Find a unit vector perpendicular to each of the vector (\(\vec{a}\) +\(\vec{b}\)) and (\(\vec{a}\)  -\(\vec{b}\)), where \(\vec{a}\)=3\(\hat{i}\) + 2\(\hat{j}\) + 2 \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) +2\(\hat{j}\) -2 \(\hat{k}\) 

±(2/3)\(\hat{i}\) ± (2/3)\(\hat{j}\)  ± (1/3) \(\hat{k}\)

We have, \(\vec{a}\)=3\(\hat{i}\) + 2\(\hat{j}\) + 2 \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) +2\(\hat{j}\) -2 \(\hat{k}\) 

(\(\vec{a}\) + \(\vec{b}\))= 4\(\hat{i}\)+  4\(\hat{j}\)

(\(\vec{a}\) -\(\vec{b}\) ) = 2\(\hat{i}\) + 4 \(\hat{j}\)

(\(\vec{a}\) +\(\vec{b}\)) x (\(\vec{a}\)  -\(\vec{b}\)) = 16\(\hat{i}\) - 16\(\hat{j}\) - 8 \(\hat{k}\)

 |(\(\vec{a}\) +\(\vec{b}\)) x (\(\vec{a}\)  -\(\vec{b}\))| = √(16)² + (-16)² + (-8)²

|(\(\vec{a}\) +\(\vec{b}\)) x (\(\vec{a}\)  -\(\vec{b}\))| = 24

Hence, unit vector perpendicular to each of the vector (\(\vec{a}\) +\(\vec{b}\)) x (\(\vec{a}\)  -\(\vec{b}\)) is given by the relation = ±{((\(\vec{a}\) +\(\vec{b}\)) x (\(\vec{a}\)  -\(\vec{b}\)) }/ |(\(\vec{a}\) +\(\vec{b}\)) x (\(\vec{a}\)  -\(\vec{b}\))| 

                                                                                                                                                = ±{16\(\hat{i}\) - 16\(\hat{j}\) - 8 \(\hat{k}\)}/24 

                                                                                                                                                = ±(2/3)\(\hat{i}\)  ± (2/3)\(\hat{j}\)  ± (1/3) \(\hat{k}\)