벡터와 공간 전체 문제

The area of a parallelogram is the length of its base time its height.

We can choose the base to be $\mathbf{a}$. Then, if $\theta$ is the angle between its sides $\mathbf{a}$ and $\mathbf{b}$, its height is $|\mathbf{b}|\sin\theta$. So

$$\text{area} = |\mathbf{a}||\mathbf{b}|\sin\theta=|\mathbf{a}\times\mathbf{b}|$$

The volume of a parallelepiped is the area of its base time its height. We can choose the base to be the parallelogram determined by the vectors $\mathbf{b}$ and $\mathbf{c}$. It has area $|\mathbf{b}\times\mathbf{c}|$. The vector $\mathbf{b}\times\mathbf{c}$ is perpendicular to the base.

Denote by $\theta$ the angle between $\mathbf{a}$ and the perpendicular $\mathbf{b}\times\mathbf{c}$. The height of the parallelepiped is $|\mathbf{a}| |\cos\theta|$. So

$$\text{volume} = |\mathbf{a}|\, |\cos\theta|\, |\mathbf{b}\times\mathbf{c}| =|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|$$

This statement is false. The two numbers $\mathbf{a}\cdot\mathbf{b}$, $\mathbf{a}\cdot\mathbf{c}$ are equal if and only if $\mathbf{a}\cdot(\mathbf{b}-\mathbf{c})= 0$. This in turn is the case if and only if $\mathbf{a}$ is perpendicular to $\mathbf{b}-\mathbf{c}$ (under the convention that $\mathbf{0}$ is perpendicular to all vectors). For example, if $\mathbf{a}=\left< 1,0,0\right>$, $\mathbf{b}=\left< 0,1,0\right> ,\ \mathbf{c}=\left< 0,0,1\right>$, then $\mathbf{b}-\mathbf{c}=\left< 0,1,-1\right>$ is perpendicular to $\mathbf{a}$ so that $\mathbf{a}\cdot\mathbf{b}=\mathbf{a}\cdot\mathbf{c}$.

This statement is true. In the event that $\mathbf{b}$ and $\mathbf{c}$ are parallel, $\mathbf{b}\times\mathbf{c}=\mathbf{0}$ so that $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\mathbf{0}=0\mathbf{b}+0\mathbf{c}$, so we may assume that $\mathbf{b}$ and $\mathbf{c}$ are not parallel. Then as $\alpha$ and $\beta$ run over $\mathbb{R}$, the vector $\alpha \mathbf{b}+\beta \mathbf{c}$ runs over the plane that contains the origin and the vectors $\mathbf{b}$ and $\mathbf{c}$. Call this plane $P$. Because $\mathbf{d}=\mathbf{b}\times\mathbf{c}$ is nonzero and perpendicular to both $\mathbf{b}$ and $\mathbf{c}$, $P$ is the plane that contains the origin and is perpendicular to $\mathbf{d}$. As $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\mathbf{a}\times\mathbf{d}$ is always perpendicular to $\mathbf{d}$, it lies in $P$.

Here is one answer of many.

Setting $t=0$ in the first equation shows that $(1,9)$ is on the first line. To see that $(1,9)$ is also on the second line, we substitute $x=1$, $y=9$ into the second equation to give

$$\left< 1-9,9-13\right>=t\left< 1,\tfrac12\right>\qquad \text{or}\qquad \left< -8,-4\right>=t\left< 1,\tfrac12\right>$$

This equation is satisfied when $t=-8$. So $(1,9)$ is on both lines.

Setting $t=0$ in the second equation shows that $(9,13)$ is on the second line. To see that $(9,13)$ is also on the first line, we substitute $x=9$, $y=13$ into the first equation to give

$$\left< 9-1,13-9\right>=t\left< 8,4\right>\qquad \text{or}\qquad \left< 8,4\right>=t\left< 8,4\right>$$

This equation is satisfied when $t=1$. So $(9,13)$ is on both lines.

Since both lines pass through $(1,9)$ and $(9,13)$, the lines are identical.

(a) If $\kappa (s)\equiv 0$, then $\frac{\mathrm{d}\hat{\mathbf{T}}}{\mathrm{d}s}=\kappa (s)\hat{\mathbf{N}}(s)\equiv 0$ so that $\hat{\mathbf{T}}$ is a constant. As a result $\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}s}(s)=\hat{\mathbf{T}}$ and $\mathbf{r}(s) =s\hat{\mathbf{T}}+\mathbf{r}(0)$ so that the curve is the straight line with direction vector $\hat{\mathbf{T}}$ that passes through $\mathbf{r}(0)$.

(b) If $\tau(s)\equiv 0$, then $\frac{\mathrm{d}\hat{\mathbf{B}}}{\mathrm{d}s}=-\tau(s)\hat{\mathbf{N}}(s)\equiv 0$ so that $\hat{\mathbf{B}}$ is a constant. As $\hat{\mathbf{T}}(s)\perp\hat{\mathbf{B}}$,

$$\frac{\mathrm{d}}{\mathrm{d}s} (\mathbf{r}(s)-\mathbf{r}(0))\cdot\hat{\mathbf{B}} =\hat{\mathbf{T}}(s)\cdot\hat{\mathbf{B}}=0$$

and $(\mathbf{r}(s)-\mathbf{r}(0))\cdot\hat{\mathbf{B}}$ must be a constant. The constant must be zero (set $s=0$), so $(\mathbf{r}(s)-\mathbf{r}(0))\cdot\hat{\mathbf{B}}=0$ and $\mathbf{r}(s)$ always lies in the plane through $\mathbf{r}(0)$ with normal vector $\hat{\mathbf{B}}$.

(c) Parametrize the curve by arc length. Define the “centre of curvature” at $s$ by

$$\mathbf{r}_c(s)=\mathbf{r}(s)+\frac{1}{\kappa (s)}\hat{\mathbf{N}}(s)$$

Since $\kappa (s)=\kappa _0$ is a constant and $\tau(s)\equiv 0$,

$$\frac{\mathrm{d}}{\mathrm{d}s}\mathbf{r}_c(s) =\hat{\mathbf{T}}(s)+\frac{1}{\kappa _0}\big[\tau(s)\hat{\mathbf{B}}-\kappa (s)\hat{\mathbf{T}}\big] =\hat{\mathbf{T}}(s)+\frac{1}{\kappa _0}\big[0\hat{\mathbf{B}}-\kappa _0\hat{\mathbf{T}}\big] =0$$

Thus $\mathbf{r}_c(s)=\mathbf{r}_c$ is a constant and $\big|\mathbf{r}(s)-\mathbf{r}_c\big|=\frac{1}{\kappa _0}$ lies on the sphere of radius $\frac{1}{\kappa _0}$ centred on $\mathbf{r}_c$. Since $\tau(s)\equiv 0$, the curve also lies on a plane, so it is a circle.