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) 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

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

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.

(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)

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

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

Since $t$ can be any real number, these equation describe the same line. They're both valid. For example, the point given by the first parametric equation with $t=7$, namely $\mathbf c + 7\mathbf d$, is exactly the same as the point given by the second parametric equation with $t=-7$, namely $\mathbf c -(-7)\mathbf d$.

$\left< d_x,d_y\right>$ is the direction of the line, so it can be any non-zero scalar multiple of $\left< 9,7\right>$.

$\left< x_0,y_0\right>$ can be any point on the line. Describing these is the same as describing the line itself. We're trying to find all doubles $\left< x_0,y_0\right>$ that obey

$$\begin{aligned} \begin{cases} x_0-3&=9t\\ y_0-5&=7t \end{cases} \end{aligned}$$

for some real number $t$. That is,

$$\begin{aligned} t=\frac{x_0-3}{9}&=\frac{y_0-5}{7}\\ 7(x_0-3)&=9(y_0-5)\\ 7x_0+24&=9y_0 \end{aligned}$$

Any of these steps could specify the possible values of $\left< x_0,y_0\right>$. Say, they can be any pair satisfying $7x_0+24=9y_0$.

We are looking for a vector that is perpendicular to $z=0$ and hence is parallel to $\hat{\mathbf{k}}$. To be parallel of $\hat{\mathbf{k}}$, the vector has to be of the form $c\,\hat{\mathbf{k}}$ for some real number $c$. For the vector to be nonzero, we need $c\ne 0$ and for the vector to be different from $\hat{\mathbf{k}}$, we need $c\ne 1$. So three possible choices are $-\hat{\mathbf{k}}$, $2\,\hat{\mathbf{k}}$, $7.12345\,\hat{\mathbf{k}}$.

(a) If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(0,0,0)$, namely $\left< x,y,z\right>$, lie in the plane. So the vector $\left< x,y,z\right>$ must be perpendicular to $\left< 1,2,3\right>$ and

$$0=\left< x,y,z\right>\cdot \left< 1,2,3\right> =x+2y+3z$$

(b) If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(0,0,1)$, namely $\left< x,y,z-1\right>$, lie in the plane. So the vector $\left< x,y,z-1\right>$ must be perpendicular to $\left< 1,1,3\right>$ and

$$0=\left< x,y,z-1\right>\cdot \left< 1,1,3\right> =x+y+3(z-1) \iff x+y+3z=3$$

(c) If both $(1,2,3)$ and $(1,0,0)$ are on the plane, then both the head and the tail of the vector from $(1,2,3)$ to $(1,0,0)$, namely $\left< 0,2,3\right>$, lie in the plane. So the vector $\left< 0,2,3\right>$ must be perpendicular to $\left< 4,5,6\right>$. As

$$\left< 0,2,3\right> \cdot \left< 4,5,6\right> = 28\ne 0$$

the vector $\left< 0,2,3\right>$ is not perpendicular to $\left< 4,5,6\right>$. So there is no plane that passes through both $(1,2,3)$ and $(1,0,0)$ and has normal vector $\left< 4,5,6\right>$.

(d) If both $(1,2,3)$ and $(0,3,4)$ are on the plane, then both the head and the tail of the vector from $(1,2,3)$ to $(0,3,4)$, namely $\left< 1,-1,-1\right>$, lie in the plane. So the vector $\left< 1,-1,-1\right>$ must be perpendicular to $\left< 2,1,1\right>$. As

$$\left< 1,-1,-1\right> \cdot \left< 2,1,1\right> = 0$$

the vector $\left< 1,-1,-1\right>$ is indeed perpendicular to $\left< 2,1,1\right>$. So there is a plane that passes through both $(1,2,3)$ and $(0,3,4)$ and has normal vector $\left< 2,1,1\right>$. We now just have to build its equation.

If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(1,2,3)$, namely $\left< x-1,y-2,z-3\right>$, lie in the plane. So the vector $\left< x-1,y-2,z-3\right>$ must be perpendicular to $\left< 2,1,1\right>$ and

$$0=\left< x-1,y-2,z-3\right>\cdot \left< 2,1,1\right> =2(x-1)+(y-2)+(z-3) \iff 2x+y+z=7$$

As a check, note that both $(x,y,z)=(1,2,3)$ and $(x,y,z)=(0,3,4)$ obey the equation $2x+y+z=7$.