02. 도형의 방정식Equations of Lines, Planes, and Surfaces

Solution. The line through $P_0(-2,0,4)$ with direction vector $\mathbf{v}=(2,4,-2)$ has parametric equations

Solution. The direction vector is $\overrightarrow{PQ}=Q-P=(1-(-3),\,-1-2,\,4-(-3))=(4,-3,7)$.

(1) The line through $P(-3,2,-3)$ with direction vector $(4,-3,7)$ has parametric equations $x=-3+4t,\ y=2-3t,\ z=-3+7t\ (t\in\mathbb{R})$.

(2) Since $t=0$ gives $x=-3,\ y=2,\ z=-3$, i.e. $P$, and $t=1$ gives $x=-3+4=1,\ y=2-3=-1,\ z=-3+7=4$, i.e. $Q$, the segment $PQ$ is given by the same equations as in (1) restricted to $0\leq t\leq 1$.

Solution. The $xy$-plane consists of points with $z=0$, so $z=3t=0$ gives $t=0$. Then $x=1+2(0)=1$ and $y=-1-0=-1$, so the point of intersection is $(1,-1,0)$.

Solution. The unit vector in the direction $(1,1,1)$ is $\mathbf{u}=\dfrac{1}{\sqrt3}(1,1,1)$, so the velocity vector at speed 60 m/s is $\mathbf{v}=60\mathbf{u}=\dfrac{60}{\sqrt3}(1,1,1)=20\sqrt3\,(1,1,1)$. Since the helicopter starts at the origin, its position vector after $t$ seconds is

(1) At $t=10$: $\mathbf{r}(10)=200\sqrt3\,(1,1,1)=(200\sqrt3,\,200\sqrt3,\,200\sqrt3)$.

(2) After $t$ seconds, the position is $\mathbf{r}(t)=20\sqrt3\,t\,(1,1,1)$, i.e., $x=y=z=20\sqrt3\,t$.

Solution. The line $L$ passes through $P_0=(1,3,0)$ (at $t=0$) with direction vector $\mathbf{v}=(1,-1,2)$. Let $\overrightarrow{P_0S}=S-P_0=(0,-2,5)$. By the point-to-line distance formula,

Since $\overrightarrow{P_0S}\times\mathbf{v}=(0,-2,5)\times(1,-1,2)=\big((-2)(2)-(5)(-1),\ (5)(1)-(0)(2),\ (0)(-1)-(-2)(1)\big)=(1,5,2)$, we get $|\overrightarrow{P_0S}\times\mathbf{v}|=\sqrt{1+25+4}=\sqrt{30}$, and $|\mathbf{v}|=\sqrt{1+1+4}=\sqrt6$. Hence

Solution.

(1) The plane with normal vector $\mathbf{v}=(2,4,-2)$ through $(-3,0,7)$ is $2(x+3)+4(y-0)-2(z-7)=0$, i.e., $2x+4y-2z+20=0$. Dividing by 2 gives $x+2y-z=-10$.

(2) Let $\overrightarrow{AB}=(2,0,-1)$ and $\overrightarrow{AC}=(0,3,-1)$. The normal vector of the plane is

Since the plane passes through $A(0,0,1)$, its equation is $3(x-0)+2(y-0)+6(z-1)=0$, i.e., $3x+2y+6z=6$.

(3) A parallel plane shares the normal vector $(3,1,1)$, so the plane through $(1,-1,3)$ is $3(x-1)+1(y+1)+1(z-3)=0$, i.e., $3x+y+z=5$.

Solution. The normal vectors of the two planes are $\mathbf{n_1}=(3,-6,-2)$ and $\mathbf{n_2}=(2,1,-2)$, so the direction vector of the line of intersection is

To find a point on the line, set $z=0$: then $3x-6y=15$ and $2x+y=5$. Substituting $y=5-2x$ into the first equation gives $3x-6(5-2x)=15 \Rightarrow 15x=45 \Rightarrow x=3,\ y=-1$. So the line passes through $(3,-1,0)$, and its parametric equations are

Solution. Substituting the point $\left(\dfrac{8}{3}+2t,\,-2t,\,1+t\right)$ on the line into the equation of the plane gives

so $t=-1$. Then the point of intersection is $x=\dfrac{8}{3}-2=\dfrac{2}{3},\ y=2,\ z=0$, i.e., $\left(\dfrac{2}{3},2,0\right)$.

Proof.

(1) A point on $L_1$ is $(2t+1,3t+2,4t+3)$, and a point on $L_2$ is $(s+2,2s+4,-4s-1)$. For the lines to meet, there must exist $t,s$ satisfying

From the first and third equations, $s=2t-1$ and $t+s=-1$, so $t+(2t-1)=-1 \Rightarrow 3t=0 \Rightarrow t=0,\ s=-1$. Substituting into the second equation gives $3(0)-2(-1)=2$, which holds, so $t=0,\ s=-1$ satisfies all three equations. Hence the lines intersect, and setting $t=0$ gives the point of intersection $(1,2,3)$.

(2) The direction vectors of $L_1,L_2$ are $\mathbf{d_1}=(2,3,4)$ and $\mathbf{d_2}=(1,2,-4)$, so the normal vector of the plane containing both lines is

Since the plane passes through the point of intersection $(1,2,3)$ found in (1), its equation is $-20(x-1)+12(y-2)+1(z-3)=0$, i.e., $-20x+12y+z=7$.

Solution. By the point-to-plane distance formula,

Solution. The normal vectors of the two planes are $\mathbf{n_1}=(3,-6,-2)$ and $\mathbf{n_2}=(2,1,-2)$. Since $\mathbf{n_1}\cdot\mathbf{n_2}=3(2)+(-6)(1)+(-2)(-2)=6-6+4=4$, $|\mathbf{n_1}|=\sqrt{9+36+4}=\sqrt{49}=7$, and $|\mathbf{n_2}|=\sqrt{4+1+4}=\sqrt9=3$,

Hence the angle between the two planes is $\theta=\cos^{-1}\dfrac{4}{21}$.

Solution. Translating the parabola $y=x^2\ (z=0)$ in the $xy$-plane parallel to the $z$-axis leaves the $x,y$-coordinates unchanged while $z$ ranges over all real numbers. Hence the resulting cylinder is given, with no condition on $z$, by

In other words, a point $(x,y,z)$ lies on this cylinder if and only if $y=x^2$; slicing by any plane $z=k$ gives the same parabola $y=x^2$.

Solution. Since $y$ does not appear in the equation $x^2+4z^2=16$, the rulings of this cylinder are parallel to the $y$-axis. Hence we may take the trace in the plane $y=0$ (the $xz$-plane) as the generating curve.

which is an ellipse in the $xz$-plane.

Solution. Examining the trace of each surface in the plane $z=k$ (and similar planes) gives the following shapes.

(1) Ellipsoid. It meets the $x,y,z$-axes at $(\pm a,0,0),(0,\pm b,0),(0,0,\pm c)$. The trace in $z=k\ (|k|\leq c)$ is the ellipse $\dfrac{x^2}{a^2(1-k^2/c^2)}+\dfrac{y^2}{b^2(1-k^2/c^2)}=1$, shrinking to a point at $k=\pm c$. Hence it is a closed surface centered at the origin.

(2) Elliptical paraboloid. The vertex is at the origin. The trace in $z=k\geq0$ is the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{k}{c}$, growing as $k$ increases, with no real trace for $z\lt0$ (since $c\gt0$). Hence it is a bowl opening in the positive $z$-direction.

(3) Elliptical cone. The vertex is at the origin. The trace in $z=k$ is the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{k^2}{c^2}$, whose radius grows proportionally to $|k|$; two nappes, symmetric for $z\gt0$ and $z\lt0$, meet at the origin.

(4) Hyperboloid of one sheet. The trace in $z=k$ is always the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1+\dfrac{k^2}{c^2}$ (never empty); the radius is smallest (the waist) at $k=0$ and increases as $|k|$ grows. Hence it is a single connected surface.

(5) Hyperboloid of two sheets. The trace in $z=k$ is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{k^2}{c^2}-1$, which exists only for $|k|\geq c$ and is empty for $|k|\lt c$. Hence the surface consists of two separate sheets, with vertices at $(0,0,\pm c)$.

(6) Hyperbolic paraboloid. The trace in $x=0$ is the upward parabola $z=\dfrac{c}{b^2}y^2$, and the trace in $y=0$ is the downward parabola $z=-\dfrac{c}{a^2}x^2$. Hence the surface curves upward in one direction and downward in the other through the origin, forming a saddle shape.

Solution. Completing the square gives

This is an ellipsoid centered at $(1,-2,0)$ with semi-axes $2,2,1$ in the $x,y,z$ directions, respectively (the equatorial cross-section is a circle of radius 2, while the axis in the $z$-direction is shorter).

Solution. With the substitution $x=au,\ y=bv,\ z=cw$, the ellipsoid $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$ becomes the unit ball $u^2+v^2+w^2\leq1$, and the Jacobian is $\left|\dfrac{\partial(x,y,z)}{\partial(u,v,w)}\right|=abc$. Since the volume of the unit ball is $\dfrac{4}{3}\pi$, the volume of the region enclosed by the ellipsoid is

Solution. The cross-section of the hyperboloid of one sheet at height $z=k\ (0\leq k\leq h)$ is

an ellipse with semi-axes $a\sqrt{1+k^2/c^2}$ and $b\sqrt{1+k^2/c^2}$, so its area is

Hence the volume is