Line of intersection of Two Planes

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SUMMARY

The line of intersection of the planes defined by the equations 2x+y-2z=5 and 3x-6y-2z=15 can be determined using parametric equations. The correct parametric equations are x=7t+5, y=t, and z=(15t+5)/2. The initial approach using the cross product of the normal vectors was incorrect, leading to erroneous results. The correct method involves eliminating z by manipulating the equations directly.

PREREQUISITES
  • Understanding of parametric equations in three-dimensional space
  • Knowledge of vector operations, specifically cross products
  • Familiarity with solving simultaneous equations
  • Basic algebraic manipulation of linear equations
NEXT STEPS
  • Study the method of finding intersections of planes in three-dimensional geometry
  • Learn about vector calculus and its applications in geometry
  • Explore the use of software tools like CalcPlot3D for visualizing geometric concepts
  • Investigate the implications of normal vectors in determining plane orientations
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Students studying geometry, mathematics educators, and anyone interested in understanding the intersection of planes in three-dimensional space.

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Homework Statement

Please disregard, sign error corrected in the cross product

Determiner the line of intersection of the following two planes. Write the parametric equations for this line.

2x+y-2z=5
3x-6y-2z=15

Homework Equations

The Attempt at a Solution


First I crossed my normal vectors from the given equations: (2,1,-2)cross(3,-6,-2)= (-14,-2,15)

Then I solved my simultaneous equations starting with y=0, which gave me x=10 and z= (15/2).

To get into the parametric equation I took the point (10,0,15/2) + t(-14,-2,15), which gave me
x=10-14t
y=-2t
z=15t+(15/2)

Viewing this data in calcplot 3d shows that my line of intersection is no where near my planes.
 
Last edited:
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Error found in cross product (-14,-2,-15). Problem solved.
 
I'm not sure why you took the cross product of the normal vectors. Simply subtracting the first equation from the second eliminates z and gives x- 7y= 5.
That is, x= 7y+ 5 and putting that back into the first equation, 2x+y-2z= 2(7y+ 5)+ y- 2z= 15y+ 10- 2z= 5. 2z= 15y+ 5.

Parametric equations for the line of intersection are x= 7t+ 5, y= t, z= (15t+ 5)/2.
 

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