Finding an equation for a plane containing 3 points

yecko · Sep 30, 2017

Homework Statement

Online homework prompt: Find an equation of a plane containing the three points (-4, 0, 2), (-5, -5, -2), (-5, -4, 0) in which the coefficient of x is -6, followed by an answer box and = 0.

Homework Equations

https://answers.yahoo.com/question/index?qid=20120204205204AAEvo8V

The Attempt at a Solution

Handwritten working on lined paper: points A(-4, 0, 2), B(-5, -5, -2), C(-5, -4, 0); AB = <-1, -5, -4>, AC = <-1, -4, -2>; cross product AB x AC set up as a 3x3 determinant giving 2i + 2j - 1k; plane 6x + 6y - 3z = D, with D = 6(-4) - 3(2) = -30, rearranged to -6x - 6y + 3z + 30 = 0 and -6x - 6y + 3z - 30 = 0.

This is my attempt following the solution in the yahoo answer, yet it is incorrect...
Can anyone tell me if I have made mistake in any of the manipulation?
Any help is appreciated~~
Thanks!

Orodruin · Sep 30, 2017

You should double check your algebra.

yecko · Sep 30, 2017

I just multiplied the whole equation by -3 and manipulated all the signs... after i checked several times, i still can't find what is wrong...

Orodruin · Sep 30, 2017

Your algebra. You are not adding and multiplying correctly.

Write out every step.

ehild · Sep 30, 2017

yecko said:

I just multiplied the whole equation by -3 and manipulated all the signs... after i checked several times, i still can't find what is wrong...

Check the determinant.

Mark44 · Oct 1, 2017

@yecko, when you calculate a cross product, it's a good idea to make sure that the vector you get is actually perpendicular to the two input vectors you're working with. This can be done very quickly, using the dot product. In the work you did, your calculated cross-product is <2, 2, -1>. This vector isn't perpendicular to either of the other two vectors, which means your calculation is incorrect.

Finding an equation for a plane containing 3 points

Homework Statement

Homework Equations

The Attempt at a Solution

Discussion

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