Unit Vectors — Question 3

Question 3

A negative scale reverses a unit direction

Let $\hat u=(3\hat\imath-4\hat\jmath)/5$ and $\vec B=(-15\ \mathrm m)\hat u$. A student calls $-15$ m the magnitude of $\vec B$.

1. Find the components and actual magnitude of $\vec B$.

2. Find the unit vector in the direction of $\vec B$.

3. Explain what happens for a general real multiplier $c$ in $\vec C=c\hat u$, including $c=0$.

Question 3 - Solution

A negative scale reverses a unit direction

(a) Distribute the signed coefficient.

$$\vec B=-15(3/5,-4/5)=(-9,12)\,\mathrm m,\quad\boxed{B=\sqrt{81+144}=15\,\mathrm m}$$ $\vec B=(-9\hat\imath+12\hat\jmath)\ \mathrm m$, with magnitude $\sqrt{81+144}=15$ m. A magnitude is nonnegative. The minus sign belongs to the multiplier and encodes reversal, not a negative length.

(b) Normalize the resulting vector.

$$\hat b=\frac{(-9,12)}{15}=(-3/5,4/5)=-\hat u$$ $\hat b=\vec B/(15\ \mathrm m)=-3\hat\imath/5+4\hat\jmath/5=-\hat u$. Its length is one and it points northwest, opposite to the southeast direction of $\hat u$.

(c) Separate sign, size and degeneracy.

$$|c\hat u|=|c|,\quad\widehat{c\hat u}=\operatorname{sgn}(c)\hat u\ (c\ne0);\quad c=0\Rightarrow\vec C=\vec0$$ For $c\ne0$, $|\vec C|=|c|$ and its unit direction is $\operatorname{sgn}(c)\hat u$. Positive $c$ preserves direction and negative $c$ reverses it. When $c=0$, the vector is zero and has no unique direction; it cannot be normalized by dividing by its magnitude.