14.1 Points and planes

The fill patterns are only included to distinguish different parts of the diagram.

The $xz$ plane is filled with vertical lines; the $yz$ plane is crosshatched; and the $xy$ plane is solid.

The left bottom triangle vertex is $(1,0,0)$; the right bottom triangle vertex is $(0,1,0)$; the top triangle vertex is $(0,0,1)$.

The $xz$ plane is filled with vertical lines; the $yz$ plane is crosshatched; and the $xy$ plane is solid.

The left bottom triangle vertex is $(1,0,0)$; the right bottom triangle vertex is $(0,1,0)$; the top triangle vertex is $(0,0,1)$.

Section 14.1 gives the equation for a sphere.

(a) The sphere of radius 3 centered on $(1,-2,0)$.

(b) The interior of the sphere of radius 3 centered on $(1,-2,0)$.

(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 $(1,-2,0)$.

This is a review question to get you thinking about $\mathbb R^2$ in a way that will help you get used to $\mathbb R^3$.

(a) $x=y$ 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) $x^2+y^2=4$ is the circle with centre $(0,0)$ and radius 2. It is sketched in the figure on the left below.

(d) $x^2+y^2=2y$ is the circle with centre $(0,1)$ and radius 1. It is sketched in the figure on the right above.

(e) $x^2+y^2<2y$ is the set of points that are strictly inside the circle with centre $(0,1)$ and radius 1. It is the shaded region (not including the dashed circle) in the sketch below.

(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),

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.

Compare to Question 3. To visualize what's going on, it can help to consider what shapes you'd get if $z$ were a constant.

If you're struggling to visualize $\mathbb R^3$, section 14.1.1 in the text shows you how to fold a model of its first octant.

(a) The set $z=x$ is 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) $x^2+y^2+z^2=4$ 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.

(c) $x^2+y^2+z^2=4$, $z=1$ is the 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

(d) $x^2+y^2=4$ 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.

(e) $z=x^2+y^2$ is a paraboloid consisting of 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) 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) $x^2+y^2+z^2$ 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.

(c) $x^2+y^2+z^2=4$, $z=1$ 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

(d) 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.

(e) 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.

From the text, the distance from the point $(x,y,z)$ to the point $(x',y',z')$ is

$\sqrt{67}$

From the text, the distance from the point $(x,y,z)$ to the point $(x',y',z')$ is

So, our distance is

From the text, the distance from the point $(x,y,z)$ to the $xy$-plane is $|z|$.

9

From the text, the distance from the point $(x,y,z)$ to the $xy$-plane is $|z|$. In this case, 9.

From the text, the distance from the point $(x,y,z)$ to the point $(x',y',z')$ is

100 metres is one-tenth of a kilometre.

$\sqrt{5.01}$ km

From the text, the distance from the point $(x,y,z)$ to the $xy$-plane is $|z|$. Let the nest be the origin $(0,0,0)$ with the $z$-axis pointing north, the $x$-axis pointing south, and the $y$-axis pointing east. Then the bird's coordinates after flying are $(-1,2,0.1)$. So, its distance from its nest is

From the text, the distance from the point $(x,y,z)$ to the point $(x',y',z')$ is

Given the distance and the $x$ and $y$ coordinates, you can solve for the $z$ coordinate.

1 km

Let the nest be the origin $(0,0,0)$ with the $z$-axis pointing north, the $x$-axis pointing south, and the $y$-axis pointing east. From the text, the distance from the point $(x,y,z)$ to the $xy$-plane (which, in this case, is the ground) is $|z|$. Then the bird's coordinates after flying are $(-2,2,z)$. So,

So, the bird is 1 km above the ground. (Or, possibly, 1 km below it.)

At which part of the journey are you actually getting farther away from the wall?

2 km

The first 2 km of the journey bring you 2 km away from the wall. Walking parallel to the wall neither increases nor decreases your distance to the wall. Similarly, moving vertically neither increases nor decreases your distance to the wall. So, the murder hornets are 2 km from the wall.

If we wanted to impose a coordinate system, we could place the wall as the $xz$ axis, with $z$ being the vertical direction, and the origin the place where you started walking. Then the murder hornets are at the point $(1,2,0.003)$. The distance from $(x,y,z)$ to the $xz$ axis is $|y|$. In this case, 2 km.

The isobar is a curve of the form $x^2-2cx+y^2=1$, where $c$ is a constant. These describe circles – figure out what their centres and radii are.

For each fixed $c$, the isobar $p(x,y)=c$ is the curve $x^2-2cx+y^2=1$, or equivalently, $(x-c)^2+y^2=1+c^2$. This is a circle with centre $(c,0)$ and radius $\sqrt{1+c^2}$, which for large $c$ is just a bit bigger than $c$.

The sphere has radius 3 and is centered on $(1,2,-1)$.

Let $(x,y,z)$ be a point in $P$. The distances from $(x,y,z)$ to $(3,-2,3)$ and to $(3/2,1,0)$ are

respectively. To be in $P$, $(x,y,z)$ must obey

This is a sphere of radius 3 centered on $(1,2,-1)$.

This centre must be equidistant from the three vertices.

The circumscribing circle has centre $(\bar x,\bar y)$ and radius $r$ with $\bar x=\frac{a}{2}$, $\bar y=\frac{b^2+c^2-ab}{2c}$ and $r=\sqrt{\big(\frac{a}{2}\big)^2+\big(\frac{b^2+c^2-ab}{2c}\big)^2}$.

Call the centre of the circumscribing circle $(\bar x,\bar y)$. This centre must be equidistant from the three vertices. So

or, subtracting $\bar x^2+\bar y^2$ from the three equal expressions,

which implies

The radius is the distance from the vertex $(0,0)$ to the centre $(\bar x,\bar y)$, which is $\sqrt{\big(\frac{a}{2}\big)^2+\big(\frac{b^2+c^2-ab}{2c}\big)^2}$.

From the text, the distance from the point $(x,y,z)$ to the point $(x',y',z')$ is

Also from the text, the distance from the point $(x,y,z)$ to the $xy$-plane is $|z|$. Use a similar thought process to find the distance from a point $(x,y,z)$ to the plane $z=-1$.

$x^2+y^2=4z$ The surface is a paraboloid consisting of a stack of horizontal circles, starting with a point at the origin and with radius increasing vertically. The circle in the plane $z=z_0$ has radius $2\sqrt{z_0}$.

The distance from $P$ to the point $(0,0,1)$ is $\sqrt{x^2+y^2+(z-1)^2}$. The distance from $P$ to the specified plane is $|z+1|$. Hence the equation of the surface is

All points on this surface have $z\ge 0$. The set of points on the surface that have any fixed value, $z_0\ge 0$, of $z$ consists of a circle that is centred on the $z$–axis, is parallel to the $xy$-plane and has radius $2\sqrt{z_0}$. The surface consists of a stack of these circles, starting with a point at the origin and with radius increasing vertically. The surface is a paraboloid and is sketched below.