벡터와 공간 전체 문제

The parallelogram determined by the vectors $\mathbf{a}$ and $\mathbf{b}$ has vertices $\mathbf{0},\ \mathbf{a},\ \mathbf{b}$ and $\mathbf{a}+\mathbf{b}$. As $t$ varies from $0$ to $1$, $t(\mathbf{a}+\mathbf{b})$ traverses the diagonal from $\mathbf{0}$ to $\mathbf{a}+\mathbf{b}$. As $s$ varies from $0$ to $1$, $\mathbf{a}+s(\mathbf{b}-\mathbf{a})$ traverses the diagonal from $\mathbf{a}$ to $\mathbf{b}$. These two straight lines meet when $s$ and $t$ are such that

$$t(\mathbf{a}+\mathbf{b})=\mathbf{a}+s(\mathbf{b}-\mathbf{a})$$

or

$$(t+s-1)\mathbf{a}=(s-t)\mathbf{b}$$

Assuming that $\mathbf{a}$ and $\mathbf{b}$ are not parallel (i.e. the parallelogram has not degenerated to a line segment), this is the case only when $t+s-1=0$ and $s-t=0$. That is, $s=t=\tfrac{1}{2}$. So the two lines meet at their midpoints.

We may choose our coordinate axes so that $A=(0,0,0)$, $B=(s,0,0)$, $C=(s,s,0)$, $D=(0,s,0)$ and $A'=(0,0,s)$, $B'=(s,0,s)$, $C'=(s,s,s)$, $D'=(0,s,s)$.

(a) Then

$$\begin{aligned} |A'C'|&=\big|\left< s,s,s\right> -\left< 0,0,s\right> \big| &&=\big|\left< s,s,0\right> \big|&=\sqrt{2}\,s\\ |A'B|&=\big|\left< s,0,0\right> -\left< 0,0,s\right> \big| &&=\big|\left< s,0,-s\right> \big|&=\sqrt{2}\,s\\ |A'D|&=\big|\left< 0,s,0\right> -\left< 0,0,s\right> \big| &&=\big|\left< 0,s,-s\right> \big|&=\sqrt{2}\,s\\ |C'B|&=\big|\left< s,0,0\right> -\left< s,s,s\right> \big| &&=\big|\left< 0,-s,-s\right> \big|&=\sqrt{2}\,s\\ |C'D|&=\big|\left< 0,s,0\right> -\left< s,s,s\right> \big| &&=\big|\left< -s,0,-s\right> \big|&=\sqrt{2}\,s\\ |BD|&=\big|\left< 0,s,0\right> -\left< s,0,0\right> \big| &&=\big|\left< -s,s,0\right> \big|&=\sqrt{2}\,s \end{aligned}$$

(b) $E=\tfrac{1}{2}(s,s,s)$ so that $EA=\left< 0,0,0\right> -\tfrac{1}{2}\left< s,s,s\right> =-\tfrac{1}{2}\left< s,s,s\right>$ and $EC=\left< s,s,0\right> -\tfrac{1}{2}\left< s,s,s\right> =\tfrac{1}{2}\left< s,s,-s\right>$. $$\cos\theta =\frac{-\left< s,s,s\right> \cdot\left< s,s,-s\right>} {|\left< s,s,s\right> |\,|\left< s,s,-s\right> |} =\frac{-s^2}{3s^2} =-\frac{1}{3} \qquad \implies\quad \theta=109.5^\circ$$

By definition,

$$\begin{aligned} \mathbf{b}\times\mathbf{c} \ =\ &(b_2c_3-b_3c_2)\hat{\boldsymbol{\imath}}-(b_1c_3-b_3c_1)\hat{\boldsymbol{\jmath}} +(b_1c_2-b_2c_1)\hat{\mathbf{k}} \\ \\ \text{so that the left and right hand sides are} \\ \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) \ =\ &\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}}\\ a_1&a_2&a_3\\ b_2c_3-b_3c_2&-b_1c_3+b_3c_1&b_1c_2-b_2c_1\end{matrix}\right]\\ \ =\ &\hat{\boldsymbol{\imath}}\,[a_2(b_1c_2-b_2c_1)-a_3(-b_1c_3+b_3c_1)]\\ {-}&\hat{\boldsymbol{\jmath}}\,[a_1(b_1c_2-b_2c_1)-a_3(b_2c_3-b_3c_2)]\\ {+}&\hat{\mathbf{k}}\,[a_1(-b_1c_3+b_3c_1)-a_2(b_2c_3-b_3c_2)]&{\rm (lhs)}\\ (\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c} \ =\ &(a_1c_1+a_2c_2+a_3c_3)(b_1\hat{\boldsymbol{\imath}}+b_2\hat{\boldsymbol{\jmath}}+b_3\hat{\mathbf{k}}) -(a_1b_1+a_2b_2+a_3b_3)(c_1\hat{\boldsymbol{\imath}}+c_2\hat{\boldsymbol{\jmath}}+c_3\hat{\mathbf{k}})\\ {=}\ & \hat{\boldsymbol{\imath}}\,[a_1b_1c_1+a_2b_1c_2+a_3b_1c_3-a_1b_1c_1-a_2b_2c_1-a_3b_3c_1] \\ {+}&\hat{\boldsymbol{\jmath}}\,[a_1b_2c_1+a_2b_2c_2+a_3b_2c_3-a_1b_1c_2-a_2b_2c_2-a_3b_3c_2] \\ {+}&\hat{\mathbf{k}}\,[a_1b_3c_1+a_2b_3c_2+a_3b_3c_3-a_1b_1c_3-a_2b_2c_3-a_3b_3c_3]\\ {=}\ & \hat{\boldsymbol{\imath}}\,[a_2b_1c_2+a_3b_1c_3-a_2b_2c_1-a_3b_3c_1] \\ {+}&\hat{\boldsymbol{\jmath}}\,[a_1b_2c_1+a_3b_2c_3-a_1b_1c_2-a_3b_3c_2] \\ {+}&\hat{\mathbf{k}}\,[a_1b_3c_1+a_2b_3c_2-a_1b_1c_3-a_2b_2c_3] &{\rm (rhs)} \end{aligned}$$

(lhs) and (rhs) are the same.

(a) $AA'=\left< 4,0,1\right>$ and $BB'=\left< 4,0,1\right>$ are opposite sides of the quadrilateral $AA'B'B$. They have the same length and direction. The same is true for $AB=\left< -1,3,0\right>$ and $A'B'=\left< -1,3,0\right>$. So $AA'B'B$ is a parallelogram. Because, $AA'\cdot AB=\left< 4,0,1\right> \cdot\left< -1,3,0\right> =-4\ne 0$, the neighbouring edges of $AA'B'B$ are not perpendicular and so $AA'B'B$ is not a rectangle.

Similarly, the quadilateral $ACC'A'$ has opposing sides $AA'=\left< 4,0,1\right> =CC'=\left< 4,0,1\right>$ and $AC=\left< -1,0,4\right> =A'C'=\left< -1,0,4\right>$ and so is a parallelogram. Because $AA'\cdot AC=\left< 4,0,1\right> \cdot\left< -1,0,4\right> = 0$, the neighbouring edges of $ACC'A'$ are perpendicular, so $ACC'A'$ is a rectangle.

Finally, the quadilateral $BCC'B'$ has opposing sides $BB'=\left< 4,0,1\right> =CC'=\left< 4,0,1\right>$ and $BC=\left< 0,-3,4\right> =B'C'=\left< 0,-3,4\right>$ and so is a parallelogram. Because $BB'\cdot BC=\left< 4,0,1\right> \cdot\left< 0,-3,4\right> = 4\ne 0$, the neighbouring edges of $BCC'B'$ are not perpendicular, so $BCC'B'$ is not a rectangle.

(b) The length of $AA'$ is $|\left< 4,0,1\right> |=\sqrt{16+1}=\sqrt{17}$.

(c) The area of a triangle is one half its base times its height. That is, one half times $|AB|$ times $|AC|\sin\theta$, where $\theta$ is the angle between $AB$ and $AC$. This is precisely $\tfrac{1}{2} |AB\times AC|=\tfrac{1}{2}|\left< -1,3,0\right> \times\left< -1,0,4\right> | =\tfrac{1}{2} |\left< 12,4,3\right>|=\frac{13}{2}$.

(d) The volume of the prism is the area of its base $ABC$, times its height, which is the length of $AA'$ times the cosine of the angle between $AA'$ and the normal to $ABC$. This coincides with $\tfrac{1}{2} \left< 12,4,3\right> \cdot\left< 4,0,1\right> =\tfrac{1}{2}(48+3)=\frac{51}{2}$, which is one half times the length of $\left< 12,4,3\right>$ (the area of $ABC$) times the length of $\left< 4,0,1\right>$ (the length of $AA'$) times the cosine of the angle between $\left< 12,4,3\right>$ and $\left< 4,0,1\right>$ (the angle between the normal to $ABC$ and $AA'$).

Choose our coordinate axes so that the vertex opposite the face of area $D$ is at the origin. Denote by $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ the vertices opposite the sides of area $A$, $B$ and $C$ respectively. Then the face of area $A$ has edges $\mathbf{b}$ and $\mathbf{c}$ so that $A=\tfrac{1}{2} |\mathbf{b}\times\mathbf{c}|$. Similarly $B=\tfrac{1}{2}|\mathbf{c}\times\mathbf{a}|$ and $C=\tfrac{1}{2}|\mathbf{a}\times \mathbf{b}|$. The face of area $D$ is the triangle spanned by $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ so that

$$\begin{aligned} D&=\tfrac{1}{2}|(\mathbf{b}-\mathbf{a})\times(\mathbf{c}-\mathbf{a})|\\ &=\tfrac{1}{2}|\mathbf{b}\times \mathbf{c}-\mathbf{a}\times\mathbf{c}-\mathbf{b}\times\mathbf{a}|\\ &=\tfrac{1}{2}|\mathbf{b}\times \mathbf{c}+\mathbf{c}\times\mathbf{a}+\mathbf{a}\times\mathbf{b}| \end{aligned}$$

By hypothesis, the vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are all perpendicular to each other. Consequently the vectors $\mathbf{b}\times \mathbf{c}$ (which is a scalar times $\mathbf{a}$), $\mathbf{c}\times\mathbf{a}$ (which is a scalar times $\mathbf{b}$) and $\mathbf{a}\times\mathbf{b}$ (which is a scalar times $\mathbf{c}$) are also mutually perpendicular. So, when we multiply out

$$D^2=\frac{1}{4}\big[\mathbf{b}\times \mathbf{c}+\mathbf{c}\times\mathbf{a}+\mathbf{a}\times\mathbf{b}\big]\cdot\big[\mathbf{b}\times \mathbf{c}+\mathbf{c}\times\mathbf{a}+\mathbf{a}\times\mathbf{b}\big]$$

all the cross terms vanish, leaving

$$D^2=\frac{1}{4}\big[(\mathbf{b}\times \mathbf{c})\cdot(\mathbf{b}\times \mathbf{c}) +(\mathbf{c}\times\mathbf{a})\cdot(\mathbf{c}\times\mathbf{a}) +(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{a}\times\mathbf{b})\big]=A^2+B^2+C^2$$

The face opposite $\mathbf{p}$ is the triangle with vertices $\mathbf{q}$, $\mathbf{r}$ and $\mathbf{s}$. The centroid of this triangle is $\frac{1}{3}(\mathbf{q}+\mathbf{r}+\mathbf{s})$. The direction vector of the line through $\mathbf{p}$ and the centroid $\frac{1}{3}(\mathbf{q}+\mathbf{r}+\mathbf{s})$ is $\frac{1}{3}(\mathbf{q}+\mathbf{r}+\mathbf{s})-\mathbf{p}$. The points on the line through $\mathbf{p}$ and the centroid $\frac{1}{3}(\mathbf{q}+\mathbf{r}+\mathbf{s})$ are those of the form

$$\mathbf{x}=\mathbf{p}+ t\left[\frac{1}{3}(\mathbf{q}+\mathbf{r}+\mathbf{s})-\mathbf{p}\right]$$

for some real number $t$. Observe that when $t=\frac{3}{4}$ $$\mathbf{p}+ t\left[\frac{1}{3}(\mathbf{q}+\mathbf{r}+\mathbf{s})-\mathbf{p}\right] =\frac{1}{4}(\mathbf{p}+\mathbf{q}+\mathbf{r}+\mathbf{s})$$ so that $\frac{1}{4}(\mathbf{p}+\mathbf{q}+\mathbf{r}+\mathbf{s})$ is on the line. The other three lines have vector parametric equations

$$\begin{aligned} \mathbf{x}&=\mathbf{q}+ t\left[\frac{1}{3}(\mathbf{p}+\mathbf{r}+\mathbf{s})-\mathbf{q}\right] \\ \mathbf{x}&=\mathbf{r}+ t\left[\frac{1}{3}(\mathbf{p}+\mathbf{q}+\mathbf{s})-\mathbf{r}\right] \\ \mathbf{x}&=\mathbf{s}+ t\left[\frac{1}{3}(\mathbf{p}+\mathbf{q}+\mathbf{r})-\mathbf{s}\right] \end{aligned}$$

When $t=\frac{3}{4}$, each of the three right hand sides also reduces to $\frac{1}{4}(\mathbf{p}+\mathbf{q}+\mathbf{r}+\mathbf{s})$ so that $\frac{1}{4}(\mathbf{p}+\mathbf{q}+\mathbf{r}+\mathbf{s})$ is also on each of these three lines.