| --- |
| 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. |
