Introduction to Three Dimensional Geometry

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 11.

Where exactly is the fly?

A fly is sitting somewhere in your classroom. To tell a friend where, "two metres from the left wall" is not enough, because the fly could be on the floor or near the ceiling. You need a second distance from the front wall and a third from the floor. On a sheet of paper two numbers fix a point. In a room, you need three. This chapter turns that idea into a system of coordinates and then asks the first natural question: how far apart are two points in space?

Axes, planes and octants

Take the floor and two adjacent walls of a room. They meet at one corner and are mutually perpendicular. Call the corner the origin $O$. The three lines along which the planes meet are the coordinate axes: the x-axis $X'OX$, the y-axis $Y'OY$ and the z-axis $Z'OZ$. Each pair of axes spans one coordinate plane: the XY-plane, the YZ-plane and the ZX-plane.

The convention is that the page is the XY-plane and the z-axis stands up through it. Above the XY-plane is positive $z$, in front of the YZ-plane is positive $x$ and to the right of the ZX-plane is positive $y$.

The three planes cut space into eight octants, the 3-D cousin of the four quadrants. Octant I is the all-positive one, $(+,+,+)$. The sign pattern of the other seven is listed below.

A memory trick. The z-sign splits the table in two halves: octants I to IV lie above the XY-plane ($z > 0$) and V to VIII lie below it. Inside each half, the signs of $(x, y)$ follow the quadrant order $(+,+)$, $(-,+)$, $(-,-)$, $(+,-)$. So the octant number is the quadrant of $(x, y)$, plus 4 if $z$ is negative.

Pause and predict: which octant contains $(-5, -2, 3)$?

Check your answer: Octant III. The $(x, y)$ part $(-,-)$ is the third quadrant, and $z$ is positive, so you add nothing. Reading the z-sign first, as many students do, is how octant questions go wrong.

What each coordinate measures

To find the coordinates of a point $P$, draw a plane through $P$ parallel to each coordinate plane. They cut the axes at $A$, $B$, $C$, and $x = OA$, $y = OB$, $z = OC$. The consequence is the single most tested fact in the chapter.

The coordinate belongs to the plane that does not carry its letter: $x$ is measured from the YZ-plane, not from the x-axis. The distance is always positive, so a point $(4, -5, 7)$ is $4$ units from the YZ-plane, $5$ units from the ZX-plane and $7$ units from the XY-plane, even though $y$ itself is negative.

Setting coordinates to zero puts a point on a particular axis or plane.

Notice that the equation of a coordinate plane is the one coordinate that is zero: the XY-plane is $z = 0$, the YZ-plane is $x = 0$ and the ZX-plane is $y = 0$. An axis needs two such equations at once.

The distance formula, and why it holds

Take $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$. Through $P$ and $Q$ draw planes parallel to the coordinate planes. They form a rectangular box with $PQ$ as its space diagonal, and the edges of the box have lengths $\lvert x_2 - x_1 \rvert$, $\lvert y_2 - y_1 \rvert$ and $\lvert z_2 - z_1 \rvert$.

Now use Pythagoras twice. First on the base of the box, the face diagonal squared is the sum of two edge squares. Then on the right triangle formed by that face diagonal, the vertical edge and $PQ$. Together they give $PQ^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$.

$$PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$

Because every difference is squared, the order of the points and the signs of the differences do not matter. A distance is never negative, so $PQ = QP \ge 0$. For the distance of $Q(x, y, z)$ from the origin, put $P = O$:

$$OQ = \sqrt{x^2 + y^2 + z^2}$$

Set $z_1 = z_2 = 0$ and you get back the plane formula, so nothing you learnt in Straight Lines is lost.

Worked example: a plain distance. Find the distance between $P(1, -3, 4)$ and $Q(-4, 1, 2)$.

Common wrong step. Dropping the $z$ term because the question "looks like" a plane problem, which gives $\sqrt{41}$. A close second is forgetting that $(-5)^2 = +25$ and writing $-25$.

Worked example: are three points collinear? Show that $P(-2, 3, 5)$, $Q(1, 2, 3)$ and $R(7, 0, -1)$ are collinear.

So the three points lie on one line, with $Q$ between $P$ and $R$.

Common wrong step. Adding the squares: $14 + 56 = 70 \ne 126$, and then declaring the points non-collinear. Adding squares is the right angle test, not the collinearity test. For collinearity, add the actual lengths.

Worked example: what kind of triangle? Are $A(0, 7, 10)$, $B(-1, 6, 6)$ and $C(-4, 9, 6)$ the vertices of a right-angled triangle?

The triangle is right-angled and also isosceles, since $AB = BC$.

Common wrong step. Testing $AB^2 + CA^2$ against $BC^2$, i.e. picking the wrong side as the hypotenuse. The hypotenuse is always the longest side, so find it first. The same table decides isosceles (two equal squares) and equilateral (three equal squares).

Beyond the textbook. For the exam: the formula for points in space is the plane formula with one more term. In reality: nothing stops at three. In $n$ dimensions the distance is still the square root of the sum of the squared coordinate differences. Machine learning uses exactly this to compare data points with hundreds of features, and a GPS receiver uses the 3-D version to turn its distances from several satellites into your position. Why it matters: one idea, proved once with a box, powers all of these.

Distances from planes and axes

Because $x$, $y$, $z$ are distances from the planes, the distance from a plane is just the modulus of one coordinate. Distances from an axis need the distance formula: the foot of the perpendicular from $P(a, b, c)$ to the x-axis is $(a, 0, 0)$, so the distance is $\sqrt{b^2 + c^2}$.

The rule for an axis: leave out the coordinate that carries the axis's own letter. So for $(3, -4, 5)$ the distance from the x-axis is $\sqrt{16 + 25} = \sqrt{41}$, not $3$.

Pause and predict: what is the distance of $(3, -4, 12)$ from the z-axis?

Check your answer: $\sqrt{3^2 + (-4)^2} = 5$. Leave out $z$ and keep $x$ and $y$. The tempting wrong answer is $12$, which is the distance from the XY-plane, not from the z-axis.

Loci: points that satisfy a distance condition

A locus is the set of all points obeying a condition. When the condition is about distances, write $P = (x, y, z)$, square both sides to kill the roots, and simplify. The squared terms $x^2$, $y^2$ and $z^2$ usually cancel when two distances are set equal.

Worked example: equidistant from two points. Find the equation of the set of points $P$ with $PA = PB$, where $A(1, 2, 3)$ and $B(3, 2, -1)$.

The locus is the plane $x - 2z = 0$. It is the perpendicular bisector plane of $AB$, and it passes through the midpoint $(2, 2, 1)$ of $AB$: $2 - 2(1) = 0$.

Common wrong step. Expecting a curved surface because the starting equation has squares. When the two distances are equal, the squares cancel and what is left is linear, which is a plane. A sphere appears only when one distance is fixed to a constant.

Beyond the textbook. For the exam: "equidistant from two points" gives a plane, found by squaring and cancelling. In reality: the same idea is used to locate a position from reference points. On a flat map, being equally far from three towers pins you to one point. In space, equal distance from two points gives a whole plane, and you need four reference points to reach a single point. Why it matters: it explains why locating something in space needs one more reference point than on a map.

Centroid of a triangle

The centroid of a triangle is the point where its medians meet. In space it is found by averaging each coordinate separately.

$$G = \left( \dfrac{x_1 + x_2 + x_3}{3},\ \dfrac{y_1 + y_2 + y_3}{3},\ \dfrac{z_1 + z_2 + z_3}{3} \right)$$

Worked example: a missing vertex. The centroid of triangle $ABC$ is $G(1, 1, 1)$, with $A(3, -5, 7)$ and $B(-1, 7, -6)$. Find $C$.

So $C = (1, 1, 2)$.

Common wrong step. Setting the vertex coordinates equal to the centroid's, or forgetting to multiply by $3$ before subtracting. Always turn "average equals" into "sum equals three times".

Parallelogram or rectangle?

Worked example: sides and diagonals. The points $A(1, 2, 3)$, $B(-1, -2, -1)$, $C(2, 3, 2)$ and $D(4, 7, 6)$ are the vertices of a parallelogram $ABCD$. Show that it is not a rectangle.

Common wrong step. Stopping after finding the opposite sides equal and calling it a rectangle, or testing only one diagonal. Rectangle means equal diagonals as well as equal opposite sides.

Beyond the textbook. For the exam: equal opposite sides and unequal diagonals mean a parallelogram that is not a rectangle. In reality: in space, four points can have equal opposite sides without lying in one plane, which gives a skew quadrilateral and not a parallelogram. The question here is safe because the four points are given as a parallelogram. Why it matters: in 3-D, equal lengths alone never prove that a shape is flat.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Before: Straight Lines and Conic Sections use the plane distance formula, which is the special case $z_1 = z_2 = 0$ of the formula here.
• Next in Class XI: Limits and Derivatives starts calculus, the second big idea of Class XI.
• Class XII: Three Dimensional Geometry builds lines and planes on these coordinates, and Vector Algebra reads $\sqrt{x^2+y^2+z^2}$ as the length of a position vector.