Finding equation of plane containing triangle

In summary, the normal vector in the equation for the plane containing the triangle BCD is making the later problems difficult to solve.
  • #1
Yitin
8
0
Homework Statement

B(-6,-7,-3) C(2,7,3) D(-4,-1,-11)
Find the equations of the plane containing the triangle BCD.

The attempt at a solution
I've gotten to the end and got an equation, but the normal vector it has is just making me really paranoid.

Vector BC = [2+6, 7+7, 3+3] = [8,14,6]
Vector BD = [-4+6,-1+7,-11+3] = [2,6,-8]

BC x BD = (-112 -36)i - (-64 -12)j + (-64 -28)k
Normal vector = [-148, 76, -92]


Equation:
-148(x+6) +76(y+7) -92(z+3) =0
-148x +76y +92z -632 = 0


The normal vector is what's bothering me. Some later problems require it, and it's making all of them look kinda gross as well, so I just wanted to make sure I didn't make any mistakes.
 
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  • #2
Yitin said:
Homework Statement

B(-6,-7,-3) C(2,7,3) D(-4,-1,-11)
Find the equations of the plane containing the triangle BCD.

The attempt at a solution
I've gotten to the end and got an equation, but the normal vector it has is just making me really paranoid.

Vector BC = [2+6, 7+7, 3+3] = [8,14,6]
Vector BD = [-4+6,-1+7,-11+3] = [2,6,-8]

BC x BD = (-112 -36)i - (-64 -12)j + (-64 -28)k
Normal vector = [-148, 76, -92]
For a quick check, take the dot product of your normal vector with BC and then with BD. If you get 0 for both, your normal is perpendicular to both.
Yitin said:
Equation:
-148(x+6) +76(y+7) -92(z+3) =0
-148x +76y +92z -632 = 0


The normal vector is what's bothering me. Some later problems require it, and it's making all of them look kinda gross as well, so I just wanted to make sure I didn't make any mistakes.
 
  • #3
Alright, so I did that and neither of them were 0.
Now I have no idea where I went wrong.
 
  • #4
I think you have a mistake in your cross product. When I dot BC with the normal I don't get 0.

BC = <8, 14, 6> = 2<4, 7, 3>
N = <-148, 76, -92> = 4<-37, 19, -23>

BC ## \cdot ## N = 2<4, 7, 3> ## \cdot ## 4<-37, 19, -23>
= 8 * <4, 7, 3> ## \cdot ## <-37, 19, -23>
= 8 * 4(-37) + 7(19) + 3(-23)
= 8 * (-148 + 133 - 69) ## \neq ## 0
 
  • #5
I think I found the problem. When I set up the cross product, I accidentally put int a -8 instead of a 6.

N should be <-148, 76, 20>
And N ⋅ BC does = 0, as well as N ⋅ BD

Thank you for your help Mark44.
 

1. What is the general equation for a plane?

The general equation for a plane is Ax + By + Cz + D = 0, where A, B, and C are the coefficients of the x, y, and z variables, respectively, and D is a constant term.

2. How can I find the equation of a plane containing a given triangle?

To find the equation of a plane containing a given triangle, you can use the cross product of two vectors that lie on the plane. These vectors can be found by subtracting the coordinates of two points on the triangle. Then, you can use the general equation for a plane and plug in the values of the coefficients and one of the points to find the equation.

3. How many points do I need to find the equation of a plane containing a triangle?

You need at least three non-collinear points to find the equation of a plane containing a triangle. These points can be any three points on the triangle, as long as they are not all on the same line.

4. Can I use any three points on the triangle to find the equation of the plane?

Yes, as long as the points are not all on the same line. If the points are collinear, then they do not define a unique plane, and you will not be able to find the equation.

5. Is there a faster way to find the equation of a plane containing a triangle?

Yes, if you know the normal vector of the plane, you can use the point-normal form of the equation, which is (x - x0) · n = 0, where x0 is a point on the plane and n is the normal vector. This method only requires one point and the normal vector instead of three points, making it faster and more efficient.

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