Finding the Equation of a Plane in a 3D Space

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SUMMARY

The equation of the plane defined by the normal vector n = i - k and passing through the origin (0, 0, 0) is given by the formula A(x - x0) + B(y - y0) + C(z - z0) = 0. Substituting the normal vector components, the equation simplifies to x - z = 0. This confirms that the plane contains points such as (0, y, 0) and (x, 0, x), which lie along the edges of the square defined in the xz-plane and along the positive y-axis.

PREREQUISITES
  • Understanding of vector notation and operations
  • Familiarity with the equation of a plane in 3D space
  • Basic knowledge of geometry, specifically regarding squares and planes
  • Ability to interpret normal vectors and their significance in defining planes
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  • Study the derivation of the plane equation from normal vectors
  • Explore the geometric interpretation of planes in 3D space
  • Learn about the implications of normal vectors in determining plane orientation
  • Investigate applications of plane equations in computer graphics and modeling
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Students studying geometry, mathematics educators, and anyone interested in understanding the properties of planes in three-dimensional space.

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Homework Statement



S is the square of side 2 with one vertex at the origin, one edge along the positive y-axis, one edge in the xz-plane with x ≥ 0, z ≥ 0, and the normal is n = i - k .

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The Attempt at a Solution



Can someone help me how to find the equation of this plane?
 
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One vertex is the origin, (0, 0, 0), and you know the normal vector. What more do you need? The plane containing the point (x0, y0, z0), with normal vector <A, B, C>, has equation A(x- x0)+ B(y- y0)+ C(z- z0)= 0.
 
yes.. I tried (0,0,0) and I got x - z = 0 ? is this true?
 
Strictly speaking, a "square" is not a plane, it lies in a plane, so this is a strangely worded question!

Certainly, the origin, the corner (0,0,0) satisfies x- z= 0, the edge along the positive y- axis, (0, y, 0) satisfies x- z= 0, and the edge in the xz-plane, satisfying x- z= 0, (x, 0, x) is at right angles to (0, y, 0) (because <0, y, 0>.<x, 0, x>= 0). What more do you want?
 
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