Multiplying Vectors — Question 5

Question 5

The cross product includes a sense of rotation

Vector A has length 6 m and points east. Vector B has length 4 m and points $30^\circ$ north of east. Positive z points out of the page. Use $|\vec A\times\vec B|=AB\sin\theta$ and the right-hand rule.

1. Find the full vector $\vec A\times\vec B$ and the area of the triangle formed by the two arrows.

2. Find $\vec B\times\vec A$. Explain why the parallelogram area does not change when the order changes.

3. If B instead points $30^\circ$ south of east, find the new cross product. What happens when B points exactly west?

Question 5 - Solution

The cross product includes a sense of rotation

(a) Use the ordered turn from A toward B.

$$\boxed{\vec A\times\vec B=6(4)\sin30^\circ\hat k=12\hat k\,\mathrm{m^2}},\quad\mathcal A_\triangle=\tfrac12(12)=6\,\mathrm{m^2}$$ The magnitude is $6(4)\sin30^\circ=12\ \mathrm{m^2}$. The turn is counterclockwise in the page, so $\vec A\times\vec B=12\hat k\ \mathrm{m^2}$. This magnitude is the parallelogram area; the triangle area is half as large, $6\ \mathrm{m^2}$.

(b) Reverse the vector sense, not the area.

$$\vec B\times\vec A=-12\hat k\,\mathrm{m^2},\quad|\vec B\times\vec A|=|\vec A\times\vec B|=12\,\mathrm{m^2}$$ $\vec B\times\vec A=-12\hat k\ \mathrm{m^2}$. A geometric area is nonnegative and depends on the magnitude, which is unchanged. The cross product additionally records an oriented normal to the plane.

(c) Inspect both sign and limiting angle.

$$(\vec A\times\vec B)_{-30^\circ}=-12\hat k\,\mathrm{m^2},\quad(\vec A\times\vec B)_{180^\circ}=\vec0$$ Below east, the ordered turn is clockwise, giving $-12\hat k\ \mathrm{m^2}$. Due west gives $\theta=180^\circ$ and zero cross product. A zero cross product has no normal direction; parallel or antiparallel nonzero vectors span zero area.