벡터와 공간 전체 문제

Let $(x,y,z)$ be a point in $P$. The distances from $(x,y,z)$ to $(3,-2,3)$ and to $(3/2,1,0)$ are

$$\sqrt{(x-3)^2+(y+2)^2+(z-3)^2}\quad\text{ and }\quad \sqrt{(x-3/2)^2+(y-1)^2+z^2}$$

respectively. To be in $P$, $(x,y,z)$ must obey

$$\begin{aligned} \sqrt{(x-3)^2+(y+2)^2+(z-3)^2}&=2\sqrt{(x-3/2)^2+(y-1)^2+z^2} \\ (x-3)^2+(y+2)^2+(z-3)^2&=4(x-3/2)^2+4(y-1)^2+4z^2 \\ x^2-6x+9+y^2+4y+4+z^2-6z+9&=4x^2-12x+9+4y^2-8y+4+4z^2 \\ 3x^2-6 x+3y^2-12y+3z^2+6z-9&=0 \\ x^2-2 x+y^2-4y+z^2+2z-3&=0 \\ (x-1)^2+(y-2)^2+(z+1)^2&=9 \end{aligned}$$

This is a sphere of radius 3 centered on $(1,2,-1)$.

Call the vertices of the triangle $A$, $B$ and $C$ with $C$ being the vertex that joins the two sides. We can always choose our coordinate system so that $C$ is at the origin. Let $\mathbf{a}$ be the vector from $C$ to $A$ and $\mathbf{b}$ be the vector from $C$ to $B$.

• Then the vector from $C$ to the midpoint of the side from $C$ to $A$ is $\tfrac{1}{2}\mathbf{a}$ and
• the vector from $C$ to the midpoint of the side from $C$ to $B$ is $\tfrac{1}{2}\mathbf{b}$ so that
• the vector joining the two midpoints is $\tfrac{1}{2}\mathbf{b}-\tfrac{1}{2}\mathbf{a}$.

As the vector from $A$ to $B$ is $\mathbf{b}-\mathbf{a}=2\big[\tfrac{1}{2}\mathbf{b}-\tfrac{1}{2}\mathbf{a}\big]$, the line joining the midpoints is indeed parallel to the third side and half its length.

The derivative of $L$ is

$$\begin{aligned} \frac{\mathrm{d}L}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\big(\mathbf{r}(t)\times\mathbf{r}'(t)\big) =\mathbf{r}'(t)\times\mathbf{r}'(t)+\mathbf{r}(t)\times\mathbf{r}''(t) =\mathbf{r}'(t)\times\mathbf{r}'(t)+\mathbf{r}(t)\times\big(\rho(t)\mathbf{r}(t)\big) \end{aligned}$$

Both terms vanish because the cross product of any two parallel vectors is zero. So $\frac{\mathrm{d}L}{\mathrm{d}t}=0$ and $L(t)$ is independent of $t$.