Introduction to Three-dimensional Geometry

To fix a point in a plane, two mutually perpendicular lines (the x-axis and y-axis) are enough. To fix a point in space, a third line perpendicular to both is needed. Through a fixed point O, called the origin, three mutually perpendicular lines are drawn — the x-axis, y-axis and z-axis. Together these form the coordinate axes, and the system is a rectangular (Cartesian) coordinate system in three dimensions.

Any two of these axes determine a plane called a coordinate plane:

• The x-axis and y-axis determine the XY-plane.
• The y-axis and z-axis determine the YZ-plane.
• The z-axis and x-axis determine the ZX-plane.

These three coordinate planes divide the whole of space into eight parts, called octants, labelled Octant I to Octant VIII. Which octant a point falls in is decided purely by the signs of its three coordinates.

Worked Example. Determine the octant in which the point P(−2, 3, −5) lies.

Here x = −2 (negative), y = 3 (positive), z = −5 (negative), so the sign pattern is (−, +, −). Comparing with the standard octant table, this sign pattern corresponds to Octant VI. Hence P(−2, 3, −5) lies in Octant VI.

Every point P in space corresponds to a unique ordered triple of real numbers (x, y, z), called its coordinates. Here x, y and z are the (signed) perpendicular distances of P from the YZ-plane, the ZX-plane and the XY-plane respectively. Geometrically, P(x, y, z) is located at the corner opposite O of a rectangular box with one corner at the origin and edges of length |x|, |y|, |z| parallel to the three axes.

Some special positions are worth remembering:

• A point on the x-axis has the form (x, 0, 0); on the y-axis, (0, y, 0); on the z-axis, (0, 0, z).
• A point on the XY-plane has the form (x, y, 0); on the YZ-plane, (0, y, z); on the ZX-plane, (x, 0, z).
• The origin O has coordinates (0, 0, 0).

Worked Example. For the point P(3, −4, 5), find the coordinates of the feet of the perpendiculars drawn from P to (i) the XY-plane, (ii) the x-axis, and (iii) the YZ-plane.

(i) The foot on the XY-plane keeps the x- and y-coordinates and sets z = 0, giving (3, −4, 0).

(ii) The foot on the x-axis keeps only the x-coordinate and sets y = z = 0, giving (3, 0, 0).

(iii) The foot on the YZ-plane keeps the y- and z-coordinates and sets x = 0, giving (0, −4, 5).

The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in space is found by applying the Pythagoras theorem twice along the edges of the rectangular box joining them:

PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]

This is a direct extension of the two-dimensional distance formula, with one extra squared term for the z-coordinates. When one point is the origin O(0, 0, 0), the formula reduces to OP = √(x₁² + y₁² + z₁²).

Worked Example. Show that the triangle with vertices A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2) is equilateral.

AB = √[(2−1)² + (3−2)² + (1−3)²] = √[1 + 1 + 4] = √6

BC = √[(3−2)² + (1−3)² + (2−1)²] = √[1 + 4 + 1] = √6

CA = √[(1−3)² + (2−1)² + (3−2)²] = √[4 + 1 + 1] = √6

Since AB = BC = CA = √6, triangle ABC is equilateral.

Let P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be two points in space, and let R(x, y, z) be a point on segment PQ that divides it internally in the ratio m : n, that is, PR : RQ = m : n. Using similar triangles formed by dropping perpendiculars from P, Q, R onto each coordinate plane, the coordinates of R work out to be

x = (m·x₂ + n·x₁)/(m + n), y = (m·y₂ + n·y₁)/(m + n), z = (m·z₂ + n·z₁)/(m + n)

Each coordinate of R is obtained by combining the corresponding coordinates of P and Q in the ratio m : n, exactly as in the two-dimensional case.

Worked Example. Find the coordinates of the point that divides the line segment joining P(4, −2, 6) and Q(10, −16, −18) internally in the ratio 7 : 5.

Here m = 7, n = 5, so m + n = 12.

x = (7×10 + 5×4)/12 = (70 + 20)/12 = 90/12 = 15/2

y = (7×(−16) + 5×(−2))/12 = (−112 − 10)/12 = −122/12 = −61/6

z = (7×(−18) + 5×6)/12 = (−126 + 30)/12 = −96/12 = −8

So the required point is (15/2, −61/6, −8).

Two important results follow directly from the section formula for internal division.

Midpoint. Putting m = n = 1 (ratio 1:1) in the section formula gives the midpoint M of P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):

M = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2)

Centroid of a triangle. For a triangle with vertices A(x₁,y₁,z₁), B(x₂,y₂,z₂), C(x₃,y₃,z₃), the median from A meets the midpoint D of BC, and the centroid G divides AD internally in ratio 2 : 1 from A. Applying the section formula gives

G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3)

Worked Example. Find the centroid of the triangle with vertices A(3, −5, 7), B(−1, 7, −6), C(1, 1, 2), and verify it using the median from A.

Directly, G = ((3−1+1)/3, (−5+7+1)/3, (7−6+2)/3) = (3/3, 3/3, 3/3) = (1, 1, 1).

Verification: the midpoint of BC is D = ((−1+1)/2, (7+1)/2, (−6+2)/2) = (0, 4, −2). Since G divides AD in ratio 2:1 from A, G = (2×D + 1×A)/3 = ((2×0+3)/3, (2×4+(−5))/3, (2×(−2)+7)/3) = (1, 1, 1), which matches.

Three points A, B, C in space are collinear (lie on one straight line) if one of them divides the segment joining the other two in some real ratio. This can be tested in two equivalent ways.

• Section-formula method: assume the middle point divides the segment joining the other two in ratio k : 1, find k using the x-coordinates, then check that the same k also satisfies the y- and z-coordinates.
• Distance method: compute all three pairwise distances; if the sum of the two smaller distances equals the largest, the points are collinear.

Worked Example. Show that the points A(2, −3, 4), B(−1, 2, 1) and C(0, 1/3, 2) are collinear.

Suppose C divides AB internally in ratio k : 1. Using the x-coordinates,

0 = (k×(−1) + 1×2)/(k+1) ⟹ −k + 2 = 0 ⟹ k = 2

Check with y-coordinates: (k×2 + 1×(−3))/(k+1) = (2×2 − 3)/3 = 1/3, which matches C's y-coordinate.

Check with z-coordinates: (k×1 + 1×4)/(k+1) = (2 + 4)/3 = 2, which matches C's z-coordinate.

Since the same value k = 2 satisfies all three coordinates, C divides AB in ratio 2 : 1, so A, B, C are collinear.