Chapter 3

Analytical Solid Geometry

We have learned points and lines in two-dimensional rectangular coordinate system. In this chapter, we will extend the system to three dimensions.

3.1     Coordinates of a Point in Space

In the plane, each point is associated with an order paired of real number. In space, each point is associated with an ordered triple of real numbers. Through a fixed point, called the origin O, draw three mutually perpendicular lines: the x-axis, they-axis and the z-axis. A point P in space is determined by an ordered triple (x, y, z) of real numbers as shown in the diagram. These numbers x, y, z are called the coordinate of P.

fig-3.1

The xy-plane consists of the x-axis and the y-axis, the z-axis is perpendicular to the xy-plane. The coordinates of the points on the xy-plane are of form (x, y, 0).

The yz-plane consists of the y-axis and z-axis, and the x-axis is perpendicular to the yz-plane. The coordinates of the points on the yz-plane are of the form (0, y, z).

fig-3.2

The zx-plane consists of the x-axis and the z-axis, and the y-axis is perpendicular to the zx-plane. The coordinates of the points on the zx-plane are of the form (x, 0, z).

The equation of xy-plane, yz-plane and zx-plane are as shown in the following table.

Plane Equation Coordinate
xy-plane z = 0 (x, y, 0)
yz-plane x = 0 (0, y, z)
zx-plane y = 0 (x, 0, z

Planes parallel to xy-plane: Equation of any plane parallel to the xy-plane is z = c, and the coordinates of the points on that plane are of the form (x, y, c).
fig-3.3

Planes parallel to yz-plane: Equation of any plane parallel to the xy-plane is x = a, and the coordinates of the points on that plane are of the form (a, y, z).

Planes parallel to zx-plane: Equation of any plane parallel to the xy-plane is y = b, and the coordinates of the points on that plane are of the form (x, b, z).
table-3.2

fig 3.4

fig 3.5

fig 3.6


Lines perpendicular to xy-plane: Equation of any line perpendicular to the xy-plane and passing through the point (a, b, c) is x = a, y = b, and the coordinates of the points on that line are of the form (a, b, z).
fig 3.7

Lines perpendicular to yz-plane: Equation of any line perpendicular to the yz-plane and passing through the point (a, b, c) is y = b, z = c, and the coordinates of the points on that line are of the form (x, b, c).

Lines perpendicular to zx-plane: Equation of any line perpendicular to the zx-plane and passing through the point (a, b, c) is z = c, x = a, and the coordinates of the points on that line are of the form (a, y, c).

Example 1.
Find the equation of the line through the point (-3, 5, 7) and perpendicular to
(a) xy-plane         (b) yz-plane         (c) zx-plane

Find the point of intersection of the line and plane.

Solution
(a) The equation of the line through the point (-3, 7, 5) and perpendicular to
x = -3, y = 5 or (-3, 5, z)

The point of intersection of the line and xy-plane is (-3, 5, 0).

(b) The equation of the line through the point (-1, 5, 7) and perpendicular to yz-plane is
y = 5, z = 7 or (x, 5, 7)
The point of intersection of the line and yz-plane is (0, 5, 7).

(c) The equation of the line through the point (-3, 5, 7) and perpendicular to zx-plane is
z = 7, x = -3 or (-3, y, 7).

The point of intersection of the line and zx-plane is (-3, 0, 7).

Note that
  • equation of x-axis: y = 0, z = 0 
  • equation of y-axis: x = 0, z = 0  
  • equation of z-axis: x = 0, y = 0  

Distance between two points: From the right triangle PAB, we have PB2 = AB2 + PA2.
fig 3.7

Then from the right triangle PBQ, we have

PQ2 = PB2 + BQ2
        = AB2 + PA2 + BQ2
        = (x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2

So, the distance between P(x1, y1, z1) and Q(x2, y2, z2) is

PQ = √ (x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2

Exercise 3.1
  1. Find the equation of the plane containing the point (1, -2, 3) and parallel to
  2. (a) xy-plane         (b) yz-plane         (c) zx-plane.

  3. Find the equation of the line through the point (2, 3, -4) and perpendicular to
  4. (a) xy-plane         (b) yz-plane         (c) zx-plane.
    Find the point of intersection of the line and plane.

  5. Find the distance between the points (2, -3, 5) and (7, 5, -2).

  6. Show that the points (-1, 2, 5), (1, 1, 6) and (0, 5, 6) form a right triangle.

  7. Show that the points (1, -1, 2), (3, -2. 3) and (5, -3, 4) are colliner.


Analytical Solid Geometry

Answers

1.   Find the equation of the plane containing the point (1, -2, 3) and parallel to
(a) xy-plane         (b) yz-plane         (c) zx-plane.

Solution
(a) The equation of the plane containing the point (1, -2, 3) and parallel to xy-plane is 3.

(b) The equation of the plane containing the point (1, -2, 3) and parallel to yz-plane is 1.

(c) The equation of the plane containing the point (1, -2, 3) and parallel to zx-plane is -2.

2.  Find the equation of the line through the point (2, 3, -4) and perpendicular to (a) xy-plane         (b) yz-plane         (c) zx-plane.
Find the point of intersection of the line and plane.

Solution
(a) The equation of the line through the point (2, 3, -4) and perpendicular to xy-plane is x = 2,   y = 3   or (2, 3, z).
The point of intersection of the line and xy-plane is (2, 3, 0).

(b) The equation of the line through the point (2, 3, -4) and perpendicular to yz-plane is y = 3,   z = -4 or   (x, 3, -4).
The point of intersection of the line and yz-plane is (0, 3, -4).

(c) The equation of the line through the point (2, 3, -4) and perpendicular to zx-plane is z = -4,   x = 2 or   (2, y, -4).
The point of intersection of the line and zx-plane is (2, 0, -4).

3.  Find the distance between the points (2, -3, 5) and (7, 5, -2).

Solution
Let P = (2, -3, 5) and   Q = (7, 5, -2)
The distance between two points = √ (x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2
PQ = (7 - 2)2 + (5 - (-3))2 + (-2 - 5)2
= 52 + 82 + (-7)2
= 25 + 64 + 49
= 138
4.   Show that the points (-1, 2, 5), (1, 1, 6) and (0, 5, 6) form a right triangle.

Solution
Let P = (-1, 2, 5)
Q = (1, 1, 6)
R = (0, 5, 6)

The distance between two points = √ (x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2
PQ = (1 - (-1))2 + (1 - 2)2 + (6 - 5)2
= 22 + (-1)2 + 12
= 6
QR = (0 - 1)2 + (5 - 1)2 + (6 - 6)2
= 1 + 16 + 0
= 17
RP = (-1 - 0)2 + (2 - 5)2 + (5 - 6)2
= 1 + 9 + 1
= 11

(RP)2 + (PQ)2 = (√ 6 ) 2 + (√ 11 ) 2
= 6 + 11
= 17
(QR)2 = (√ 17 ) 2
= 17
∴ (QR)2 = (RP)2 + (PQ)2
∴ The point P, Q and R form a right triangle.

5.   Show that the point (1, -1, 2), (3, -2, 3) and (5, -3, 4) are collinear.

Solution
Let A = (1, -1, 2)
B = (3, -2, 3)
C = (5, -3, 4)
The distance between two points = √ (x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2
AB = (3 - 1)2 + (-2 - (-1))2 + (3 - 2)2
= 22 + (-1)2 + 12
= 6
BC = (5 - 3)2 + (-3 - (-2))2 + (4 - 3)2
= 22 + (-1)2 + 12
= 6
CA = (1 - 5)2 + (-1 - (-3))2 + (2 - 4)2
= (-4)2 + 22 + (-2)2
= 16 + 4 + 4
= 24
= 4 × 6
= 2 √ 6

AB + BC = 6 + 6
= 2 √ 6
∴ The point A, B and C are collinear.

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