The direction vector of the line of intersection of the planes is perpindicular to both planes therefore perpindicular to the normals
of the 2 planes.
Also the cross product of the 2 normals is perpindicular to both planes.
Find parametric equations of the line of intersection of the two planes
-8 x - 2 y + 3 z = 9
and
5 x + 4 y + 2 z = -8
. Assign your direction vector, as a Maple list, to the name DirectionVector.
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This is a repeat question that i cant figure out. If you take the crossproduct o both vectors, that returns you the normal vector which is apraeel correct? The answer i got is [-16,31,-22], but the computer keeps telling me that its not parallel to a correct direction vector for the line. Happy labor day everyone.
[-16,31,-22] is the correct direction vector
If you still don't believe what I' saying there is Another way of seeing this verify:
(-10/11,-19/22,0) and (0,-21/8,5/4) are on the line of intersection
Compute the vector between these 2 points and you get
a mutiple of [-16,31,-22] namely 5/88 [-16,31,-22] again verifying
[-16,31,-22] is the correct direction vector.
Thnk about what I said: (don't rely always on what the computer says)
The direction vector of the line of intersection of the planes is perpindicular to both planes therefore perpindicular to the normals
of the 2 planes.
Also the cross product of the 2 normals is perpindicular to both planes.
This will be my last post on this matter