Intersection of line through a plane

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SUMMARY

The intersection of the line through the points (−2, 3, 2) and (3, 5, −1) with the plane defined by the equation x + y − 2z = 6 occurs at the point (19/13, 57/13, -1/13). The parametric equations for the line are derived using the vector r = r0 + tv, where the vector between the two points is <5, 2, -3>. The value of t that satisfies the intersection is 9/13, confirmed by substituting back into the plane equation.

PREREQUISITES
  • Understanding of parametric equations in three-dimensional space
  • Knowledge of vector representation and operations
  • Familiarity with plane equations in the form Ax + By + Cz = D
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study the derivation of parametric equations from two points in 3D space
  • Learn how to verify points against plane equations
  • Explore vector algebra and its applications in geometry
  • Investigate the concept of line-plane intersections in three-dimensional geometry
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Students studying geometry, particularly those focusing on three-dimensional space, as well as educators looking for examples of line-plane intersections in mathematical discussions.

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Homework Statement


Where does the line through (−2, 3, 2) and (3, 5, −1) intersect the plane x + y − 2z = 6?

Homework Equations

The Attempt at a Solution


i used r = r0 + tv

the vector between the 2 given points is <5,2,-3>

r = (-2,3,2) + t<5,2-3>

x = -2 + 5t y=3+2t z=2 - 3t

plugging these into the plane given: -2+5t + 3+2t -2(2-3t)= 6

solving for t , t should equal 9/13 if my algebra is correct.

plugging into the parametric equations , the point of intersection should be (19/13 , 57/13, -1/13)

can someone check all this?
 
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goonking said:

Homework Statement


Where does the line through (−2, 3, 2) and (3, 5, −1) intersect the plane x + y − 2z = 6?

Homework Equations

The Attempt at a Solution


i used r = r0 + tv

the vector between the 2 given points is <5,2,-3>

r = (-2,3,2) + t<5,2-3>

x = -2 + 5t y=3+2t z=2 - 3t

plugging these into the plane given: -2+5t + 3+2t -2(2-3t)= 6

solving for t , t should equal 9/13 if my algebra is correct.

plugging into the parametric equations , the point of intersection should be (19/13 , 57/13, -1/13)

can someone check all this?

It is correct. Nice work.:oldcool:
 
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goonking said:

Homework Statement


Where does the line through (−2, 3, 2) and (3, 5, −1) intersect the plane x + y − 2z = 6?

Homework Equations

The Attempt at a Solution


i used r = r0 + tv

the vector between the 2 given points is <5,2,-3>

r = (-2,3,2) + t<5,2-3>

x = -2 + 5t y=3+2t z=2 - 3t

plugging these into the plane given: -2+5t + 3+2t -2(2-3t)= 6

solving for t , t should equal 9/13 if my algebra is correct.

plugging into the parametric equations , the point of intersection should be (19/13 , 57/13, -1/13)

can someone check all this?
It's not that hard to check it for yourself, and it's a good habit to get into. You can do this by verifying that the point you found is on both the line and on the plane.

To verify that the point is on the line, find the value of t so that 19/13 = -2 + 5t, 57/13 = 3 + 2t, and -1/13 = 2 - 3t. The same value of t should work in all three equations.

To verify that the same point is on the plane, confirm that 19/13 + 57/13 - 2(-1/13) = 6.
 
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