Introduction to Three-Dimensional Geometry

Introduction to 3D Geometry

Board: CBSE | Class: Class 11

Comprehensive study notes for Introduction to 3D Geometry by Ajay Yadav (Math King of Katargam).

Key Concepts

Coordinate Axes

Three mutually perpendicular axes: x, y, z. Origin at intersection. A point P = (x,y,z) in space. Octants: 8 regions divided by coordinate planes (xyz signs determine octant).

Distance Formula

Distance between P(x1,y1,z1) and Q(x2,y2,z2): d = √((x&#8322-x&#8321)&sup2+(y&#8322-y&#8321)&sup2+(z&#8322-z&#8321)&sup2).

Section Formula

Point dividing PQ in ratio m:n internally: ((mx2+nx1)/(m+n), (my2+ny1)/(m+n), (mz2+nz1)/(m+n)). Midpoint: ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2).

Centroid of Triangle

G = ((x1+x2+x₃)/3, (y1+y2+y₃)/3, (z1+z2+z₃)/3).

Centroid of Tetrahedron

G = ((x1+x2+x₃+x₄)/4, (y1+y2+y₃+y₄)/4, (z1+z2+z₃+z₄)/4).

Important Formulas

Distance 3Dd = √((x&#8322-x&#8321)&sup2+(y&#8322-y&#8321)&sup2+(z&#8322-z&#8321)&sup2)
Section((mx2+nx1)/(m+n), ...)
Midpoint((x1+x2)/2, (y1+y2)/2, (z1+z2)/2)

Solved Examples

Example 1: Find distance between (1,2,3) and (4,5,6).

Solution: d = √((3)&sup2+(3)&sup2+(3)&sup2) = √27 = 3√3.

Example 2: Find midpoint of (1,2,3) and (7,8,9).

Solution: ((1+7)/2, (2+8)/2, (3+9)/2) = (4,5,6).

Example 3: Find centroid of triangle with vertices (1,1,1), (2,3,4), (3,5,7).

Solution: ((1+2+3)/3, (1+3+5)/3, (1+4+7)/3) = (2,3,4).

Practice Questions

  1. Find distance from origin to (3,4,5).
  2. Find ratio in which (1,2,3) divides line joining (0,0,0) and (2,4,6).
  3. Show that points (1,2,3), (4,5,6), (7,8,9) are collinear.
  4. Find the centroid of tetrahedron with vertices (0,0,0), (a,0,0), (0,b,0), (0,0,c).
  5. Find the coordinates of point equidistant from (1,0,0), (0,1,0), (0,0,1), and origin.

Download PDF

Click here to download PDF notes.

Video Lessons

Watch video explanations on our Videos page.

Scroll to Top