Answer to Question #342837 in Analytic Geometry for guy

Question

Three points with position vectors, b and c are said to be colinear. If the parallelogram with adjacent sides a - b and a - c has zero geometry area. Use this fact to check whether or not the following triples of points are collinear

(a) (2,2,3), (6,1,5) (2,4,3)

(b) (2,3,3), (3,7,5), (0,-5,-1)

(c) (1,3,2), (4,2,1), (1,0,2)

Answer 1

(a)

$$\vec{a}-\vec{b}=(-4, 1, -2)$$

$$\vec{a}-\vec{c}=(0, -2, 0)$$

$$(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -4 & 1 & -2 \\ 0 & -2 & 0 \end{vmatrix}$$

$$=2\begin{vmatrix} \vec{i} & \vec{k} \\ -4 & -2 \end{vmatrix}=-4\vec{i}+8\vec{k}$$

$$\big|(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})\big|=\sqrt{(-4)^2+(8)^2}=4\sqrt{5}\not=0$$

Three points are not collinear.

(b)

$$\vec{a}-\vec{b}=(-1, -4, -2)$$

$$\vec{a}-\vec{c}=(2, 8, 4)$$

$$(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & -4 & -2 \\ 2 & 8 & 4 \end{vmatrix}$$

$$=2\begin{vmatrix} \vec{j} & \vec{k} \\ -4 & -2 \end{vmatrix}-8\begin{vmatrix} \vec{i} & \vec{k} \\ -1 & -2 \end{vmatrix}+4\begin{vmatrix} \vec{i} & \vec{j} \\ -1 & -4 \end{vmatrix}$$

$$=-4\vec{j}+8\vec{k}+16\vec{i}-8\vec{k}-16\vec{i}+4\vec{j}=\vec{0}$$

$$\big|(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})\big|=\sqrt{(-4)^2+(8)^2}=0$$

Three points are collinear.

(c)

$$\vec{a}-\vec{b}=(-3, 1, 1)$$

$$\vec{a}-\vec{c}=(0, -3, 0)$$

$$(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -3 & 1 &1 \\ 0 & -3 & 0 \end{vmatrix}$$

$$=-3\begin{vmatrix} \vec{i} & \vec{k} \\ -3 & 1 \end{vmatrix}=-3\vec{i}-9\vec{k}$$

$$\big|(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})\big|=\sqrt{(-3)^2+(-9)^2}$$

$$=3\sqrt{10}\not=0$$

Three points are not collinear.