toandhsp

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12 years, 172 days

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Thank you very much. 

Thank you very much. 

Thank you very much.

Thank you very much.

Thank you very much. The code is based on advanced mathematics. want a code based on elementary mathematics. I solve the problem by hand following:

Put n = (a, b, c) is a normal vector of (P). Then, the equation of (P) has the form

a*(x + 1) + b*(y -3) + c(z + 6) = 0. 

Another way, vector n is perpendicular to vector AB, therfore b = 3a - 4c. We get

a*(x + 1) + (3a - 4c)*(y - 3) + c(z + 6) = 0.

The distance form C(1; -1; 7) (C is center of the given sphere) to the plane (P) equal to 3. 

abs(-10a + 29c)/sqrt(a^2 + (3a-4c)^2 +c^2) = 3, therefore a = 2c or a = 172c/5.

With a = 2c, we choose  c = 1, a = 2. The equation (P) is 2x + 2y +z + 2 = 0.

Similarly, we find a second plane.




Thank you very much. The code is based on advanced mathematics. want a code based on elementary mathematics. I solve the problem by hand following:

Put n = (a, b, c) is a normal vector of (P). Then, the equation of (P) has the form

a*(x + 1) + b*(y -3) + c(z + 6) = 0. 

Another way, vector n is perpendicular to vector AB, therfore b = 3a - 4c. We get

a*(x + 1) + (3a - 4c)*(y - 3) + c(z + 6) = 0.

The distance form C(1; -1; 7) (C is center of the given sphere) to the plane (P) equal to 3. 

abs(-10a + 29c)/sqrt(a^2 + (3a-4c)^2 +c^2) = 3, therefore a = 2c or a = 172c/5.

With a = 2c, we choose  c = 1, a = 2. The equation (P) is 2x + 2y +z + 2 = 0.

Similarly, we find a second plane.




Thank you very much.

Thank you very much.

It's great. Thank you very much.

It's great. Thank you very much.

I want a code so that the distance from A to (P) by k times the distance from B to (P).

Thank you very much.

Thank you very much.

Can you explain M:=x*A+y*B+z*C and x+y+z = 1. I don't understant. Thank you.

Can you explain M:=x*A+y*B+z*C and x+y+z = 1. I don't understant. Thank you.

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