מצאו את נקודת החיתוך של הישר הנורמלי למשטח 

2⋅z3+2⋅x⋅y⋅z+4⋅y3+4⋅x3=1422⋅𝑧3+2⋅𝑥⋅𝑦⋅𝑧+4⋅𝑦3+4⋅𝑥3=142

בנקודה 

P(2,2,3)𝑃(2,2,3)

עם המישור 

3⋅z+2⋅y+3⋅x+2=0


To find the intersection point of the normal line to the surface \(2z^3 + 2xyz + 4y^3 + 4x^3 = 142\) at the point \(P(2,2,3)\) with the plane \(3z + 2y + 3x + 2 = 0\), follow these steps:

1. **Calculate the gradient of the surface at \(P(2,2,3)\):**

   The gradient vector \(\nabla f(x,y,z)\) of the surface \(f(x,y,z) = 2z^3 + 2xyz + 4y^3 + 4x^3\) is given by:
   \[
   \nabla f(x,y,z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)
   \]

   Calculate the partial derivatives:
   \[
   \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (2z^3 + 2xyz + 4y^3 + 4x^3) = 2yz + 12x^2
   \]
   \[
   \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (2z^3 + 2xyz + 4y^3 + 4x^3) = 2xz + 12y^2
   \]
   \[
   \frac{\partial f}{\partial z} = \frac{\partial}{\partial z} (2z^3 + 2xyz + 4y^3 + 4x^3) = 6z^2 + 2xy
   \]

   Evaluate these at \(P(2,2,3)\):
   \[
   \left. \frac{\partial f}{\partial x} \right|_{(2,2,3)} = 2(2)(3) + 12(2)^2 = 12 + 48 = 60
   \]
   \[
   \left. \frac{\partial f}{\partial y} \right|_{(2,2,3)} = 2(2)(3) + 12(2)^2 = 12 + 48 = 60
   \]
   \[
   \left. \frac{\partial f}{\partial z} \right|_{(2,2,3)} = 6(3)^2 + 2(2)(2) = 54 + 8 = 62
   \]

   Thus, the gradient vector at \(P(2,2,3)\) is:
   \[
   \nabla f(2,2,3) = (60, 60, 62)
   \]

2. **Determine the equation of the normal line at \(P(2,2,3)\):**

   The normal line can be written in parametric form as:
   \[
   (x, y, z) = (2, 2, 3) + t(60, 60, 62)
   \]
   This gives:
   \[
   x = 2 + 60t, \quad y = 2 + 60t, \quad z = 3 + 62t
   \]

3. **Find the intersection with the plane \(3z + 2y + 3x + 2 = 0\):**

   Substitute the parametric equations of the line into the plane equation:
   \[
   3(3 + 62t) + 2(2 + 60t) + 3(2 + 60t) + 2 = 0
   \]
   Simplify:
   \[
   3(3 + 62t) + 2(2 + 60t) + 3(2 + 60t) + 2 = 0
   \]
   \[
   9 + 186t + 4 + 120t + 6 + 180t + 2 = 0
   \]
   Combine like terms:
   \[
   21 + 486t = 0
   \]
   Solve for \(t\):
   \[
   486t = -21
   \]
   \[
   t = -\frac{21}{486} = -\frac{7}{162}
   \]

4. **Calculate the intersection point:**

   Substitute \(t = -\frac{7}{162}\) back into the parametric equations:
   \[
   x = 2 + 60\left(-\frac{7}{162}\right) = 2 - \frac{420}{162} = 2 - \frac{210}{81} = 2 - \frac{70}{27}
   \]
   Simplify \(x\):
   \[
   x = 2 - \frac{70}{27} = \frac{54}{27} - \frac{70}{27} = \frac{-16}{27}
   \]
   Similarly, for \(y\):
   \[
   y = 2 - \frac{420}{162} = 2 - \frac{210}{81} = 2 - \frac{70}{27} = \frac{-16}{27}
   \]
   And for \(z\):
   \[
   z = 3 + 62\left(-\frac{7}{162}\right) = 3 - \frac{434}{81} = \frac{243}{81} - \frac{434}{81} = \frac{-191}{81}
   \]

So, the intersection point is:
\[
\left( \frac{-16}{27}, \frac{-16}{27}, \frac{-191}{81} \right)
\]

Would you like more details or have any questions? Here are five questions you might find useful:

1. How to find the gradient vector of a multivariable function?
2. What are the steps to write the parametric form of a normal line to a surface?
3. How to solve for the intersection of a line with a plane?
4. What is the significance of the normal vector in geometry?
5. Can you explain the method of parametric equations for lines in 3D?

**Tip:** When solving problems involving normals and intersections, always verify your final parametric equations to ensure accuracy in the results.

Learn how to find the intersection point of the normal line to a surface at a given point with a plane. Detailed step-by-step solution involving gradient calculation, parametric equations of the normal line, and solving for intersection with a plane.

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