1. ## Normal Vector Help

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.

2. 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.

3. Originally Posted by Calculus26
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.

Im still confused on how to use that information to find a direction vector that is parallel to the correct direction vector for the line. here is what my computer is telling me:

"Your Direction vector is not parallel to a correct Direction vector for the line."

4. [-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