} + zx \, \mathbf{j} + xy \, \mathbf{k}}{6}
\]
Calculate the dot product:
\[
D_{\vec{v}} \phi = \frac{1}{6\sqrt{14}} \left( 3yz + zx + 2xy \right)
\]
Substitute \( x = 3 \), \( y = 1 \), \( z = 2 \):
\[
D_{\vec{v}} \phi = \frac{1}{6\sqrt{14}} \left( 3 \cdot 1 \cdot 2 + 3 \cdot 2 \cdot 3 + 2 \cdot 1 \cdot 3 \right)
\]
\[
D_{\vec{v}} \phi = \frac{1}{6\sqrt{14}} \left( 6 + 18 + 6 \right)
\]
\[
D_{\vec{v}} \phi = \frac{1}{6\sqrt{14}} \cdot 30 = \frac{5}{\sqrt{14}}
\]
Therefore, the directional derivative of \( \phi = \sqrt{x^2 + y^2 + z^2} \) at the point \( (3, 1, 2) \) in the direction of \( \vec{v} = yz \, \mathbf{i} + zx \, \mathbf{j} + xy \, \mathbf{k} \) is \( \frac{5}{\sqrt{14}} \).