Directional derivative of \[\phi \] = 2xz – y2 at the point (1, 3, 2) becomes maximum in the direction of: A. \[{\rm{4\hat i}} + 2{\rm{\hat j}} – 3{\rm{\hat k}}\] B. \[{\rm{4\hat i}} – 6{\rm{\hat j}} + 2{\rm{\hat k}}\] C. \[{\rm{2\hat i}} – 6{\rm{\hat j}} + 2{\rm{\hat k}}\] D. \[{\rm{4\hat i}} – 6{\rm{\hat j}} – 2{\rm{\hat k}}\]

”[{ m{4hat
” option2=”\[{\rm{4\hat i}} – 6{\rm{\hat j}} + 2{\rm{\hat k}}\]” option3=”\[{\rm{2\hat i}} – 6{\rm{\hat j}} + 2{\rm{\hat k}}\]” option4=”\[{\rm{4\hat i}} – 6{\rm{\hat j}} – 2{\rm{\hat k}}\]” correct=”option3″]

The correct answer is $\boxed{\text{(B)}}$.

The directional derivative of a function $\phi$ at a point $P$ in the direction of a unit vector $\hat{u}$ is given by:

$$D_{\hat{u}} \phi(P) = \nabla \phi(P) \cdot \hat{u}$$

where $\nabla \phi(P)$ is the gradient of $\phi$ at $P$.

In this case, we have:

$$\phi(x, y, z) = 2xz – y^2$$

$$\nabla \phi(x, y, z) = (2x + 2z) \hat{i} + (2y) \hat{j} – 2y \hat{k}$$

Substituting $x = 1$, $y = 3$, and $z = 2$, we get:

$$\nabla \phi(1, 3, 2) = (4 + 4) \hat{i} + (6) \hat{j} – (4) \hat{k} = 4 \hat{i} – 6 \hat{j} + 2 \hat{k}$$

The directional derivative of $\phi$ at the point $(1, 3, 2)$ in the direction of $\hat{u}$ is then:

$$D_{\hat{u}} \phi(1, 3, 2) = (4 \hat{i} – 6 \hat{j} + 2 \hat{k}) \cdot \hat{u}$$

The maximum value of the directional derivative occurs when $\hat{u}$ is in the same direction as $\nabla \phi(1, 3, 2)$. Therefore, the maximum value of the directional derivative is:

$$\max_{\hat{u}} D_{\hat{u}} \phi(1, 3, 2) = (4 \hat{i} – 6 \hat{j} + 2 \hat{k}) \cdot (4 \hat{i} – 6 \hat{j} + 2 \hat{k}) = 24$$

Therefore, the directional derivative of $\phi$ at the point $(1, 3, 2)$ becomes maximum in the direction of $\boxed{\text{(B)}}$.