Multiplying Vectors — Question 1

Question 1

Three products have different output types

In a right-handed Cartesian basis, $\vec A=(2\hat\imath-\hat\jmath)$ m, $\vec B=(3\hat\imath+4\hat\jmath)$ m and $c=-2$. Define $\vec A\cdot\vec B=A_xB_x+A_yB_y$ and $\vec A\times\vec B=(A_xB_y-A_yB_x)\hat k$ for planar inputs.

1. Compute $c\vec A$, $\vec A\cdot\vec B$, and $\vec A\times\vec B$, giving each output type and units.

2. Which of the expressions $c\vec A+\vec B$, $\vec A+(\vec A\cdot\vec B)$, and $\vec A+(\vec A\times\vec B)$ are physically meaningful as written?

3. Find the products after reversing B. Does changing product order have the same effect?

Question 1 - Solution

Three products have different output types

(a) Identify the operation before calculating.

$$c\vec A=(-4,2)\,\mathrm m,\quad\vec A\cdot\vec B=6-4=2\,\mathrm{m^2},\quad\boxed{\vec A\times\vec B=(8+3)\hat k=11\hat k\,\mathrm{m^2}}$$ $c\vec A=(-4\hat\imath+2\hat\jmath)$ m is a vector. $\vec A\cdot\vec B=6-4=2\ \mathrm{m^2}$ is a scalar. $\vec A\times\vec B=(8+3)\hat k=11\hat k\ \mathrm{m^2}$ is a vector perpendicular to the input plane.

(b) Check type and dimensions.

$$c\vec A+\vec B=(-1,6)\,\mathrm m;\qquad L\ne L^2,\quad\text{vector}\ne\text{scalar}$$ Only $c\vec A+\vec B$ is valid: it adds two displacement vectors. The second expression adds a vector to a scalar and also mismatches units. The third adds vectors with units m and $\mathrm{m^2}$, so it is dimensionally invalid despite matching vector type.

(c) Separate reversal from exchange.

$$\vec A\cdot(-\vec B)=-2\,\mathrm{m^2},\quad\vec A\times(-\vec B)=-11\hat k\,\mathrm{m^2},\quad\vec B\cdot\vec A=\vec A\cdot\vec B,\ \vec B\times\vec A=-\vec A\times\vec B$$ Replacing B by $-\vec B$ changes the dot product to $-2\ \mathrm{m^2}$ and the cross product to $-11\hat k\ \mathrm{m^2}$. Exchanging A and B leaves the dot product unchanged but reverses the cross product. Dot multiplication is commutative; cross multiplication is anticommutative.