벡터의 연산

$\theta = 45^\circ$; the weight is lifted $2.93$ ft.

$\theta = 84.29^\circ$; the weight is lifted $9$ ft.

$\mathbf{a}+\mathbf{b}=\left< 3,1\right>$, $\mathbf{a}+2\mathbf{b}=\left< 4,2\right>$, $2\mathbf{a}-\mathbf{b}=\left< 3,-1\right>$

If three points are collinear, then the vector from the first point to the second point, and the vector from the first point to the third point must both be parallel to the line, and hence must be parallel to each other (i.e. must be multiples of each other).

(a) The vectors $\left< 0,3,7\right> -\left< 1,2,3\right> =\left< -1,1,4\right>$ and $\left< 3,5,11\right> -\left< 1,2,3\right> =\left< 2,3,8\right>$ are not parallel (i.e. are not multiples of each other), so the three points are not on the same line.

(b) The vectors $\left< 1,2,-2\right> -\left< 0,3,-5\right> =\left< 1,-1,3\right>$ and $\left< 3,0,4\right> -\left< 0,3,-5\right> =\left< 3,-3,9\right>$ are parallel (i.e. are multiples of each other), so the three points are on the same line.

$\text{proj}_{\hat{\boldsymbol{\imath}}}\mathbf{a}=(\mathbf{a}\cdot\hat{\boldsymbol{\imath}})\hat{\boldsymbol{\imath}} =a_1\hat{\boldsymbol{\imath}}$ and $\text{proj}_{\hat{\boldsymbol{\jmath}}}\mathbf{a}=(\mathbf{a}\cdot\hat{\boldsymbol{\jmath}})\hat{\boldsymbol{\jmath}} =a_2\hat{\boldsymbol{\jmath}}$.

The vector from $(1,2,3)$ to $(4,0,5)$ is $\left< 3,-2,2\right>$. The vector from $(1,2,3)$ to $(3,6,4)$ is $\left< 2,4,1\right>$. The dot product between these two vectors is $\left< 3,-2,2\right>\cdot\left< 2,4,1\right>=0$, so the vectors are perpendicular and the triangle does contain a right angle.

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

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

(a)

$$\begin{aligned} \hat{\boldsymbol{\imath}}\times\hat{\boldsymbol{\jmath}}&=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}}\\ 1&0&0\\ 0&1&0\end{matrix}\right] &&=\hat{\boldsymbol{\imath}}(0\times 0-0\times 1) -\hat{\boldsymbol{\jmath}}(1\times 0-0\times 0) +\hat{\mathbf{k}}(1\times 1-0\times 0) \\ &=\hat{\mathbf{k}}\\[0.1in] \hat{\boldsymbol{\jmath}}\times\hat{\mathbf{k}}&=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}}\\ 0&1&0\\ 0&0&1\end{matrix}\right] &&=\hat{\boldsymbol{\imath}}(1\times 1-0\times 0) -\hat{\boldsymbol{\jmath}}(0\times 1-0\times 0) +\hat{\mathbf{k}}(0\times 0-1\times 0)\\ &=\hat{\boldsymbol{\imath}}\\[0.1in] \hat{\mathbf{k}}\times\hat{\boldsymbol{\imath}}&=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}}\\ 0&0&1\\ 1&0&0\end{matrix}\right] &&=\hat{\boldsymbol{\imath}}(0\times 0-1\times 0) -\hat{\boldsymbol{\jmath}}(0\times 0-1\times 1) +\hat{\mathbf{k}}(0\times 0-0\times 1)\\ &=\hat{\boldsymbol{\jmath}} \end{aligned}$$

(b)

$$\begin{aligned} \mathbf{a}\cdot(\mathbf{a}\times\mathbf{b}) &=a_1\big(a_2b_3-a_3b_2\big) -a_2\big(a_1b_3-a_3b_1\big) +a_3\big(a_1b_2-a_2b_1\big) &&=0\\ \mathbf{b}\cdot(\mathbf{a}\times\mathbf{b}) &=b_1\big(a_2b_3-a_3b_2\big) -b_2\big(a_1b_3-a_3b_1\big) +b_3\big(a_1b_2-a_2b_1\big) &&=0 \end{aligned}$$

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 $P$.

None. The given equation is nonsense. The left hand side is a number while the right hand side is a vector.

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 $\left< 4,10,6\right>-\left< 3,6,9\right>=\left< 1,4,-3\right>$.