벡터와 공간 전체 문제

(a) The point $(x,y,z)$ satisfies $x^2 +y^2+z^2= 2x-4y+4$ if and only if it satisfies $x^2-2x +y^2+4y+z^2= 4$, or equivalently $(x-1)^2 +(y+2)^2+z^2=9$. Since $\sqrt{(x-1)^2 +(y+2)^2+z^2}$ is the distance from $(1, -2, 0)$ to $(x,y,z)$, our point satisfies the given equation if and only if its distance from $(1,-2,0)$ is three. So the set is the sphere of radius 3 centered on $(1,-2,0)$.

(b) As in part (a), $x^2 +y^2+z^2< 2x-4y+4$ if and only if $(x-1)^2 +(y+2)^2+z^2<9$. Hence our point satifies the given inequality if and only if its distance from $(1,-2,0)$ is strictly smaller than three. The set is the interior of the sphere of radius 3 centered on $(1,-2,0)$.

(a) $x=y$ is a straight line and passes through the points $(0,0)$ and $(1,1)$. So it is the straight line through the origin that makes an angle $45^\circ$ with the $x$– and $y$–axes. It is sketched in the figure on the left below.

(b) $x+y=1$ is the straight line through the points $(1,0)$ and $(0,1)$. It is sketched in the figure on the right above.

(c) $x^2+y^2$ is the square of the distance from $(0,0)$ to $(x,y)$. So $x^2+y^2=4$ is the circle with centre $(0,0)$ and radius 2. It is sketched in the figure on the left below.

(d) The equation $x^2+y^2=2y$ is equivalent to $x^2+(y-1)^2=1$. As $x^2+(y-1)^2$ is the square of the distance from $(0,1)$ to $(x,y)$, $x^2+(y-1)^2=1$ is the circle with centre $(0,1)$ and radius 1. It is sketched in the figure on the right above.

(e) As in part (d),

$$x^2+y^2<2y \iff x^2+y^2-2y<0 \iff x^2+y^2-2y+1<1 \iff x^2+(y-1)^2<1$$

As $x^2+(y-1)^2$ is the square of the distance from $(0,1)$ to $(x,y)$, $x^2+(y-1)^2<1$ is the set of points whose distance from $(0,1)$ is strictly less than $1$. That is, it is the set of points strictly inside the circle with centre $(0,1)$ and radius 1. That set is the shaded region (not including the dashed circle) in the sketch below.

(a) For each fixed $y_0$, $z=x,\ y=y_0$ is a straight line that lies in the plane, $y=y_0$ (which is parallel to the plane containing the $x$ and $z$ axes and is a distance $y_0$ from it). This line passes through $x=z=0$ and makes an angle $45^\circ$ with the $xy$–plane. Such a line (with $y_0=0$) is sketched in the figure below. The set $z=x$ is the union of all the lines $z=x,\ y=y_0$ with all values of $y_0$. As $y_0$ varies $z=x,\ y=y_0$ sweeps out the plane which contains the $y$–axis and which makes an angle $45^\circ$ with the $xy$–plane. Here is a sketch of the part of the plane that is in the first octant.

(b) $x+y+z=1$ is the plane through the points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$. Here is a sketch of the part of the plane that is in the first octant.

(c) $x^2+y^2+z^2$ is the square of the distance from $(0,0,0)$ to $(x,y,z)$. So $x^2+y^2+z^2=4$ is the set of points whose distance from $(0,0,0)$ is $2$. It is the sphere with centre $(0,0,0)$ and radius 2. Here is a sketch of the part of the sphere that is in the first octant.

(d) $x^2+y^2+z^2=4$, $z=1$ or equivalently $x^2+y^2=3$, $z=1$, is the intersection of the plane $z=1$ with the sphere of centre $(0,0,0)$ and radius 2. It is a circle in the plane $z=1$ that has centre $(0,0,1)$ and radius $\sqrt{3}$. The part of the circle in the first octant is the heavy quarter circle in the sketch

(e) For each fixed $z_0$, $x^2+y^2=4$, $z=z_0$ is a circle in the plane $z=z_0$ with centre $(0,0,z_0)$ and radius $2$. So $x^2+y^2=4$ is the union of $x^2+y^2=4,\ z=z_0$ for all possible values of $z_0$. It is a vertical stack of horizontal circles. It is the cylinder of radius $2$ centered on the $z$–axis. Here is a sketch of the part of the cylinder that is in the first octant.

(f) For each fixed $z_0\ge 0$, the curve $z = x^2 + y^2,\ z=z_0$ is the circle in the plane $z=z_0$ with centre $(0,0,z_0)$ and radius $\sqrt{z_0}$. As $z = x^2 + y^2$ is the union of $z = x^2 + y^2,\ z=z_0$ for all possible values of $z_0\ge 0$, it is a vertical stack of horizontal circles. The intersection of the surface with the $yz$–plane is the parabola $z=y^2$. Here is a sketch of the part of the paraboloid that is in the first octant.

(a) The $z$ coordinate of any point is the signed distance from the point to the $xy$-plane. So the distance from $(2,1,3)$ to the $xy$-plane is $|3|=3$.

(b) The $y$ coordinate of any point is the signed distance from the point to the $xz$-plane. So the distance from $(2,1,3)$ to the $xz$-plane is $|1|=1$.

(c) The distance from $(2,1,3)$ to $(x,0,0)$ is

$$\sqrt{(2-x)^2 + (1-0)^2 + (3-0)^2} = \sqrt{(x-2)^2 + 10}$$

(d) Since $(x-2)^2\ge 0$, the distance $\sqrt{(x-2)^2 + 10}$ is minimized when $x=2$. Alternatively,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \sqrt{(x-2)^2 + 10} =\frac{x-2}{\sqrt{(x-2)^2 + 10}} =0 &\iff x=2 \end{aligned}$$

So the point on the $x$-axis that is closest to $A$ is $(2,0,0)$.

(e) As $(2,0,0)$ is the point on the $x$-axis that is nearest $(2,1,3)$, the distance from $A$ to the $x$-axis is

$$\sqrt{(2-2)^2 + (1-0)^2 + (3-0)^2} = \sqrt{1^2+3^2} = \sqrt{10}$$

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

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)

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