Multiplying Vectors — Question 10

Question 10

Check whether two product measurements are compatible

Two nonzero vectors have exact magnitudes $A=5$ m and $B=4$ m. A report gives $\vec A\cdot\vec B=12\ \mathrm{m^2}$ and $|\vec A\times\vec B|=16\ \mathrm{m^2}$. The relative direction angle lies in $[0^\circ,180^\circ]$.

1. Check the measurements jointly and find the angle between the vectors.

2. If the cross magnitude were instead $18\ \mathrm{m^2}$ with all other entries unchanged, could the report describe real vectors?

3. If only the original cross magnitude 16 were known, find every possible angle. What extra information in the original report removes the ambiguity?

Question 10 - Solution

Check whether two product measurements are compatible

(a) Use complementary sine and cosine information.

$$12^2+16^2=400=(5\cdot4)^2,\quad\boxed{\theta=\cos^{-1}(12/20)=53.130^\circ}$$ The identity $(\vec A\cdot\vec B)^2+|\vec A\times\vec B|^2=A^2B^2$ gives $12^2+16^2=400=(5\cdot4)^2$, so the data are compatible. Since $\cos\theta=12/20=0.6$ and $\sin\theta=16/20=0.8$, $\theta\approx53.130^\circ$.

(b) Individual bounds are not sufficient.

$$12^2+18^2=468\ne400,\qquad C_{\mathrm{compatible}}=\sqrt{400-144}=16\,\mathrm{m^2}$$ Both 12 and 18 are individually less than $AB=20$, but $12^2+18^2=468\ne400$. The two products cannot belong to the same pair of vectors with the stated magnitudes. The scalar product fixes the compatible cross magnitude to $\sqrt{400-144}=16\ \mathrm{m^2}$.

(c) Retain the supplementary-angle branch.

$$\sin\theta=.8\Rightarrow\theta=53.130^\circ\text{ or }126.870^\circ,\quad\cos\theta>0\text{ selects the first}$$ Knowing only $\sin\theta=0.8$ permits $53.130^\circ$ and $126.870^\circ$. They have dot products $+12$ and $-12\ \mathrm{m^2}$ respectively. The positive sign in the original dot-product measurement selects the acute angle. Neither scalar measurement alone specifies the plane normal in space.