some help please:

i need to solve this problem in ten ways: prove that the following points in R^3 ,are collinear ,i.e,they are located on a straight line : A=(2,1,4), B=(1,-1,2), C =(3,3,6).

#HelpMe!

Easy Math Editor

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Comments

actually i don't agree with you Mr/ jatin yadav, because doctor gave us seven ways to solve it and i still need three ways to complete ten :( .

Aisha Adham - 7 years, 6 months ago

OK , Mr/ Calvin Lin i know 7 ways or 8, and they are : 1-Length's method 2-Cross product method 3-Box product 4-Slope method 5-Rank method 6-Buchhate formula 7-shoelace method (surerya's method) :) 8-reduce shoelace method .( but it is just abridgment to number (8)). and that is what i know ......

Aisha Adham - 7 years, 6 months ago

Here's another method. Show that $A$ is the midpoint of line segment $BC$ .

Calvin Lin Staff - 7 years, 6 months ago

What have you tried?

Since you want 10 ways, how many have you already come up with? If you list them out, others will be more likely to chime in with different approaches that they have used.

Calvin Lin Staff - 7 years, 6 months ago

I believe that there can not exist $10$ ways for this one, and i can think of only 1, i.e. equating the direction cosines of the line segments.

jatin yadav - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$