-->

Three Dimensional Geometry

Question
CBSEENMA12033323

Show that the line joining the middle points of two sides of a triangle is parallel to the third side and half of it in length.

Solution
Let A (x, y1, z1). B (x2, y2, z2), C (x3, y3, z3) be the vertices of Δ ABC and D, E, F be mid-points of BC, CA and AB respectively.
             therefore space space space straight E space is space open parentheses fraction numerator straight x subscript 3 plus straight x subscript 1 over denominator 2 end fraction comma space fraction numerator straight y subscript 3 plus straight y subscript 1 over denominator 2 end fraction comma space fraction numerator straight z subscript 3 plus straight z subscript 1 over denominator 2 end fraction close parentheses
and straight F space is space open parentheses fraction numerator straight x subscript 1 plus straight x subscript 2 over denominator 2 end fraction comma space fraction numerator straight y subscript 1 plus straight y subscript 2 over denominator 2 end fraction comma space fraction numerator straight z subscript 1 plus straight z subscript 2 over denominator 2 end fraction close parentheses

Direction-ratios of BC are
x– x2, y– y2, z3 – z2.
Directions-ratios of FE are
          fraction numerator straight x subscript 3 plus straight x subscript 1 over denominator 2 end fraction minus fraction numerator straight x subscript 1 plus straight x subscript 2 over denominator 2 end fraction comma space space fraction numerator straight y subscript 3 plus straight y subscript 1 over denominator 2 end fraction minus fraction numerator straight y subscript 1 plus straight y subscript 2 over denominator 2 end fraction comma space space fraction numerator straight z subscript 3 plus straight z subscript 1 over denominator 2 end fraction minus fraction numerator straight z subscript 1 plus straight z subscript 2 over denominator 2 end fraction
or        fraction numerator straight x subscript 3 minus straight x subscript 2 over denominator 2 end fraction comma space space fraction numerator straight y subscript 3 minus straight y subscript 2 over denominator 2 end fraction comma space space fraction numerator straight z subscript 3 minus straight z subscript 2 over denominator 2 end fraction
or      straight x subscript 3 minus straight x subscript 2 comma space space straight y subscript 3 minus straight y subscript 2 comma space space straight z subscript 3 minus straight z subscript 2
which are the same as that of BC
∴    FE || BC.
Also,
 FE space equals space square root of open parentheses fraction numerator straight x subscript 3 plus straight x subscript 1 over denominator 2 end fraction minus space fraction numerator straight x subscript 1 plus straight x subscript 2 over denominator 2 end fraction close parentheses squared plus open parentheses fraction numerator straight y subscript 3 plus straight y subscript 1 over denominator 2 end fraction minus fraction numerator straight y subscript 1 plus straight y subscript 2 over denominator 2 end fraction close parentheses squared space plus space open parentheses fraction numerator straight z subscript 3 plus straight z subscript 1 over denominator 2 end fraction minus fraction numerator straight z subscript 1 plus straight z subscript 2 over denominator 2 end fraction close parentheses squared end root
space space space space space equals space 1 half square root of left parenthesis straight x subscript 3 minus straight x subscript 2 right parenthesis squared plus left parenthesis straight y subscript 3 minus straight y subscript 2 right parenthesis squared plus left parenthesis straight z subscript 3 minus straight z subscript 2 right parenthesis squared end root space equals space 1 half BC
Hence the result. 

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.