91Ó°ÊÓ

The given problem states that the dot product of a unit vector with itself is 1: \( \hat{\mathbf{x}}^\prime \cdot \hat{\mathbf{x}}^\prime = 1 \), where \( \hat{\mathbf{x}}^\prime \) is a unit vector in a rotated system. We also need to verify that this property leads to \( \hat{\mathbf{x}} \cdot \hat{\mathbf{y}} = 0 \) in the original coordinate system when the primed system is rotated by \( 45^\circ \) about the \( z \)-axis.

When a coordinate system is rotated \( 45^\circ \) about the \( z \)-axis, the transformation of the unit vector \( \hat{\mathbf{x}} \) can be described by:\[\begin{align*}\hat{\mathbf{x}}^\prime &= \cos(45^\circ) \hat{\mathbf{x}} + \sin(45^\circ) \hat{\mathbf{y}}, \\hat{\mathbf{y}}^\prime &= -\sin(45^\circ) \hat{\mathbf{x}} + \cos(45^\circ) \hat{\mathbf{y}}.\end{align*}\]Here \( \cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2} \).

Substitute the rotation transformations into the dot product:\[\hat{\mathbf{x}}^\prime \cdot \hat{\mathbf{x}}^\prime = \left( \frac{\sqrt{2}}{2} \hat{\mathbf{x}} + \frac{\sqrt{2}}{2} \hat{\mathbf{y}} \right) \cdot \left( \frac{\sqrt{2}}{2} \hat{\mathbf{x}} + \frac{\sqrt{2}}{2} \hat{\mathbf{y}} \right).\]Expanding the dot product:\[= \left( \frac{\sqrt{2}}{2} \right)^2 (\hat{\mathbf{x}} \cdot \hat{\mathbf{x}}) + \left( \frac{\sqrt{2}}{2} \right)^2 (\hat{\mathbf{y}} \cdot \hat{\mathbf{y}}) + 2 \cdot \left( \frac{\sqrt{2}}{2} \right)^2 (\hat{\mathbf{x}} \cdot \hat{\mathbf{y}}).\]Since we know \( \hat{\mathbf{x}} \cdot \hat{\mathbf{x}} = \hat{\mathbf{y}} \cdot \hat{\mathbf{y}} = 1 \), the expression simplifies to:\[= \frac{1}{2} + \frac{1}{2} + (\hat{\mathbf{x}} \cdot \hat{\mathbf{y}}).\]

Since \( \hat{\mathbf{x}}^\prime \cdot \hat{\mathbf{x}}^\prime = 1 \), the expanded expression becomes:\[\frac{1}{2} + \frac{1}{2} + (\hat{\mathbf{x}} \cdot \hat{\mathbf{y}}) = 1.\]This implies \( \hat{\mathbf{x}} \cdot \hat{\mathbf{y}} = 0 \), confirming that the two vectors are orthogonal.