Multiplying Vectors — Question 6

Question 6

Compute a spatial cross product and audit it

In a right-handed orthonormal frame, $\vec A=(1,2,3)$ m and $\vec B=(2,-1,1)$ m. Use $\vec C=\vec A\times\vec B$ with components $(A_yB_z-A_zB_y,\ A_zB_x-A_xB_z,\ A_xB_y-A_yB_x)$.

1. Compute C, keeping track of all three signs.

2. Check that C is perpendicular to both inputs.

3. Check $C^2=A^2B^2-(\vec A\cdot\vec B)^2$ and find the parallelogram area.

Question 6 - Solution

Compute a spatial cross product and audit it

(a) Evaluate the three signed differences.

$$\vec C=(2+3,6-1,-1-4)=\boxed{(5,5,-5)\,\mathrm{m^2}}$$ $C_x=2(1)-3(-1)=5$, $C_y=3(2)-1(1)=5$, and $C_z=1(-1)-2(2)=-5$, each in $\mathrm{m^2}$. Hence $\vec C=(5\hat\imath+5\hat\jmath-5\hat k)\ \mathrm{m^2}$. The middle component is often a source of sign errors.

(b) Test the result against both input directions.

$$\vec C\cdot\vec A=5+10-15=0,\quad\vec C\cdot\vec B=10-5-5=0\quad\mathrm{m^3}$$ $\vec C\cdot\vec A=5+10-15=0\ \mathrm{m^3}$ and $\vec C\cdot\vec B=10-5-5=0\ \mathrm{m^3}$. Both tests are required: being perpendicular to only one input would not determine the cross direction.

(c) Verify the magnitude independently.

$$C^2=14(6)-3^2=75\,\mathrm{m^4},\quad\mathcal A=|\vec C|=5\sqrt3\,\mathrm{m^2}$$ $A^2=14\ \mathrm{m^2}$, $B^2=6\ \mathrm{m^2}$, and $\vec A\cdot\vec B=3\ \mathrm{m^2}$. Thus $A^2B^2-(\vec A\cdot\vec B)^2=84-9=75\ \mathrm{m^4}$, equal to $5^2+5^2+(-5)^2$. The area is $|\vec C|=5\sqrt3\ \mathrm{m^2}$.