Question 38: The unit vector perpendicular to each of the vectors ( a→ + b→ ) and ( a→ − b→ ) , where $\overrightarrow{a}=i+j+k$ and$\overrightarrow{\mathrm{ b}}=i+2j+3k$  is:

The correct option is:
$-\frac{1}{\sqrt{6}}\hat{i}+\frac{2}{\sqrt{6}}\hat{j}-\frac{2}{\sqrt{6}}\hat{k}$

Step-by-step Derivation:

1. Find the vectors $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$:
Given:
$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$
$\overrightarrow{b}=\hat{i}+2\hat{j}+3\hat{k}$

Now we calculate the sum of the two vectors:
$\overrightarrow{a}+\overrightarrow{b}=(1+1)\hat{i}+(1+2)\hat{j}+(1+3)\hat{k}=2\hat{i}+3\hat{j}+4\hat{k}$

Next, we calculate the difference of the two vectors:
$\overrightarrow{a}-\overrightarrow{b}=(1-1)\hat{i}+(1-2)\hat{j}+(1-3)\hat{k}=0\hat{i}-\hat{j}-2\hat{k}$

2. Find a vector perpendicular to both vectors:
A vector $\overrightarrow{c}$ perpendicular to both $(\overrightarrow{a}+\overrightarrow{b})$ and $(\overrightarrow{a}-\overrightarrow{b})$ is given by their cross product:
$\overrightarrow{c}=(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})$

Using the determinant method to calculate the cross product:
$\overrightarrow{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 0& -1& -2\end{array}\right|$

Expanding the determinant along the first row:
$\overrightarrow{c}=\hat{i}\left[\right(3\left)\right(-2)-(4\left)\right(-1\left)\right]-\hat{j}\left[\right(2\left)\right(-2)-(4\left)\right(0\left)\right]+\hat{k}\left[\right(2\left)\right(-1)-(3\left)\right(0\left)\right]$
$\overrightarrow{c}=\hat{i}[-6+4]-\hat{j}[-4-0]+\hat{k}[-2-0]$
$\overrightarrow{c}=-2\hat{i}+4\hat{j}-2\hat{k}$

3. Compute the magnitude of the perpendicular vector:
$\left|\overrightarrow{c}\right|=\sqrt{{(-2)}^2+{\left(4\right)}^2+{(-2)}^2}=\sqrt{4+16+4}=\sqrt{24}=2\sqrt{6}$

4. Find the unit vector:
The unit vector $\hat{u}$ in the direction of $\overrightarrow{c}$ is:
$\hat{u}=\frac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}=\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{2\sqrt{6}}=-\frac{1}{\sqrt{6}}\hat{i}+\frac{2}{\sqrt{6}}\hat{j}-\frac{1}{\sqrt{6}}\hat{k}$

Allowing for a minor typographical print error in the coefficient of the component $\hat{k}$ in the options (where the numerator of the coefficient of $\hat{k}$ is written as 2 instead of 1), the matching option is:
$-\frac{1}{\sqrt{6}}\hat{i}+\frac{2}{\sqrt{6}}\hat{j}-\frac{2}{\sqrt{6}}\hat{k}$