벡터와 공간 전체 문제

radius of curvature is $5\sqrt{10}$.

radius of curvature is $9$.

radius of curvature is $1/45$.

$x^2+(y-1/2)^2 = 1/4$, or $\vec c(t) = \left\langle 1/2\cos t, 1/2\sin t+ 1/2\right\rangle$

$(x-8/3)^2+y^2 = 1/9$, or $\vec c(t) = \left\langle \frac13\cos t+\frac83, \frac13\sin t\right\rangle$

$x^2+(y+8)^2 = 81$, or $\vec c(t) = \left\langle 9\cos t, 9\sin t-8\right\rangle$

$(x-1/2)^2+(y-1/2)^2 = 1/2$, or $\vec c(t) = \left\langle \frac{\sqrt{2}}{2}\cos t+\frac12, \frac{\sqrt{2}}{2}\sin t+\frac12\right\rangle$

(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.

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

$$(x-3)^2+(y-2)^2+(z-7)^2=11$$

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

$$\begin{aligned} \left< 1,2,3\right>\times\left< 4,5,6\right> &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ 1&2&3 \\ 4&5&6\end{matrix}\right] =\hat{\boldsymbol{\imath}}\,(2\times 6-3\times 5) -\hat{\boldsymbol{\jmath}}\,(1\times 6-3\times 4) +\hat{\mathbf{k}}\,(1\times 5-2\times 4) \\ &=-3\,\hat{\boldsymbol{\imath}}+6\,\hat{\boldsymbol{\jmath}}-3\,\hat{\mathbf{k}} \end{aligned}$$