Unit Vectors — Question 10

Question 10

Direction data can be dimensionally invalid

A device reports a unit direction $\hat u=(0.6\hat\imath+0.8\hat\jmath)\ \mathrm{m/s}$ and speed 15 m/s. Another operator writes $\vec v=(15\ \mathrm{m/s})\hat u$ using that report literally.

1. Check the numerical norm and the physical dimensions of the reported unit direction.

2. Give a consistent repaired direction and velocity, assuming the numerical direction coefficients were intended.

3. Could $6\hat\imath+8\hat\jmath$ be a dimensionless unit direction? Normalize it and compare.

Question 10 - Solution

Direction data can be dimensionally invalid

(a) Unit norm alone is not the whole definition.

$$\sqrt{.6^2+.8^2}=1,\quad[(15\,\mathrm{m/s})\hat u_{\mathrm{reported}}]=L^2T^{-2}\ne LT^{-1}$$ The numerical norm is $\sqrt{0.36+0.64}=1$, but the stated object has speed units and magnitude 1 m/s. It is a velocity vector, not a dimensionless unit direction. Multiplying it literally by speed would give units $\mathrm{m^2/s^2}$, not velocity.

(b) Remove the misplaced unit from direction data.

$$\boxed{\hat u=(.6,.8),\quad\vec v=15\hat u=(9,12)\,\mathrm{m/s}},\quad v=15\,\mathrm{m/s}$$ The intended direction is $\hat u=0.6\hat\imath+0.8\hat\jmath$. Then $\vec v=(9\hat\imath+12\hat\jmath)\ \mathrm{m/s}$ and $|\vec v|=15$ m/s. This repair is justified only by the stated interpretation of the device output.

(c) Check both size and units.

$$|(6,8)|=10,\quad\frac{(6,8)}{10}=(.6,.8)=\hat u$$ The dimensionless vector $6\hat\imath+8\hat\jmath$ has magnitude 10, so it is not a unit vector. Dividing by 10 gives the same $\hat u$. Being dimensionless is necessary for these direction vectors, but does not guarantee unit magnitude.