You must have learn about 2-D geometry or two-dimensional coordinate system. Here you will learn about three dimensional geometry and other related topics such as definition, formulas, problems based on it.

## Introduction to Three Dimensional Geometry

When we talk about 2 dimension geometry, it consists of two coordinates say, x and y in the plane. But when we speak about 3 dimensional geometry, it consists of three coordinates say, x, y and z.

Three-dimensional geometry deal with shapes of square, rectangle, cuboid, rhombus, trapezium, parallelogram etc, which has coordinates in three-dimensional space. These shapes or figures have dimensions-length, width and height. You can take examples of 3D geometry from real worlds also.Like, a room of a house has length, width and height as its dimensions.

### Three Dimensional Geometry Formulas

#### Distance Formula

Let P and Q are two points in space which has coordinates (\(x_{1},y_{1},z_{1}\)) and (\(x_{2},y_{2},z_{2}\)), respectively.The distance between two points P and Q is defined by then,

\(PQ=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}\)#### Section Formula

- If a point R divides the line segment joining the two points P(\(x_{1}, y_{1}, z_{1}\)) and Q(\(x_{2}, y_{2}, z_{2}\)) in the ratio of m:n, internally, then, coordinates of R are;

- If a point R divides the line segment joining the two points P(\(x_{1},y_{1},z_{1}\)) and Q(\(x_{2},y_{2},z_{2}\)) in the ratio of m:n, externally, then, coordinates of R are;

#### Mid-Point Formula

If R is the mid-point of the segment joining P(\(x_{1},y_{1},z_{1}\)) and Q(\(x_{2},y_{2},z_{2}\)), then m=n=1 and the coordinate R is given by;

\(R( \frac{x_{1}+ x_{2}}{2}, \frac{y_{1}+ y_{2}}{2}, \frac{z_{1}+ z_{2}}{2})\)#### For Centroid of Triangle

Centroid of the triangle whose vertices are (\(x_{1},y_{1},z_{1}\)), (\(x_{2},y_{2},z_{2}\)) and (\(x_{3},y_{3},z_{3}\)), the coordinate R is given by;

\(R( \frac{x_{1}+ x_{2}+ x_{3}}{3}, \frac{y_{1}+ y_{2}+ y_{3}}{3}, \frac{z_{1}+ z_{2}+ z_{3}}{3})\)These formulas are the key concepts and are also covered under three dimensional geometry Class 11 syllabus.

#### Three Dimensional Geometry Problem and Solution:

To get a clear understanding of the three dimensional formulas, you should solve three dimensional geometry problems. Let us give an example problem here.

**Problem:** Find the distance between the points P(-2,4,1) and Q(1,2,-5).

**Solution: **Here, \(x_{1}\)=-2, \(y_{1}\)=4, \(z_{1}\)=1, \(x_{2}\)=1, \(y_{2}\)=2 and \(z_{2}\)=-5.

By the distance formula, we know,

PQ=\(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}\)

PQ=\(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}\)

PQ=\(\sqrt{(1-(-2))^{2}+(2-4)^{2}+(-5-1)^{2}}\)

PQ=\(\sqrt{9+4+36}\)

PQ=7 units

Similarly, you can also solve more problems on the above-given formulas by defining the values for each quantity.

For geometry related topics, download BYJUâ€™S- The Learning App.