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

The correct option is C

$(–2,4,–3),(4,–3,–2),(–3,–2,4)$

Explanation for correct option:

Check whether the points $(–2,4,–3),(4,–3,–2),(–3,–2,4)$ are colinear:

Given $(–2,4,–3),(4,–3,–2),(–3,–2,4)$

The distance between two points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ is $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$.

Let $M=(–2,4,–3),N=(4,–3,–2),O=(–3,–2,4)$

$$\begin{gathered} MN=\sqrt{{(-2-4)}^2+{(4+3)}^2+{(-3-2)}^2} \\ =\sqrt{{(-6)}^2+7^2+{(-5)}^2} \\ =\sqrt{36+49+25} \\ =\sqrt{110}\text{unit} \end{gathered}$$

$$\begin{gathered} NO=\sqrt{{(4+3)}^2+{(-3+2)}^2+{(-2-4)}^2} \\ =\sqrt{7^2+{(-1)}^2+{(-6)}^2} \\ =\sqrt{49+1+36} \\ =\sqrt{86}\text{unit} \end{gathered}$$

and

$$\begin{gathered} MO=\sqrt{{(-2-3)}^2+{(4+2)}^2+{(-3-4)}^2} \\ =\sqrt{{(-5)}^2+6^2+{(-7)}^2} \\ =\sqrt{25+36+49} \\ =\sqrt{110}\text{unit} \end{gathered}$$

If three points $A,B,C$ are colinear then $AB+BC=AC$

For this point, we get $MN+NO\ne MO$ and $MO+MN\ne NO$ and $MO+NO\ne MN$

Therefore, these points are not colinear.

Hence, the correct option is option (c).

Explanation for incorrect options:

For option (a)

Check whether the points $(1,–1,1),(–1,1,1),(0,0,1)$ are colinear:

The distance between two points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ is $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$.

Given $(1,–1,1),(–1,1,1),(0,0,1)$

Let $A=(1,–1,1),B=(–1,1,1),C=(0,0,1)$

$$\begin{gathered} AB=\sqrt{{(-1-1)}^2+{(1+1)}^2+{(1-1)}^2} \\ =\sqrt{{(-2)}^2+2^2} \\ =\sqrt{4+4} \\ =\sqrt{8} \\ =2\sqrt{2}\text{unit} \end{gathered}$$

$$\begin{gathered} BC=\sqrt{{(0+1)}^2+{(0-1)}^2+{(1-1)}^2} \\ =\sqrt{1+1} \\ =\sqrt{2}\text{unit} \end{gathered}$$

and

$$\begin{gathered} AC=\sqrt{{(1-0)}^2+{(-1-0)}^2+{(1-1)}^2} \\ =\sqrt{1+1} \\ =\sqrt{2}\text{unit} \end{gathered}$$

If three points $A,B,C$ are colinear then $AB+BC=AC$

For this point we get $AC+BC=AB$

Therefore, these points are colinear.

For option (b)

Check whether the points $(1,2,3),(3,2,1),(2,2,2)$ are colinear:

Given $(1,2,3),(3,2,1),(2,2,2)$

The distance between two points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ is $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$.

Let $G=(1,2,3),H=(3,2,1),I=(2,2,2)$

$$\begin{gathered} GH=\sqrt{{(1-3)}^2+{(2-2)}^2+{(3-1)}^2} \\ =\sqrt{{(-2)}^2+0+2^2} \\ =\sqrt{4+4} \\ =\sqrt{8} \\ =2\sqrt{2}\text{unit} \end{gathered}$$

$$\begin{gathered} HI=\sqrt{{(3-2)}^2+{(2-2)}^2+{(1-2)}^2} \\ =\sqrt{1+1} \\ =\sqrt{2}\text{unit} \end{gathered}$$

and

$$\begin{gathered} GI=\sqrt{{(1-2)}^2+{(2-2)}^2+{(3-2)}^2} \\ =\sqrt{1+1} \\ =\sqrt{2}\text{unit} \end{gathered}$$

If three points $A,B,C$ are colinear then $AB+BC=AC$

For this point we get $HI+GI=GH$

Therefore, these points are colinear.

For option (D)

Check whether the points $(2,0,–1),(3,2,–2),(5,6,–4)$ are colinear:

Given $(2,0,–1),(3,2,–2),(5,6,–4)$

The distance between two points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ is $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$.

Let $X=(2,0,–1),Y=(3,2,–2),Z=(5,6,–4)$

$$\begin{gathered} XY=\sqrt{{(2-3)}^2+{(0-2)}^2+{(-1+2)}^2} \\ =\sqrt{{(-1)}^2+{(-2)}^2+{\left(1\right)}^2} \\ =\sqrt{1+4+1} \\ =\sqrt{6}\text{unit} \end{gathered}$$

$$\begin{gathered} YZ=\sqrt{{(3-5)}^2+{(2-6)}^2+{(-2+4)}^2} \\ =\sqrt{{(-2)}^2+{(-4)}^2+{\left(2\right)}^2} \\ =\sqrt{4+16+4} \\ =\sqrt{24} \\ =2\sqrt{6}\text{unit} \end{gathered}$$

and

$$\begin{gathered} XZ=\sqrt{{(2-5)}^2+{(0-6)}^2+{(-1+4)}^2} \\ =\sqrt{{(-3)}^2+{(-6)}^2+3^2} \\ =\sqrt{9+36+9} \\ =3\sqrt{6}\text{unit} \end{gathered}$$

If three points $A,B,C$ are colinear then $AB+BC=AC$

For this point, we get $XY+YZ=XZ$

Therefore, these points are colinear.

Hence, the correct option is option (c).