Finding Orthogonal Projection of a Curve in the yz-Plane

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SUMMARY

The discussion focuses on finding the orthogonal projection of a curve defined by the intersection of two surfaces: x = y² + z² - 1 and x - 2y + 4z = 0. The key insight is that to project onto the yz-plane, one simply sets x to 0, resulting in the curve's representation in the form (0, y, z). The participants clarify that once the yz-curve is determined by eliminating x, no additional projection formula is necessary.

PREREQUISITES
  • Understanding of scalar equations and curves in three-dimensional space.
  • Familiarity with surface equations and their intersections.
  • Knowledge of vector projections and dot product calculations.
  • Basic comprehension of coordinate planes, specifically the yz-plane.
NEXT STEPS
  • Study the method of finding intersections of surfaces in three-dimensional geometry.
  • Learn about vector projections and their applications in various coordinate planes.
  • Explore the implications of setting coordinates to zero in geometric projections.
  • Investigate the use of parametric equations to represent curves in three-dimensional space.
USEFUL FOR

Mathematicians, physics students, and anyone involved in three-dimensional geometry or vector calculus will benefit from this discussion.

hungryhippo
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How would you approach a question where you're given a curve in terms of a scalar equation, and asked to find the orthogonal projection of this curve in the yz-plane

You know that the curve is the intersection of the surfaces of:

x=y^2+z^2 --1
x-2y+4z=0 --2

From here, I would just substitute equation 1 into 2 for x, to find the resulting curve

I know that a yz-plane indicates that the x-coordinate will always be 0 , so (0,y,z)
For scalar projections, you can find it as just

(a) dot (b) / (length of a)

I'm not sure if what I'm thinking so far is correct, and extremely unsure on the projection part.

I really need help on this :eek::confused:

Thanks in advance for any advice
 
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You are probably just over thinking this whole thing. Once you've found a yz curve by eliminating x, you are done, right? Projection into the yz plane just means ignore the x value. You don't need a projection formula for this case.
 

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