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  • Task: Find the equation of the plane containing the intersection line of E1 and E2 and the point P

$$E_1:\,\, 2x + y + 5z = 31$$

$$E_2:\,\, -4x + 5y + 4z = 50\,\,\,,\,\,P( -5 \,,\, 2 \,,\, 3 )$$

  • Question: Is my approach for the following problem valid? If yes, why doesn't it yield the right result?

1.: Combine the plane equations $E_1$ and $E_2$, canceling out a variable

--> $$(2 * E_1) + E_2 \longrightarrow 7y + 14z= 112 \Longrightarrow y + 2z = 16$$

2.: Use the new 2D line equation above to find the y and z values of two points on the intersection line, using one of the original plane equations to find the corresponding x value

$$ P_1( x \,,\, 0 \,,\, 8 ) \longrightarrow E_1: 2x + 0 + 40 = 31 \Longrightarrow A( -4.5 \,,\, 0 \,,\, 8 )\Longrightarrow $$ $$P_2( x \,,\, 2 \,,\, 7 ) \longrightarrow E_1:\,\, 2x + 2 + 35 = 31 \Longrightarrow B( -3 \,,\, 2 \,,\, 7 )$$

3.: Get two vectors containing the new plane: A to B and A to P

$AB = B - A \longrightarrow AB( 1.5 | 2 | -1 )$

$AP = P - A \longrightarrow AP( 9.5 | 1 | -10 )$

4.: Use the cross product of $AB$ and $AP$ to get the normal vector of the searched plane

My result: $n(19 | -5.5 | 17.5)$

Solution for the plane equation as given by teacher: $-2x + 2y + z = 17$

  • Where did I go wrong?
  • 1
    I found that P is not in the plane given in your teacher's solution.2012-04-21
  • 0
    @Siming: Damnit, I entered a wrong point P - fixed it now2012-04-21
  • 1
    Your work was then based off the incorrect $P$. Recalculate $P-A$ and the cross product. It will lead to the teacher's answer.2012-04-21
  • 0
    I'm a little worried about Step 1. $y+2z=16$ is not the equation of a line, it's the equation of a plane.2012-08-30

1 Answers 1