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iomtt6076
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Homework Statement
Find the equation of the plane through the points on the x, y, and z-axes where x=a, y=b, z=c, respectively.
Homework Equations
If <A,B,C> is a vector perpendicular to the plane, then A(x-a)+B(y-b)+C(z-c)=0 is the equation of the plane.
The Attempt at a Solution
I know this is extremely simple, but I can't get the right answer even using different methods.
Method 1:
The vectors <0,b,-c> and <a,-b,0> are in the plane. The cross product <0,b,-c> X <a,-b,0> gives <-bc,-ac,-ab>, which is normal to the plane. Then the equation of the plane is (-bc)(x-a)+(-ac)(y-b)+(-ab)(z-c)=0, and simplifying I get (x/a)+(y/b)+(z/c)=3. Wrong.
Method 2:
Substituting the 3 points (a,0,0), (0,b,0), (0,0,c) into A(x-a)+B(y-b)+C(z-c)=0 gives
-Bb-Cc=0
-Aa-Cc=0
-Aa-Bb=0.
Thus C=-Aa/c, B=-Aa/b and substituting this back into A(x-a)+B(y-b)+C(z-c)=0 and simplifying I get (-x/a)+(y/b)+(z/c)=1, which is different than the answer from Method 1 and still wrong.