source: ncert
topic: geometry
class_level: class_11
chapter: ch12
difficulty: intermediate
Class 11 | Ch12: Introduction to Three Dimensional Geometry COORDINATES IN 3D: Point P=(x,y,z). Three mutually perpendicular axes: x,y,z. OCTANTS: 3D space divided into 8 octants by coordinate planes. DISTANCE FORMULA: Distance between P(x1,y1,z1) and Q(x2,y2,z2): PQ=sqrt((x2-x1)^2+(y2-y1)^2+(z2-z1)^2). Distance from origin: OP=sqrt(x^2+y^2+z^2). SECTION FORMULA (Internal Division): Point dividing PQ in ratio m:n internally: R=((mx2+nx1)/(m+n), (my2+ny1)/(m+n), (mz2+nz1)/(m+n)). MIDPOINT: M=((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). COORDINATE PLANES: xy-plane: z=0. yz-plane: x=0. xz-plane: y=0. Distance from xy-plane: |z|. From yz-plane: |x|. From xz-plane: |y|. SOLVED EXAMPLES: Example 1: Distance between A(1,2,3) and B(4,-2,6). AB=sqrt((4-1)^2+(-2-2)^2+(6-3)^2)=sqrt(9+16+9)=sqrt(34). Example 2: Find point dividing A(2,1,3) and B(5,4,6) in ratio 2:1. P=((2×5+1×2)/3,(2×4+1×1)/3,(2×6+1×3)/3)=(12/3,9/3,15/3)=(4,3,5). Example 3: Show A(0,7,-10), B(1,6,-6), C(4,9,-6) form a right triangle. AB=sqrt(1+1+16)=sqrt(18). BC=sqrt(9+9+0)=sqrt(18). AC=sqrt(16+4+16)=sqrt(36)=6. AB^2+BC^2=18+18=36=AC^2. Right angle at B. COMMON MISTAKES: 3D distance has THREE terms under square root (not two). Section formula extends naturally from 2D — same structure but three coordinates. Midpoint: average each coordinate SEPARATELY.