Intersection of line and plane

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SUMMARY

The discussion focuses on finding the equation of a plane that passes through the line defined by the parametric equations \(\frac{x-2}{3}=\frac{y-3}{1}=\frac{z+1}{2}\) and is normal to the plane described by the equation \(x + 4y - 3z + 7 = 0\). The intersection point of the line and the plane is established as M(2,3,-1). The conditions for the coefficients of the plane's equation are derived, leading to the equations \(2A + 3B - C + D = 0\) and \(A + 4B - 3C = 0\). The discussion seeks clarification on the third condition necessary for determining the plane's equation.

PREREQUISITES
  • Understanding of parametric equations of a line
  • Knowledge of plane equations in three-dimensional space
  • Familiarity with vector normality conditions
  • Basic linear algebra concepts for solving systems of equations
NEXT STEPS
  • Study the derivation of plane equations from line equations
  • Learn about vector normality and its application in geometry
  • Explore methods for solving systems of linear equations
  • Investigate the geometric interpretation of intersection points in 3D space
USEFUL FOR

Students studying geometry, particularly in three-dimensional space, as well as educators and tutors assisting with plane and line intersection problems in mathematics.

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Homework Statement



Find the equation of the plane which passes through the line [itex]\frac{x-2}{3}=\frac{y-3}{1}=\frac{z+1}{2}[/itex]
and it is normal to the plane x+4yy-3z+7=0

Homework Equations





The Attempt at a Solution



I find the intersection point of the plane and the line. It is M(2,3,-1). Also I got the condition (A,B,C)(3,1,2)=0

[tex]\left\{\begin{matrix}<br /> 2A+3B-C+D=0 & \\ <br /> A+4B-3C=0 & <br /> \end{matrix}\right.[/tex]

What is the third condition ?
 
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Sorry I must fix something. Up in there is x+4y-3z+7=0, and (A,B,C)(1,4,-3)=0.
 
Is (A,B,C)(3,1,2)=0 the third condition?
 

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