Question

Points A$(1,2,3),B(-1,-2,-1),C(2,3,2)andD(4,7,6)$ are the vertices of _____

• A. Rectangle
• B. Square
• C. Parallelogram
• D. None of these

Answer 1

The correct option is C

Parallelogram

Let's check for the length of the sides of the points

AB = $\sqrt{(-1-1{)}^2+(-2-2{)}^2+(-1-3{)}^2}$ = $\sqrt{4+16+16}$ =$6$

BC = $\sqrt{(2+1{)}^2+(3+2{)}^2+(2+1{)}^2}$ = $\sqrt{9+25+9}$ = $\sqrt{43}$

CD = $\sqrt{(4-2{)}^2+(7-3{)}^2+(6-2{)}^2}$ = $\sqrt{4+16+16}$ = $6$

AD = $\sqrt{(1-4{)}^2+(2-7{)}^2+(3-6{)}^2}$ = $\sqrt{9+25+9}$ = $\sqrt{43}$

since,AB = CD and BC = AD

Either ABCD can be rectangle or parallelogram.Let's check for diagonals.since ,lengths of the diagonals are equal for rectangle.

we,have AC = $\sqrt{(2-1{)}^2+(3-2{)}^2+(2-3{)}^2}$ = $\sqrt{1+1+1}$ = $\sqrt{3}$

BD = $\sqrt{(4+2{)}^2+(7+2{)}^2+(6+1{)}^2}$ = $\sqrt{25+81+49}$ = $\sqrt{155}$

since AC $\ne$ BD ABCD is NOT a rectangle.

For parallelogram diagonals should bisect each other.

Let the intersection point of the diagonal is M

M should be midpoint of both the diagonal AC & BD

Midpoint of AC is M$(\frac{1+2}{2},\frac{2+3}{2},\frac{3+2}{2})\equiv (\frac{3}{2},\frac{5}{2},\frac{5}{2})$

Midpoint of BD is M$(\frac{4-1}{2},\frac{7-2}{2},\frac{6-1}{2})\equiv (\frac{3}{2},\frac{5}{2},\frac{5}{2})$

Midpoints of diagonals is same,AB = CD,BC = BD and AC $\ne$ BD.

ABCD are the vertices of a parallelogram.