Equation a plane through 3 points.

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SUMMARY

The discussion focuses on finding the equation of a plane through three points: A=(1,2,3), B=(0,1,3), and C=(2,14). The user initially miscalculated the normal vector, obtaining (1,1,3) instead of the correct tangent vector (1,1,0). The solution involves calculating two tangent vectors from the given points and then using the cross product to derive the normal vector, which is essential for formulating the plane's equation.

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  • Understanding of vector equations of a plane
  • Knowledge of calculating cross products
  • Familiarity with the concept of normal vectors
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  • Study the derivation of the equation of a plane from a normal vector
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Homework Statement


Find an equation of the plane through the points: A=(1,2,3), B=(0,1,3) and C=(2,14)


Homework Equations


r= r[0]+tv
vector equation of a plane: a(x-x[/0)+ b(y-y[/0]) + c(z-z[/0])=0


The Attempt at a Solution


Since I am given three points and no vector, I am a bit confused on what to do. I tried to compute the normal vector, and got:
(1,2,3)-(0,1,3)
n=<1,1,3>

Any corrections/suggestions on what to do next?
 
Physics news on Phys.org
(1,2,3)-(0,1,3) is (1,1,0) not (1,1,3). And that's a tangent vector, not a normal vector. Hint: you can find a normal vector if you can find two different tangent vectors and take the cross product.
 

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