Introduction to Three Dimensional Geometry

In this chapter: the three coordinate axes and coordinate planes, octants and the signs of coordinates in each, coordinates of a point in space, the distance formula, and (for JEE and MHT‑CET) the section formula, midpoint and centroid.

Axes, planes and octants

Take three mutually perpendicular lines through a point O, the origin: the x-axis (XOX′), the y-axis (YOY′) and the z-axis (ZOZ′). Each pair of axes fixes a plane:

• the XY-plane, on which every point has z = 0;
• the YZ-plane, on which every point has x = 0;
• the ZX-plane, on which every point has y = 0.

These three coordinate planes divide space into eight parts called octants.

Coordinates of a point in space

Through a point P draw three planes parallel to the coordinate planes, meeting the x-, y- and z-axes at A, B and C. If OA = x, OB = y and OC = z (with signs according to direction), then P has coordinates (x, y, z). Equivalently:

• x is the signed distance of P from the YZ-plane;
• y is the signed distance of P from the ZX-plane;
• z is the signed distance of P from the XY-plane.

So the origin is (0, 0, 0); a point on the x-axis is (x, 0, 0); a point on the z-axis is (0, 0, z); a point in the XY-plane is (x, y, 0). The foot of the perpendicular from (a, b, c) to the YZ-plane is (0, b, c), and similarly for the other planes.

Signs of coordinates in the octants

The pattern is easy to rebuild: octants I to IV lie above the XY-plane (z > 0) and follow the signs of the four quadrants of plane geometry in order; octants V to VIII lie below it and repeat the same x, y pattern. For example, (−3, 1, −2) has x < 0, y > 0, z < 0 and lies in octant VI.

Distance between two points

For P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), apply Pythagoras twice, once in a horizontal plane and once in a vertical one:

Two related results come up often. The distance of P(x, y, z) from the x-axis is √(y² + z²) (drop the coordinate of the axis itself), and its distance from the XY-plane is simply |z|.

Worked example: (a) Find the distance between (1, −3, 4) and (−4, 1, 2). (b) Show that A(−2, 3, 5), B(1, 2, 3) and C(7, 0, −1) are collinear.
Solution: (a) √[(−5)² + 4² + (−2)²] = √(25 + 16 + 4) = √45 = 3√5.
(b) AB = √(9 + 1 + 4) = √14, BC = √(36 + 4 + 16) = √56 = 2√14 and AC = √(81 + 9 + 36) = √126 = 3√14. Since AB + BC = AC, the three points lie on one line.

The distance formula also settles shape questions. To check that three points form a right triangle, verify that the square of the longest side equals the sum of the squares of the other two; for an isosceles triangle, look for two equal sides.

It also gives loci. The set of points P with PA = PB, for A(3, 4, 5) and B(−1, 3, −7), satisfies (x − 3)² + (y − 4)² + (z − 5)² = (x + 1)² + (y − 3)² + (z + 7)². The squared terms cancel and you are left with 8x + 2y + 24z + 9 = 0, a plane: the perpendicular bisector plane of AB.

Section formula, midpoint and centroid

These are in the JEE and MHT‑CET syllabus and are used constantly in Class 12 vectors. The point R dividing the join of P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in the ratio m : n is

Worked example: Find the point that divides the join of P(1, −2, 3) and Q(3, 4, −5) internally in the ratio 2 : 3.
Solution: With m = 2 and n = 3: x = (2 × 3 + 3 × 1)/5 = 9/5, y = (2 × 4 + 3 × (−2))/5 = 2/5, z = (2 × (−5) + 3 × 3)/5 = −1/5. The point is (9/5, 2/5, −1/5).

A plane that divides a segment is handled the same way: to find where the YZ-plane cuts the join of two points, write the general point in the ratio k : 1 and set its x-coordinate equal to 0.

Common mistakes: (1) Mixing up which coordinate is zero on which plane: z = 0 on the XY-plane, not on the "Z-plane". (2) Getting the octant wrong because the table was memorised rather than rebuilt from the quadrant pattern. (3) Pairing m with x₁ in the section formula; m multiplies the coordinates of the second point. (4) Writing the distance from the x-axis as |x|; it is √(y² + z²). (5) Forgetting to square the negative differences carefully, for example writing (−5)² as −25.

Exam focus

• Identifying the octant of a point and the coordinates of feet of perpendiculars on the axes and planes.
• Distance formula for side lengths, collinearity, and right-angled or isosceles triangles.
• Locus problems using the distance formula, which lead to planes or spheres.
• Section formula, midpoint and centroid, including the ratio in which a coordinate plane divides a segment.

Practice questions