Finding equation of plane containing triangle

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    Plane Triangle
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Homework Help Overview

The discussion revolves around finding the equation of a plane containing the triangle formed by the points B(-6,-7,-3), C(2,7,3), and D(-4,-1,-11). Participants are examining the calculation of the normal vector derived from the cross product of vectors BC and BD.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of vectors BC and BD, and the subsequent cross product to find the normal vector. There is concern about the accuracy of the normal vector, particularly its implications for later problems. One participant checks the dot product of the normal vector with the original vectors to verify orthogonality.

Discussion Status

The discussion has revealed discrepancies in the calculations of the normal vector, with one participant identifying a potential error in the setup of the cross product. There is ongoing verification of the normal vector's correctness, with some participants providing checks to confirm its properties.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. The discussion reflects uncertainty regarding the calculations involved in determining the normal vector and its significance in subsequent problems.

Yitin
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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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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.
 
Alright, so I did that and neither of them were 0.
Now I have no idea where I went wrong.
 
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
 
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.
 

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