벡터의 연산

The center of the sphere is $\tfrac{1}{2}\big\{(2,1,4)+(4,3,10)\big\}=(3,2,7)$. The diameter (i.e. twice the radius) is $|(2,1,4)-(4,3,10)|=|(-2,-2,-6)|=2|(1,1,3)|=2\sqrt{11}$. So the radius of the sphere is $\sqrt{11}$ and the equation of the sphere is

$$(x-3)^2+(y-2)^2+(z-7)^2=11$$

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.

Note that

• the point on $W$ with $x=0$, $y=0$ obeys $-0+3(0)+3z=6$ and so has $z=2$
• the point on $W$ with $x=0$, $y=2$ obeys $-0+3(2)+3z=6$ and so has $z=0$
• the point on $W$ with $x=3$, $y=0$ obeys $-3+3(0)+3z=6$ and so has $z=3$
• the point on $W$ with $x=3$, $y=2$ obeys $-3+3(2)+3z=6$ and so has $z=1$

So the four corners of the parallelogram are $(0,0,2)$, $(0,2,0)$, $(3,0,3)$ and $(3,2,1)$. The vectors

$$\begin{aligned} \mathbf{d}_1&=\left< 0-0 \,,\, 2-0 \,,\, 0-2 \right> = \left< 0 \,,\, 2 \,,\, -2\right> \\ \mathbf{d}_2&=\left< 3-0 \,,\, 0-0 \,,\, 3-2 \right> = \left< 3 \,,\, 0 \,,\, 1\right> \end{aligned}$$

form two sides of the paralleogram. So the area of the parallelogram is

$$\begin{aligned} \big|\mathbf{d}_1\times\mathbf{d}_2\big| =\left|\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 0 & 2 & -2 \\ 3 & 0 & 1 \end{matrix}\right]\right| =\left| 2\,\hat{\boldsymbol{\imath}} - 6\,\hat{\boldsymbol{\jmath}} -6\hat{\mathbf{k}} \right| =\sqrt{76} =2\sqrt{19} \end{aligned}$$

(a) We want $0=\mathbf{u}\cdot\mathbf{v}=-2\alpha -10$ or $\alpha =-5$.

(b) We want $-2/\alpha =5/(-2)$ or $\alpha =0.8$.

(c) We want $\mathbf{u}\cdot\mathbf{v}=-2\alpha -10 =|\mathbf{u}|\,|\mathbf{v}|\,\cos 60^\circ =\sqrt{29}\,\sqrt{\alpha ^2+4}\,\tfrac{1}{2}$. Squaring both sides gives

$$\begin{aligned} & & 4\alpha ^2+40\alpha +100&=\frac{29}{4}(\alpha ^2+4) \\ &\implies\quad & 13\alpha ^2-160\alpha -284&=0 \\ &\implies\quad & \alpha &=\frac{160\pm\sqrt{160^2+4\times13\times284}}{26} \approx 13.88\text{ or }-1.574 \end{aligned}$$

Both of these $\alpha$'s give $\mathbf{u}\cdot\mathbf{v}<0$ so no $\alpha$ works.

(a) The component of $\mathbf{b}$ in the direction $\mathbf{a}$ is $$\mathbf{b}\cdot\frac{\mathbf{a}}{|\mathbf{a}|} =\frac{1\times 4+2\times 10+3\times 6}{\sqrt{1+4+9}} =\text{$\frac{42}{\sqrt{14}}$}$$

(b) The projection of $\mathbf{b}$ on $\mathbf{a}$ is a vector of length $42/\sqrt{14}$ in direction $\mathbf{a}/|\mathbf{a}|$, namely $\frac{42}{14}\left< 1,2,3\right>=\left< 3,6,9\right>$.

(c) The projection of $\mathbf{b}$ perpendicular to $\mathbf{a}$ is $\mathbf{b}$ minus its projection on $\mathbf{a}$, namely $\left< 4,10,6\right>-\left< 3,6,9\right>=\left< 1,4,-3\right>$.

$$\begin{aligned} \left< 1,2,3\right>\times\left< 4,5,6\right> &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ 1&2&3 \\ 4&5&6\end{matrix}\right] =\hat{\boldsymbol{\imath}}\,(2\times 6-3\times 5) -\hat{\boldsymbol{\jmath}}\,(1\times 6-3\times 4) +\hat{\mathbf{k}}\,(1\times 5-2\times 4) \\ &=-3\,\hat{\boldsymbol{\imath}}+6\,\hat{\boldsymbol{\jmath}}-3\,\hat{\mathbf{k}} \end{aligned}$$

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 $t$.