Q.Show that the points P(3,−2,4), Q(1,1,1) and R(−1,4,2) are collinear.
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3D Coordinate Geometry
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).
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 origin O(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.
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: …
Checking direction ratios of PQ=(−2,3,−3) and QR=(−2,3,1) shows they are not proportional, so the three points as printed are not collinear — most likely a coordinate misprint in the source paper (see long solution for the honest working and the likely intended correction). …
Self-verification: with the coordinates exactly as printed, P, Q, R do not turn out to be collinear. We show the genuine working honestly rather than force a fake proof.
The standard method to show three points are collinear is to check that the direction ratios of two of the segments (say PQ and QR) are proportional.
PQ=Q−P=(1−3, 1−(−2), 1−4)=(−2, 3, −3)
QR=R−Q=(−1−1, 4−1, 2−1)=(−2, 3, 1)
For collinearity we would need −2−2=33=−31. The first two ratios equal 1, but the third ratio is −31 — these are not all equal, so PQ and QR are not scalar multiples of each other.
This can also be confirmed using distances: PQ=4+9+9=22, QR=4+9+1=14, PR=16+36+4=56=214, and PQ+QR=22+14≈4.69+3.74=8.43, which is not equal to PR≈7.48 — again not collinear.
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Showing the 12 most recent of 26 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.Distance between the points P(1,−3,4) and Q(−4,1,2) is(a) 53 unit(b) 7 unit(c) 35 unit(d) 5 unit
›Reveal solutionSolution
Use the 3D distance formula d=(x2−x1)2+(y2−y1)2+(z2−z1)2.
…
- CBSE 2026Set ANNUAL1 markQ.Write True/False: The distance between points (1,0,0) and (0,1,0) is 2.
›Reveal solutionSolution
Using the 3D distance formula, the distance between (1,0,0) and (0,1,0) works out to exactly 2, confirming the statement.
Distance between (x1,y1,z1) and (x2,y2,z2) is d=(x2−x1)2+(y2−y1)2+(z2−z1)2.
…
- CBSE 2026Set ANNUAL1 markQ.Find distance between the points (4,0,0) and (−4,0,0).
›Reveal solutionSolution
Distance formula in 3D gives d=(x2−x1)2+(y2−y1)2+(z2−z1)2; here it simplifies to 8, the direct separation along the x-axis.
Points: (4,0,0) and (−4,0,0).
…
- CBSE 2026Set 1A1 markMCQQ.XOZ-plane divides the line joining the points (2,3,1) and (6,7,1) in the ratio -(1) 3:7(2) 2:7(3) −3:7(4) −2:4
›Reveal solutionSolution
Setting the y-coordinate to 0 in the section formula gives ratio −3:7.
Let the XOZ-plane (equation y=0) divide the join of (2,3,1) and (6,7,1) in the ratio k:1. The y-coordinate of the division point is …
- CBSE 2025Set ANNUAL1 markMCQQ.The coordinates of origin in three dimensional geometry are(a) (0,0)(b) (0,0,0)(c) (0,0,0,0)(d) (0,0,0,0,0)
›Reveal solutionSolution
The origin in 3D geometry is (0,0,0).
In 3D space, every point needs three coordinates (along the x, y, z axes) to be located. The origin O is the common point where all three coor …
- CBSE 2025Set ANNUAL1 markMCQQ.The distance between the points (2,3,5) and (4,3,1) is(a) 5(b) 25(c) 35(d) 5
›Reveal solutionSolution
Distance between (2,3,5) and (4,3,1) is 25.
Using the distance formula: …
- CBSE 2025Set ANNUAL1 markMCQQ.The coordinates of the mid-point of the line joining the points (4,6,8) and (−8,−6,−4) are(a) (0,0,0)(b) (−2,0,2)(c) (2,0,−2)(d) (2,0,2)
›Reveal solutionSolution
Midpoint of (4,6,8) and (−8,−6,−4) is (−2,0,2).
Midpoint formula: (2x1+x2,2y1+y2,2z1+z2).
…
- CBSE 2025Set ANNUAL1 markMCQQ.The distance between the points A(−2,1,−3) and B(4,3,−6) is(a) 0 units(b) 4 units(c) 7 units(d) 12 units
›Reveal solutionSolution
Apply the 3D distance formula directly to A(−2,1,−3) and B(4,3,−6).
…
- CBSE 2025Set ANNUAL1 markQ.Fill in the blank: A point with co-ordinates (2,3,0) lies in the ____ plane.
›Reveal solutionSolution
Any point of the form (x,y,0), with z-coordinate zero, lies in the xy-plane.
In three-dimensional geometry, the xy-plane consists of all points whose z-coordinate is 0; the yz-plane consists of points whose x-coordinate is 0; the xz-plane consists of points whose y-coordinate is …
- CBSE 2025Set ANNUAL1 markMCQQ.Match the column: Column A entry 'Distance of the point (2,4,5) from the xz plane' — find the matching value from Column B.(a) 2(b) 8(c) 32(d) 1−tan2x2tanx(e) sin2x(f) 10(g) 20(h) 1+tan2x2tanx(i) 4
›Reveal solutionSolution
The distance of any point (x,y,z) from the xz-plane equals ∣y∣; here y=4.
The xz-plane consists of all points with y=0. The perpendicular distance of a point (x,y,z) from the xz-plane is ∣y∣.
…
- CBSE 2025Set ANNUAL1 markQ.A point lies in the xz plane. What will be its y-coordinate?
›Reveal solutionSolution
The xz-plane is defined as the set of all points with y=0.
In 3D coordinate geometry, the xz-plane consists of all points of the form (x,0,z) — i.e. every point on this plane has y-coor …
- CBSE 2024Set ANNUAL1 markMCQQ.The distance of the point (1,−3,4) from x-axis is(a) 5(b) 1(c) 26(d) None of these
›Reveal solutionSolution
The distance from a point to the x-axis depends only on its y and z coordinates (the x-coordinate doesn't matter, since the foot of the perpendicular slides along the axis).
For a point (x,y,z), its perpendicular distance from the x-axis is:
d=y2+z2
…
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