91Ó°ÊÓ

The vectors mentioned are \( \hat{\mathbf{i}} + \hat{\mathbf{j}} \) and \( \hat{\mathbf{j}} + \hat{\mathbf{k}} \). These are two vectors in 3D space.

A vector that is perpendicular to both the vectors \( \hat{\mathbf{i}} + \hat{\mathbf{j}} \) and \( \hat{\mathbf{j}} + \hat{\mathbf{k}} \) can be found using the cross product of these two vectors.

The cross product \( \mathbf{v} = (\hat{\mathbf{i}} + \hat{\mathbf{j}}) \times (\hat{\mathbf{j}} + \hat{\mathbf{k}}) \) is calculated as follows:\[ (\hat{\mathbf{i}} + \hat{\mathbf{j}}) \times (\hat{\mathbf{j}} + \hat{\mathbf{k}}) = \hat{\mathbf{i}} \times \hat{\mathbf{j}} + \hat{\mathbf{i}} \times \hat{\mathbf{k}} + \hat{\mathbf{j}} \times \hat{\mathbf{j}} + \hat{\mathbf{j}} \times \hat{\mathbf{k}}. \]Since \( \hat{\mathbf{j}} \times \hat{\mathbf{j}} = 0 \), the expression simplifies to:\[ \hat{\mathbf{i}} \times \hat{\mathbf{j}} + \hat{\mathbf{i}} \times \hat{\mathbf{k}} + \hat{\mathbf{j}} \times \hat{\mathbf{k}}. \]Using standard cross product results, \( \hat{\mathbf{i}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}}, \hat{\mathbf{i}} \times \hat{\mathbf{k}} = -\hat{\mathbf{j}} \), and \( \hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}} \).

Combining all components using the standard results, we have: \[ \mathbf{v} = \hat{\mathbf{k}} - \hat{\mathbf{j}} + \hat{\mathbf{i}}. \]

To confirm if the resulting vector \( \mathbf{v} = \hat{\mathbf{k}} - \hat{\mathbf{j}} + \hat{\mathbf{i}} \) is a unit vector, we calculate its magnitude.The magnitude is found by \( \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}. \)Since this magnitude is not 1, \( \mathbf{v} \) is not a unit vector.

There are indeed infinitely many vectors perpendicular to both given vectors, as shown by the non-zero result of the cross product. However, we can not find any unit vectors native from its representation since a normalizing factor would always be required. Thus, Statement I is given as false.

The cross product \( \mathbf{a} \times \mathbf{b} \) is geometrically defined to be perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). Since \( \mathbf{a} \) and \( \mathbf{b} \) lie in the same plane, the cross-product is perpendicular to the entire plane. Hence, Statement II is true.