Derive Section formula for 3D Geometry.
Answers
Draw AL, PN, and BM perpendicular to XY plane such that AL || PN || BM as shown above.
The points L, M and N lie on the straight line formed due to the intersection of a plane containing AL, PN and BM and XY- plane.
From the point P, a line segment ST is drawn such that it is parallel to LM.
ST intersects AL externally at S and it intersects BM at T internally.
Since ST is parallel to LM and AL || PN || BM, therefore, the quadrilaterals LNPS and NMTP qualify as parallelograms.
Also, ∆ASP ~∆BTP therefore,
mn=APBP=ASBT=SL–ALBM–TM=NP–ALBM–PN=z–z1z2–z
Rearranging the above equation we get,
mz2+nz1m+n
The above procedure can be repeated by drawing perpendiculars to XZ and YZ- planes to get the x and y coordinates of the point P that divides the line segment AB in the ratio m:n internally.
x=mx2+nx1m+n,y=my2+ny1m+n