벡터의 연산

a. $\text{a}+\text{b}=3\text{i}+4\text{j},$ $\text{a}+\text{b}=\langle 3,4\rangle ;$ b. $\text{a}-\text{b}=\text{i}-2\text{j},$ $\text{a}-\text{b}=\langle 1,-2\rangle ;$ c. Answers will vary; d. $2\text{a}=4\text{i}+2\text{j},$ $2\text{a}=\langle 4,2\rangle ,$ $-\text{b}=-\text{i}-3\text{j},$ $-\text{b}=\langle -1,-3\rangle ,$ $2\text{a}-\text{b}=3\text{i}-\text{j},$ $2\text{a}-\text{b}=\langle 3,-1\rangle$

a. $\text{a}(0)=\langle 1,0\rangle ,$ $\text{a}(\pi )=\langle -1,0\rangle ;$ b. Answers may vary; c. Answers may vary

Answers may vary

The components of $\text{a}=\langle x,f(x)\rangle$ and $\text{b}=\langle x,g(x)\rangle$ are not equal, so the vectors are not equivalent.

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 $P$.

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 $t$.

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,

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