Question

Which of the following set of points are non-collinear?
A. (1, -1, 1), (-1, 1, 1), (0, 0, 1)
B. (1, 2, 3), (3, 2, 1), (2, 2, 2)
C. (-2, 4, -3), (4, -3, -2), (-3, -2, 4)
D. (2, 0, -1), (3, 2, -2), (5, 6, -4)

Answer 1

Hint: Use the distance formula to find the distances between all three points in each set of points. If the sum of any two distances equals the third side, then that set of points are collinear. The distance formula given two points $P({x_1},{y_1},{z_1})\,{\text{and }}Q({x_2},{y_2},{z_2})$ is \[\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \]

Formula used: \[{\text{Distance}}\, = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \]

Complete step by step solution: Find distances between all three points for each set of points and check if the sum of any two distances equals the distance.
Set 1: Let A be (1, -1, 1), B be (-1, 1, 1) and C be (0, 0, 1)
$AB = \sqrt {{{(1 + 1)}^2} + {{( - 1 - 1)}^2} + {{(1 - 1)}^2}} = \sqrt 8 = 2\sqrt 2 $ units
$BC = \sqrt {{{( - 1 - 0)}^2} + {{(1 - 0)}^2} + {{(1 - 1)}^2}} = \sqrt 2 $ units
$AC = \sqrt {{{(1 - 0)}^2} + {{( - 1 - 0)}^2} + {{(1 - 1)}^2}} = \sqrt 2 $ units
Clearly $BC + AC = AB$. Since the sum of two distances is equal to the third distance, this set of points are collinear.
Set 2: Let L be (1, 2, 3), M be (3, 2, 1) and N (2, 2, 2)
$LM = \sqrt {{{(1 - 3)}^2} + {{(2 - 2)}^2} + {{(3 - 1)}^2}} = \sqrt 8 = 2\sqrt 2 $ units
$MN = \sqrt {{{(3 - 2)}^2} + {{(2 - 2)}^2} + {{(1 - 2)}^2}} = \sqrt 2 $ units
$LN = \sqrt {{{(1 - 2)}^2} + {{(2 - 2)}^2} + {{(3 - 2)}^2}} = \sqrt 2 $ units
Clearly $MN + LN = LM$. Since the sum of two distances is equal to the third distance, this set of points are collinear.
Set 3: Let P be (-2, 4, -3), Q be (4, -3, -2) and R be (-3, -2, 4)
$PQ = \sqrt {{{( - 2 - 4)}^2} + {{(4 + 3)}^2} + {{( - 3 + 2)}^2}} = \sqrt {86} $ units
$QR = \sqrt {{{( - 3 - 4)}^2} + {{( - 2 + 3)}^2} + {{(4 + 2)}^2}} = \sqrt {86} $ units
$PR = \sqrt {{{( - 2 + 3)}^2} + {{(4 + 2)}^2} + {{( - 3 - 4)}^2}} = \sqrt {86} $ units
All three distances are equal and therefore, they cannot be three points on a straight line. Therefore, this set of points is non-collinear.

Therefore, the correct option is option C. (-2, 4, -3), (4, -3, -2), (-3, -2, 4)

Note: Similarly, we can find the three distances for Set 4 as follows: Let X be (2, 0, -1), Y be (3, 2, -2) and Z be (5, 6, -4)
$XY = \sqrt {{{(2 - 3)}^2} + {{(0 - 2)}^2} + {{( - 1 + 2)}^2}} = \sqrt 6 $ units
$YZ = \sqrt {{{(5 - 3)}^2} + {{(6 - 2)}^2} + {{( - 4 + 2)}^2}} = \sqrt {24} = 2\sqrt 6 $ units
$XZ = \sqrt {{{(2 - 5)}^2} + {{(0 - 6)}^2} + {{( - 1 + 4)}^2}} = \sqrt {54} = 3\sqrt 6 $ units
Clearly $XY + YZ = XZ$. Since the sum of two distances is equal to the third distance, this set of points are collinear.
It is not necessary to calculate for the fourth set of points as we already know that the third set of points is non-collinear. However, to verify our answer we can check for the fourth set too.