Unit Vectors — Question 7

Question 7

Not every hatted symbol is normalized

A worksheet defines $\hat e=2\hat\imath$ and $\hat f=2\hat\jmath$, then claims that $(3\hat e+4\hat f)\ \mathrm m$ has magnitude 5 m because the symbols have hats.

1. Find the true lengths of $\hat e$ and $\hat f$ and the true vector magnitude.

2. Replace them with genuine unit vectors, preserving the physical vector.

3. Alternatively preserve the coefficients 3 and 4 with a true unit basis. What physical vector results, and why is it different?

Question 7 - Solution

Not every hatted symbol is normalized

(a) Check definitions rather than typography.

$$|\hat e|=|\hat f|=2,\quad\vec A=(6,8)\,\mathrm m,\quad\boxed{A=\sqrt{36+64}=10\,\mathrm m}$$ $|\hat e|=|\hat f|=2$, so they are perpendicular but not unit vectors. The physical vector is $(6\hat\imath+8\hat\jmath)$ m, of magnitude 10 m. The hats are misleading; they cannot alter the definitions.

(b) Rescale the coefficients inversely.

$$\hat e_0=\hat e/2=\hat i,\quad\hat f_0=\hat f/2=\hat j,\quad\vec A=(6\hat e_0+8\hat f_0)\,\mathrm m$$ Define $\hat e_0=\hat e/2=\hat\imath$ and $\hat f_0=\hat f/2=\hat\jmath$. The same vector is $(6\hat e_0+8\hat f_0)$ m. Shrinking basis arrows by two requires doubling coefficients to keep the physical arrow unchanged.

(c) Recognize when a repair changes the object.

$$\vec A_{\mathrm{changed}}=(3\hat e_0+4\hat f_0)\,\mathrm m=\tfrac12\vec A,\quad A_{\mathrm{changed}}=5\,\mathrm m$$ Using $(3\hat e_0+4\hat f_0)$ m gives $(3\hat\imath+4\hat\jmath)$ m, magnitude 5 m. This is half the original physical vector. Correcting both notation and physical interpretation requires deciding what is held fixed: the arrow or the numerical coefficients.