The unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, where $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}$, is :

To find the unit vector perpendicular to both $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, we need to calculate their cross product and then normalize it.

Given:

$\vec{a} = \hat{i} + \hat{j} + \hat{k}$

$\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$

Let's calculate $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$:

$\vec{a} + \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k}) = 2\hat{i} + 3\hat{j} + 4\hat{k}$

$\vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = -\hat{j} - 2\hat{k}$

A vector perpendicular to both would be their cross product:

$(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})$

Computing this cross product using the determinant:

$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix}$

$= (3 \times (-2) - 4 \times (-1))\hat{i} - (2 \times (-2) - 4 \times 0)\hat{j} + (2 \times (-1) - 3 \times 0)\hat{k}$

$= (-6 + 4)\hat{i} - (-4)\hat{j} + (-2)\hat{k}$

$= -2\hat{i} + 4\hat{j} - 2\hat{k}$

To make this a unit vector, we divide by its magnitude:

Magnitude = $\sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$

Therefore, the unit vector is:

$\dfrac{1}{2\sqrt{6}}(-2\hat{i} + 4\hat{j} - 2\hat{k}) = \dfrac{1}{\sqrt{6}}(-\hat{i} + 2\hat{j} - \hat{k})$