Unit Vectors — Question 1

Question 1

Separate direction from physical size

In perpendicular east–north axes, $\vec A=(6\hat\imath+8\hat\jmath)\ \mathrm m$. The basis vectors $\hat\imath$ and $\hat\jmath$ have magnitude one, are dimensionless, and point east and north.

1. Find $|\vec A|$ and the unit vector $\hat u$ in its direction.

2. Write a 25 m displacement in the same direction and verify its length.

3. Explain why $\vec A/10$ and $\vec A/(10\ \mathrm m)$ are dimensionally different.

Question 1 - Solution

Separate direction from physical size

(a) Normalize by the magnitude.

$$A=\sqrt{6^2+8^2}=10\,\mathrm m,\quad\boxed{\hat u=(.6,.8)},\quad|\hat u|^2=.36+.64=1$$ $|\vec A|=\sqrt{6^2+8^2}=10$ m, so $\hat u=\vec A/|\vec A|=0.6\hat\imath+0.8\hat\jmath$. Its norm is $\sqrt{0.6^2+0.8^2}=1$ and it carries no length unit.

(b) Restore the requested size.

$$\vec B=25(.6,.8)=(15,20)\,\mathrm m,\quad B=\sqrt{225+400}=25\,\mathrm m$$ $\vec B=(25\ \mathrm m)\hat u=(15\hat\imath+20\hat\jmath)\ \mathrm m$. Its length is $\sqrt{225+400}=25$ m. Scaling by a positive magnitude preserves the direction.

(c) Cancel physical units as well as numerical size.

$$[\vec A/10]=L,\qquad[\vec A/(10\,\mathrm m)]=1$$ $\vec A/10$ still has units of metres and has magnitude 1 m. It is a displacement, not a dimensionless unit vector. Dividing by the physical magnitude $10$ m removes the units and isolates direction. This distinction remains important even when the numerical coefficients look identical.