Question
The points $$\left( {4,\,7,\,8} \right),\,\left( {2,\,3,\,4} \right),\,\left( { - 1,\, - 2,\,1} \right)$$ and $$\left( {1,\,2,\,5} \right)$$ are the vertices of a :
A.
parallelogram
B.
rhombus
C.
rectangle
D.
square
Answer :
parallelogram
Solution :
Let the points are $$A,\,B,\,C$$ and $$D$$ respectively.
Mid point of $$AC$$ is
$$\left( {\frac{{4 - 1}}{2},\,\frac{{7 - 2}}{2},\,\frac{{8 + 1}}{2}} \right){\text{ or }}\left( {\frac{3}{2},\,\frac{5}{2},\,\frac{9}{2}} \right)$$
Mid point of $$BD$$ is
$$\left( {\frac{{2 + 1}}{2},\,\frac{{3 + 2}}{2},\,\frac{{4 + 5}}{2}} \right){\text{ or }}\left( {\frac{3}{2},\,\frac{5}{2},\,\frac{9}{2}} \right)$$
Thus $$AC$$ and $$BD$$ bisect each other. Further,
$$\eqalign{
& AC = \sqrt {{{\left( {4 + 1} \right)}^2} + {{\left( {7 + 2} \right)}^2} + {{\left( {8 - 1} \right)}^2}} \cr
& AC = \sqrt {25 + 81 + 49} \cr
& AC = \sqrt {155} \cr
& BD = \sqrt {{{\left( {2 - 1} \right)}^2} + {{\left( {3 - 2} \right)}^2} + {{\left( {4 - 5} \right)}^2}} \cr
& BD = \sqrt {1 + 1 + 1} \cr
& BD = \sqrt 3 \cr
& \therefore \,AC \ne BD \cr} $$
Hence, $$ABCD$$ represents a parallelogram.