Find a unit vector perpendicular to each of the vector

We have,

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

$\therefore \vec{a}+\vec{b}=4 \hat{i}+4 \hat{j}, \vec{a}-\vec{b}=2 \hat{i}+4 \hat{k}$

$(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4\end{array}\right|=\hat{i}(16)-\hat{j}(16)+\hat{k}(-8)=16 \hat{i}-16 \hat{j}-8 \hat{k}$

$\therefore|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|=\sqrt{16^{2}+(-16)^{2}+(-8)^{2}}$

$=\sqrt{2^{2} \times 8^{2}+2^{2} \times 8^{2}+8^{2}}$

$=8 \sqrt{2^{2}+2^{2}+1}=8 \sqrt{9}=8 \times 3=24$

Hence, the unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ is given by the relation,

$=\pm \frac{(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})}{(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b}) \mid}=\pm \frac{16 \hat{i}-16 \hat{j}-8 \hat{k}}{24}$

$=\pm \frac{2 \hat{i}-2 \hat{j}-\hat{k}}{3}=\pm \frac{2}{3} \hat{i} \mp \frac{2}{3} \hat{j} \mp \frac{1}{3} \hat{k}$