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The equation of the plane through the points \(A = (0,1,1), B = (1,1,0)\) and \(C = (1,0,3)\) can be written as \(ax + by + cz = 8.\)
Then what's the ordered triple \((a, b, c)\)?

 Nov 8, 2019
 #1
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See here :  https://web2.0calc.com/questions/please-help_34243

 

The equation of the plane was found to be   :

 

1x + 3y +1z  - 4  = 0

 

So we can write

 

1x  + 3y  +  1z  =  4        multiplying this through by 2  we get that

 

2x + 6y + 2z  =  8

 

(a, b , c)   =  ( 2, 6 , 2)

 

 

cool cool cool

 Nov 8, 2019
edited by CPhill  Nov 8, 2019
 #2
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+1

Here's a way to do this without vectors:

 

plane equation: ax+by+cz=8

 

substituting x=0, y=1, and z=1 into the plane's equation:

a(0)+b(1)+c(1)=8

b+c=8

 

substituting the other two points in the same way:

a+b=8 ==> -a-b=-8

a+3c=8

 

add up these three equations:

b+c=8

-a-b=-8

a+3c=8

 

and you get:

(b+c)+(-a-b)+(a+3c)=(8)+(-8)+(8)

b+c-a-b+a+3c=8-8+8

(a-a)+(b-b)+(c+3c)=0+8

4c=8

c=2

 

b+c=8, and c=2, so b+2=8

b=6

 

a+b=8, and b=6, so a+6=8

a=2

 

therefore, the answer is (a, b, c) = (2, 6, 2)

 Nov 9, 2019

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