NDA Maths · 3D Geometry

Foundations: Coordinates, Distance & Section in Space

Locate points with three coordinates, then measure between them — distance, the dividing point of a segment, midpoints, centroids, and whether points line up.

All 3D Geometry notes NDA Maths strategy

Why this matters

Twenty PYQs across 2017–2026, and the launch pad for everything else in the chapter. Questions test the octant/coordinate-plane setup, the distance formula, the section formula (especially the ratio in which a coordinate plane cuts a segment), centroids, and collinearity / shape tests. Six EASY marks live here every other paper — internalise these five concepts and you bank them on sight.

Test yourself — a quick 12-question 3D Geometry recall quiz, instant score.

Concept 1 of 5: The 3D coordinate frame — axes, planes, and octants

Add a third axis (the z-axis) straight up out of the familiar xy-plane. Now every point needs three numbers

(x, y, z)

. The three axes are mutually perpendicular and meet at the origin; the three coordinate planes (XY, YZ, ZX) slice all of space into eight corner regions called octants — the 3D version of the four quadrants.

Definition

A point in space is an ordered triple $(x, y, z)$ . The three coordinate planes are:

XY-plane: all points with $z = 0$ .
YZ-plane: all points with $x = 0$ .
ZX-plane: all points with $y = 0$ .

These three planes divide space into 8 octants. The first octant holds points with all three coordinates positive. A point on an axis has its other two coordinates zero; a point on a coordinate plane has exactly one coordinate zero.

Diagram · coordinate planes & octants (drag to rotate)

Three planes (XY, YZ, ZX), each splitting space in two → 2 × 2 × 2 = 8 octants. P sits in the first octant (all coordinates positive).

Location	Condition	Example point
On the x-axis	$y = 0,\ z = 0$	$(5, 0, 0)$
On the XY-plane	$z = 0$	$(3, -2, 0)$
On the YZ-plane	$x = 0$	$(0, 4, 1)$
In the first octant	$x, y, z > 0$	$(2, 3, 4)$ The three coordinate planes (not the three axes) are what divide space — that gives 8 octants, not 6.

Zero coordinates tell you where a point sits: one zero → on a plane, two zeros → on an axis.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2020 · Q64Easy

Example 1 · 3D Geometry · Distance, Section, and Collinearity in 3D

Into how many compartments do the coordinate planes divide the space?

Concept 2 of 5: Distance between two points

The straight-line distance between two points in space is just Pythagoras done twice: take the differences in

x

y

, and

z

, square them, add, and square-root. To find a point's distance from a coordinate axis, drop the coordinate measured along that axis and take the distance of what's left.

Definition

For $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ , the distance $AB$ is the square root of the summed squared coordinate differences. Distance from a point to the x-axis is $\sqrt{y^2 + z^2}$ (ignore $x$ ); similarly $\sqrt{x^2+z^2}$ from the y-axis and $\sqrt{x^2+y^2}$ from the z-axis.

Distance between two points

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

$(x_1,y_1,z_1)$ coordinates of $A$
$(x_2,y_2,z_2)$ coordinates of $B$

Diagram · magnitude = √(x² + y²)

The components x and y are the legs of a right triangle; the vector is the hypotenuse, so |v| = √(x² + y²) = √(16 + 9) = 5. In 3-D the same idea adds a third leg: |v| = √(x² + y² + z²).

Worked example

Find the distance between

A(1, -2, 3)

and

B(4, 2, 15)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2020 · Q62Easy

Example 2 · 3D Geometry · Distance, Section, and Collinearity in 3D

What is the perpendicular distance from the point

(2,3,4)

to the line

\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}

Don't forget the square root — or any one of the three squared terms

The distance is

\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}

, not the bare sum

(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2

(that's

AB^2

). In 3D it is easy to carry only two of the three coordinate differences — Pythagoras runs over ALL THREE axes here, so the

z

-term must be included. Leaving out the

\sqrt{}

gives the squared distance; dropping a term gives a too-small answer.

Drill 6 more on distance between two points

Concept 3 of 5: Section formula — dividing a segment in a ratio

A point that splits a segment

AB

in the ratio

m:n

is a weighted average of the endpoints — the nearer endpoint gets the larger weight. The classic NDA twist asks in what ratio a coordinate plane cuts a segment: set the relevant coordinate of the dividing point to zero and solve for the ratio.

Definition

The point dividing $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ internally in ratio $m:n$ has each coordinate as the weighted mean below. For external division, replace $n$ with $-n$ . To find where the XY-plane ( $z=0$ ) cuts $AB$ , set the $z$ -coordinate of the dividing point to 0: the ratio is $-z_1 : z_2$ (equivalently $z_1 : z_2$ externally / internally depending on signs).

Internal division in ratio m : n

\left( \frac{m x_2 + n x_1}{m+n},\ \frac{m y_2 + n y_1}{m+n},\ \frac{m z_2 + n z_1}{m+n} \right)

$m:n$ ratio in which the point divides $AB$
$A, B$ the two endpoints

Diagram · section formula (internal vs external), m : n = 2 : 1

Internal: P = (m·b + n·a)/(m + n) sits between A and B. External: Q = (m·b − n·a)/(m − n) sits beyond B — the minus sign is what pushes it outside. The midpoint is the m = n case, (a + b)/2.

Worked example

In what ratio does the XY-plane divide the segment joining

(1, 2, 4)

and

(3, -1, 2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2021 · Q63Moderate

Example 3 · 3D Geometry · Distance, Section, and Collinearity in 3D

The

xy

-plane divides the line segment joining the points

(-1,3,4)

and

(2,-5,6)

The ratio m : n weights the FAR endpoint by m — don't swap the weights

For

P

dividing

AB

in ratio

m:n

, the coordinate is

\dfrac{m x_2 + n x_1}{m+n}

— the larger weight

m

multiplies

x_2

(the point

P

is NEARER to

B

). Writing

\dfrac{m x_1 + n x_2}{m+n}

silently divides in ratio

n:m

and lands you on the wrong point. Also:

m:n

is the midpoint ONLY when

m=n

— never average the endpoints for a

2:1

1:3

split.

Drill 1 more on section formula — dividing a segment in a ratio

Concept 4 of 5: Midpoint and centroid

The midpoint averages two points; the centroid of a triangle averages its three vertices. Both are just coordinate-wise means — no ratio bookkeeping. The centroid is the triangle's balance point and divides each median

2:1

Definition

The midpoint of $A$ and $B$ averages the two coordinate triples. The centroid $G$ of triangle $ABC$ averages all three vertices: each coordinate of $G$ is the mean of that coordinate over $A, B, C$ . Two median facts that come with the midpoints: the centroid divides each median $2:1$ from the vertex, and the sum of squares of the medians is three-quarters of the sum of squares of the sides, $AD^2 + BE^2 + CF^2 = \tfrac34(AB^2 + BC^2 + CA^2)$ (Apollonius: $AB^2 + AC^2 = 2AD^2 + 2BD^2$ for the median $AD$ , summed over the three medians).

Centroid of triangle ABC

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

Diagram · closed loop & centroid

Walking the edges A→B→C→A returns you to the start, so AB + BC + CA = 0. The three medians meet at the centroid G = (a + b + c)/3, the average of the vertices' position vectors.

Worked example

Find the centroid of the triangle with vertices

A(1,2,3)

B(4,-1,0)

and

C(7,2,3)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2019 · Q61Easy

Example 4 · 3D Geometry · Distance, Section, and Collinearity in 3D

The centroid of the triangle with vertices

A(2,-3,3)

B(5,-3,-4)

and

C(2,-3,-2)

is the point

Concept 5 of 5: Collinearity and shape tests

Three points are collinear when one segment is a scalar multiple of another — same direction ratios — or, equivalently, when the longest distance equals the sum of the other two. The same distance toolkit classifies triangles (right-angled, isosceles) and quadrilaterals (rectangle, parallelogram): compute side lengths and diagonals and compare.

Definition

Collinearity: $A, B, C$ are collinear iff $\overrightarrow{AB}$ and $\overrightarrow{AC}$ have proportional components, i.e. the same direction ratios. Distance check: collinear iff $AB + BC = AC$ (for $B$ between). Shapes: a triangle is right-angled where two sides satisfy Pythagoras; a parallelogram has equal, bisecting diagonals (midpoint of one diagonal = midpoint of the other); a rectangle additionally has equal diagonals. Area of a triangle in 3D: there is no $\tfrac12|x_1(y_2-y_3)+\dots|$ shortcut — use the cross product, $\text{Area} = \tfrac12\,|\overrightarrow{AB}\times\overrightarrow{AC}|$ (or Heron's formula from the three distances). For $A(1,0,0), B(0,1,0), C(0,0,1)$ : $\overrightarrow{AB}\times\overrightarrow{AC} = (1,1,1)$ , area $\tfrac{\sqrt3}{2}$ .

Worked example

If the points

A(1, -1, 2)

B(3, k, 4)

and

C(5, 3, 6)

are collinear, find

k

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2019 · Q60Moderate

Example 5 · 3D Geometry · Distance, Section, and Collinearity in 3D

If the points

(x, y, -3)

(2, 0, -1)

and

(4, 2, 3)

lie on a straight line, then what are the values of

x

and

y

respectively ?

Collinear vs coplanar vs concyclic

NDA likes asking whether four points are collinear, coplanar, or form a specific shape. Three points are ALWAYS coplanar; the real test is collinearity (proportional direction ratios). For four points, check coplanarity via the scalar triple product of three edge vectors = 0.

Drill 10 more on collinearity and shape tests

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Distance between two points
Distance between two points
$AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$
Section formula — dividing a segment in a ratio
Internal division in ratio m : n
$\left( \frac{m x_2 + n x_1}{m+n},\ \frac{m y_2 + n y_1}{m+n},\ \frac{m z_2 + n z_1}{m+n} \right)$
Midpoint and centroid
Centroid of triangle ABC
$G = \left( \frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3},\ \frac{z_1+z_2+z_3}{3} \right)$

Reference tables (1)

The 3D coordinate frame — axes, planes, and octants4 rows

Location	Condition	Example point
On the x-axis	$y = 0,\ z = 0$	$(5, 0, 0)$
On the XY-plane	$z = 0$	$(3, -2, 0)$
On the YZ-plane	$x = 0$	$(0, 4, 1)$
In the first octant	$x, y, z > 0$	$(2, 3, 4)$ The three coordinate planes (not the three axes) are what divide space — that gives 8 octants, not 6.

Zero coordinates tell you where a point sits: one zero → on a plane, two zeros → on an axis.

Watch out for (3)

Don't forget the square root — or any one of the three squared terms
→ Distance between two points
The ratio m : n weights the FAR endpoint by m — don't swap the weights
→ Section formula — dividing a segment in a ratio
Collinear vs coplanar vs concyclic
→ Collinearity and shape tests

Test yourself on 3D Geometry

20 past NDA questions from this chapter, timed at 25 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the 3D Geometry chapter test Drill all 22 questions