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10.1 Rectangular Coordinates in Space

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A two-dimensional rectangular coordinate system is represented by a plane in which every point P corresponds to exactly one ordered pair of real numbers (x,y), the coordinates of P. In space every point P corresponds to exactly one ordered triple of real numbers (x,y,z).

1 Locate Points in Space

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We begin by selecting a fixed point called the origin O. Through the origin, draw three mutually perpendicular number lines, called the coordinate axes. The coordinate axes are usually labeled the x-axis the y-axis and the z-axis. On each axis choose one direction as positive and select an appropriate scale, as shown in Figure 1.

Notice label in Figure 1 that the positive z-axis points upward, the positive y-axis points toward the right, and the positive x-axis points forward. This is known as a right-handed system because it conforms to the right-hand rule. The right-hand rule states: if the index finger of the right hand points in the direction of the positive x-axis and the other fingers point in the direction of the positive y-axis, then the thumb will point in the direction of the positive z-axis, as shown in Figure 2.

Figure 2 Right hand rule

NOTE

There is also a left-handed system and a left-handed rule; we use only a right-handed system.

As in the plane, we assign coordinates to each point P in space, but in space we use an ordered triple of real numbers (x,y,z). For example, the point (3,5,7), is the point for which 3 is the x-coordinate, 5 is the y-coordinate, and 7 is the z-coordinate. The point (3,5,7) is plotted by starting at the origin, moving 3 units along the positive x-axis, 5 units in the direction of the positive y-axis, and 7 units in the direction of the positive z-axis. See Figure 3.

Figure 3 also shows the points (3,0,0), (0,5,0), (0,0,7), (3,5,0), (3,0,7), 0,5,7) and (0,0,0), which form a box with the points (0,0,0) and (3,5,7) as opposite vertices.

Points of the form (x,0,0) lie on the x-axis, while points of the form (0,y,0) and (0,0,z) lie on the y-axis and z-axis, respectively.

Points of the form (x,y,0) lie on a plane called the xy-plane. The xy-plane is perpendicular to the z-axis and its equation is z=0. This is the plane used in the familiar two-dimensional rectangular coordinate system.

Points of the form (x,0,z) lie on the xz-plane. The xz-plane is perpendicular to the y-axis, and its equation is y=0. Finally, points of the form (0,y,z) lie on the yz-plane, which is perpendicular to the x-axis, and its equation is x=0. Collectively, these three planes are referred to as coordinate planes. See Figure 4.

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In general, for any real number k, the graph of x=k is a plane parallel to the yz-plane; the graph of y=k is a plane parallel to the xz-plane; and the graph of z=k is a plane parallel to the xy-plane. As shown in Figure 5, the graph of the equation z=5 is a plane parallel to and 5 units above the xy-plane.

NEED TO REVIEW?

The distance formula appears in Appendix A.3, p. A-16.

NOW WORK

Problem 19.

2 Find the Distance Between Two Points in Space

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The formula for the distance between two points in space is an extension of the distance formula in the plane.

To find the distance |P1 P2| between two points P1=(x1,y1,z1) and P2=(x2,y2,z2), we use the Pythagorean Theorem twice. Figure 6 shows the points P1 and P2 and a third point A=(x2,y2,z1). The first application of the Pythagorean Theorem involves observing that the triangle P1 A P2 is a right triangle where the side of length |P1 P2| is the hypotenuse. As a result, |P1 P2|=|P1 A|2+|A P2|2Pythagorean Theorem

The points P1 and A lie in a plane parallel to the xy-plane. Do you see why? To find |P1 A|, we use the distance formula in the plane: |P1A|=(x2x1)2+(y2y1)2

The points P2 and A lie on a line parallel to the z-axis, so that |A P2|=|z2z1| and |A P2|2=(z2z1)2. Combining these results, we obtain a formula for the distance between two points in space.

THEOREM Distance Formula in Space

The distance between two points P1=(x1,y1,z1) and P2=(x2,y2,z2) in space, denoted by |P1 P2|, is \bbox[5px, border:1px solid black, #F9F7ED]{ \left\vert P_{1}~P_{2}\right\vert =\sqrt{ (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}}

EXAMPLE 1Finding the Distance Between Two Points in Space

Find the distance between P_{1}=(1, 3,-2) and P_{2}=(2,-1,-3).

Solution |P_{1}~P_{2}|=\sqrt{(2-1)^{2}+(-1-3)^{2}+(-3+2)^{2}}= \sqrt{1+16+1}=\sqrt{18}=3\sqrt{2}\qquad

NOW WORK

Problem 31.

3 Find the Equation of a Sphere

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Figure 7 (x-x_0)^2 + (y-y_0)^2 \\ + (z-z_0)^2 = R^2

The collection of all points in space that are a fixed distance R from a fixed point P_{0}=( x_{0},y_{0},z_{0}) is called a sphere. See Figure 7. The fixed distance R is called the radius, and the fixed point P_{0} is called the center of the sphere. The distance from any point P=(x, y, z) on a sphere of radius R to the center point P_{0}=(x_{0},y_{0},z_{0}) is \vert PP_{0}\vert=R. The Distance Formula in space shows that \begin{equation*} \vert P\!P_{0}\vert =\sqrt{ (x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}}=R \end{equation*}

Squaring both sides gives the standard form of the equation of a sphere.

THEOREM Standard Form of the Equation of a Sphere

The equation of a sphere with center at the point P_{0}=(x_{0},y_{0},z_{0}) and radius R is \bbox[5px, border:1px solid black, #F9F7ED]{ (x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}=R^{2}}

EXAMPLE 2Finding the Equation of a Sphere

The standard form of the equation of the sphere shown in Figure 8, with radius 2 and center at (-1, 2, 0), is \begin{equation*} (x+1)^{2}+(y-2)^{2}+z^{2}=4 \end{equation*}

Figure 8 (x+1)^2+(y-2)^2+z^2=4

NOW WORK

Problem 37.

EXAMPLE 3Finding the Center and Radius of a Sphere

Show that { x^{2}+y^{2}+z^{2}+2x+4y-2z=10 }

is the equation of a sphere. Find its center and radius.

Solution We begin by writing the equation as \begin{equation*} (x^{2}+2x)+(y^{2}+4y)+(z^{2}-2z)=10 \end{equation*}

NEED TO REVIEW?

Completing the square is discussed in Appendix A.1, pp. A-2 to A-3.

and then we complete the square three times. The result is \begin{eqnarray*} (x^{2}+2x+1)+(y^{2}+4y+4)+(z^{2}-2z+1) &=&10+1+4+1 \\[4pt] (x+1)^{2}+(y+2)^{2}+(z-1)^{2} &=&16 \end{eqnarray*}

This is the equation of a sphere with radius 4 and center at (-1,-2, 1).

NOW WORK

Problem 39.