# determin arrangement of points in R3

• Jun 6th 2008, 12:40 AM
Craka
determin arrangement of points in R3
Given the points A(5,1,3), B(7,9,-1) , C(1, -15, 11) how do I determine if they are or not on a straight line?
• Jun 6th 2008, 12:46 AM
Isomorphism
Quote:

Originally Posted by Craka
Given the points A(5,1,3), B(7,9,-1) , C(1, -15, 11) how do I determine if they are or not on a straight line?

I know a couple of ways to show they are on the same line...

1) Show the area of triangle formed by these points is 0.
2) Form an equation of a line with two points and show the third satisfies it.
• Jun 6th 2008, 11:53 AM
Soroban
Hello, Craka!

Another method . . .

Quote:

Given the points: . $A(5,1,3),\;B(7,9,\text{-}1),\;C(1, \text{-}15, 11)$
how do I determine if they are or not on a straight line?

Find vector $\overrightarrow{AB}\!:\;\;\overrightarrow{AB} \:=\:\langle 7-5,\:9-1,\:\text{-}1-3\rangle \:=\:\langle 2,\:8,\:\text{-}4\rangle \;=\;2\langle 1,\:4,\:\text{-}2\rangle$

Find vector $\overrightarrow{BC}\!:\;\;\overrightarrow{BC} \:=\:\langle 1-7,\:\text{-}15-9,\:11-(\text{-}1)\rangle \:=\:\langle \text{-}6,\:\text{-}24,\:12\rangle \:=\: -6\langle1,\:4,\:\text{-}2\rangle
$

Since $\overrightarrow{AB} \parallel \overrightarrow{BC},\;\;A,B,C \text{ are collinear.}$