You already know 2D coordinate geometry — the xy-plane where every point is described by two numbers (x,y). Now imagine lifting that plane into the air. That is three-dimensional geometry.
The Intuition: Three Numbers, One Point
In the real world you rarely locate something with just two numbers. To describe where a book sits on a shelf you might say: "third shelf up, fourth book from the left, and it is the one nearest the wall." That is three pieces of information — height, sideways position, and depth.
In 3D coordinate geometry we do exactly this. We keep the familiar x and y axes (which define a flat floor) and add a third axis — the z-axis — pointing straight up. Every point in space now needs three numbers: (x,y,z).
Note
The three axes are mutually perpendicular. Picture the corner of a room: two floor edges give the x- and y-axes, and the vertical edge where the walls meet gives the z-axis.
The Precise Statement
Definition: A rectangular 3D coordinate system consists of three mutually perpendicular number lines — the x-axis, y-axis and z-axis — meeting at a common point, the originO(0,0,0). Any point P in space is uniquely represented by an ordered triple (x,y,z), where:
x = signed distance from the yz-plane,
y = signed distance from the zx-plane,
z = signed distance from the xy-plane.
P=(x,y,z)
How to Read a 3D Point
Take the point A(2,−3,4). Start at the origin. Move 2 units along the x-axis. From there move −3 units parallel to the y-axis (backward, because it is negative). From that spot move 4 units parallel to the z-axis (upward). You have reached A.
Watch out
The order matters absolutely. (2,−3,4) is not the same point as (2,4,−3). Always follow the sequence: x first, then y, then z.
The Three Coordinate Planes
Each pair of axes determines a plane:
Plane
Equation
Description
xy-plane
z=0
the floor — all points with zero height
yz-plane
x=0
one wall — all points with zero x
zx-plane
y=0
the other wall — all points with zero y
These three planes cut space into 8 octants (the 3D analogue of the four quadrants of the plane). The first octant is where x>0, y>0 and z>0.
Distance Between Two Points
This is the natural extension of the 2D distance formula. For P(x1,y1,z1) and Q(x2,y2,z2):
PQ=(x2−x1)2+(y2−y1)2+(z2−z1)2
It is just the diagonal of a rectangular box whose edges are the differences in x, y and z. The distance of P from the origin is the special case OP=x12+y12+z12.
Section Formula (Internal Division)
If R divides the segment joining P(x1,y1,z1) and Q(x2,y2,z2) internally in the ratio m:n, then:
R=(m+nmx2+nx1,m+nmy2+ny1,m+nmz2+nz1)
This is the same pattern as the 2D section formula, applied to three coordinates instead of two.
Tip
For the midpoint, set m=n=1:
M=(2x1+x2,2y1+y2,2z1+z2)
What Comes Next
Once you are comfortable with points, distances and section division, the natural next steps (in later study) are direction cosines and the equations of lines and planes in space. For now, remember the core idea: 3D coordinate geometry is 2D geometry with one extra dimension — every formula you already know simply gains a third term.
3D Coordinate Geometry is the heart of the NCERT Class 11 Mathematics chapter Introduction to Three Dimensional Geometry, matching searches such as "3D coordinate geometry formulas class 11 maths" or "distance and section formula in 3D important questions". The octants, coordinate planes, distance formula and section formula introduced here carry real weightage in CBSE Class 11 exams and form the groundwork for the Class 12 three-dimensional geometry of lines and planes, as well as the coordinate-geometry sections of JEE Main and state CETs.
Concept: Diagonals of a parallelogram bisect each other
In a parallelogram, the diagonals intersect at their mutual midpoint. Let the fourth vertex be D(x,y,z).
The diagonals are AC and BD. Their midpoint must coincide:
Midpoint of AC=(23+(−1),2−1+1,22+2)=(1,0,2)
Midpoint of BD=(21+x,22+y,2−4+z)
Equating the two midpoints:
21+x=1⟹x=1
22+y=0⟹y=−2
2−4+z=2⟹z=8
✓Final answer
The fourth vertex is D(1,−2,8).
In a parallelogram, diagonals bisect each other, so the midpoint of AC equals the midpoint of BD. Using this condition, the fourth vertex is D(1,−2,8).
The defining property of a parallelogram is that its diagonals bisect each other. This means the point where the diagonals cross is the midpoint of both diagonals. If we know three vertices A, B, and C, we can find the fourth vertex D by equating the midpoint of diagonal AC with the midpoint of diagonal BD.
Why does this work? Because the diagonals of a parallelogram always meet at their mutual midpoint, this single condition captures the entire geometry of the figure. Once we enforce that the diagonals share a common midpoint, the fourth vertex is uniquely determined.
Let's denote the unknown fourth vertex as D(x,y,z).
This gives us three equations (one for each coordinate):
21+x=1,22+y=0,2−4+z=2
Solve for x, y, and z.
From the first equation:
21+x=1⟹1+x=2⟹x=1
From the second equation:
22+y=0⟹2+y=0⟹y=−2
From the third equation:
2−4+z=2⟹−4+z=4⟹z=8
Tip
You can verify your answer by checking that AB=DC (opposite sides are parallel and equal). Here, AB=(−2,3,−6) and DC=(2,−3,6)=−AB, which confirms the parallelogram property when we account for direction.
✓Final answer
The coordinates of the fourth vertex are D(1,−2,8).
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQ
Q.The distance of point P(3,4,5) from YZ plane is:
(A) 3 units
(B) 4 units
(C) 5 units
(D) 12 units
›Reveal solutionSolution
Distance from YZ-plane =∣x∣=3.
The YZ-plane is x=0. The perpendicular distance of a point (x,y,z) from it is ∣x∣.
For P(3,4,5), the distance =∣3∣=3 units.
✓Final answer
(A) 3 units.
AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQ
Q.Distance between the points P(1,−3,4) and Q(−4,1,2) is:
(A) 35 units
(B) 45 units
(C) 3 units
(D) 4 units
›Reveal solutionSolution
PQ=25+16+4=45=35.
Distance formula in 3D:
PQ=(x2−x1)2+(y2−y1)2+(z2−z1)2.
With P(1,−3,4) and Q(−4,1,2):
PQ=(−4−1)2+(1−(−3))2+(2−4)2=(−5)2+42+(−2)2
=25+16+4=45=35 units.
✓Final answer
(A) 35 units.
AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQ
Q.If the co-ordinate of the vertices A, B, C of △ABC are (3,−5,7), (−1,7,−6) and (4,1,2) respectively. Then co-ordinates of centroid is:
(A) (1,1,2)
(B) (2,1,1)
(C) (1,1,1)
(D) (2,2,2)