Finding an equation for a plane containing 3 points

In summary, the conversation is about a person struggling with an algebra problem and seeking help on Yahoo Answers. They have tried manipulating the equation, but it is not giving them the correct answer. Another person suggests checking their algebra and making sure they are adding and multiplying correctly. They also advise the person to write out every step and double check their calculations. The conversation ends with a recommendation to check the determinant and make sure the calculated cross-product is perpendicular to the input vectors.
  • #1
yecko
Gold Member
279
15

Homework Statement


螢幕快照 2017-09-30 下午3.37.01.png


Homework Equations


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

The Attempt at a Solution


771152855.jpg


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!
 
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  • #2
You should double check your algebra.
 
  • #3
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...
 
  • #4
Your algebra. You are not adding and multiplying correctly.

Write out every step.
 
  • #5
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.
 
  • #6
@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.
 

Related to Finding an equation for a plane containing 3 points

What is the equation for a plane containing 3 points?

The equation for a plane containing 3 points is determined using the point-normal form, which is:
(x - x1)a + (y - y1)b + (z - z1)c = 0
where (x1, y1, z1) is one of the given points and (a, b, c) is the normal vector of the plane.

How do I find the normal vector of a plane from 3 points?

To find the normal vector of a plane from 3 points, you can use the cross product of two vectors formed by the given points. The resulting vector will be perpendicular to the plane and can be used as the normal vector in the point-normal form of the plane equation.

Can I use any 3 points to find the equation of a plane?

No, the 3 points must be non-collinear, meaning they cannot all lie on the same line. If the points are collinear, there are an infinite number of planes that could contain them, so an equation cannot be determined.

Is it possible to find an equation for a plane if I only have 2 points?

No, at least 3 non-collinear points are required to uniquely determine a plane. With only 2 points, there are an infinite number of planes that could contain them, so an equation cannot be determined.

Can I use the distance formula to find an equation for a plane containing 3 points?

No, the distance formula is used to calculate the distance between two points, but it cannot be used to find an equation for a plane. The point-normal form or the standard form of the plane equation should be used instead.

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