- Task: Find the equation of the plane containing the intersection line of E1 and E2 and the point P
$$E1:\,\, 2x + y + 5z = 31$$
$$E2:\,\, -4x + 5y + 4z = 50\,\,\,,\,\,P( -5 \,,\, 2 \,,\, 3 )$$
- Question: Is my approach for the following problem valid? If yes, why doesn't it yield the right result?
1.: Combine the plane equations E1 and E2, canceling out a variable
--> $$(2 * E1) + E2 \longrightarrow 7y + 14z= 112 \Longrightarrow **y + 2z = 16**$$
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
$$ P1( x \,,\, 0 \,,\, 8 ) \longrightarrow E1: 2x + 0 + 40 = 31 \Longrightarrow **A( -4.5 \,,\, 0 \,,\, 8 )**\Longrightarrow $$ $$P2( x \,,\, 2 \,,\, 7 ) \longrightarrow E1:\,\, 2x + 2 + 35 = 31 \Longrightarrow **B( -3 \,,\, 2 \,,\, 7 )**$$
3.: Get two vectors containing the new plane: A to B and A to P
AB = B - A --> AB( 1.5 | 2 | -1 )
AP = P - A --> 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?
