- 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?