Unit Vectors — Question 2

Question 2

A unit direction need not be a coordinate axis

Consider $\hat u=(\hat\imath+\hat\jmath)/\sqrt2$ and $\hat v=(-\hat\imath+\hat\jmath)/\sqrt2$ in the usual perpendicular plane. A Cartesian basis consists of two perpendicular unit directions.

1. Find the lengths and directions of $\hat u$ and $\hat v$.

2. Do they form another Cartesian basis? Explain geometrically.

3. Express the old east direction $\hat\imath$ and north direction $\hat\jmath$ in terms of $\hat u,\hat v$.

Question 2 - Solution

A unit direction need not be a coordinate axis

(a) Check size and quadrant.

$$|\hat u|=|\hat v|=\sqrt{\tfrac12+\tfrac12}=1,\quad\theta_u=45^\circ,\quad\theta_v=135^\circ$$ Each length is $\sqrt{1/2+1/2}=1$. Their standard directions are $45^\circ$ and $135^\circ$. A unit vector can point in any direction; the hats do not restrict an arrow to the original axes.

(b) Check the angle between the directions.

$$\hat u\cdot\hat v=\frac{-1+1}{2}=0,\quad\theta_v-\theta_u=90^\circ$$ The difference of direction angles is $90^\circ$, so they are perpendicular. The ordered pair $\hat u,\hat v$ is the original basis rotated counterclockwise by $45^\circ$. They form a Cartesian basis despite having two nonzero original components each.

(c) Resolve the old axes in the new basis.

$$\boxed{\hat i=\frac{\hat u-\hat v}{\sqrt2},\qquad\hat j=\frac{\hat u+\hat v}{\sqrt2}}$$ Subtraction gives $\hat u-\hat v=\sqrt2\hat\imath$ and addition gives $\hat u+\hat v=\sqrt2\hat\jmath$. Thus $\hat\imath=(\hat u-\hat v)/\sqrt2$ and $\hat\jmath=(\hat u+\hat v)/\sqrt2$. Substitution reproduces the original east and north arrows.