How to Find the Normal Vector for a Plane Perpendicular to Another?

AI Thread Summary
To find the scalar equation of a plane perpendicular to another with a normal vector of [3,1,2], one must determine a normal vector for the new plane. The vector from point A(2,-6,-1) to point B(1,2,-4) is calculated as [-1, -4, -3], which lies in the new plane. The required normal vector for the new plane can be found using the cross product of the vectors [3,1,2] and [-1,-4,-3]. The cross product will yield a vector that is perpendicular to both, providing the necessary normal vector for the scalar equation. Understanding the properties of the cross product is essential for solving this problem.
eme_girl
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Find the scalar eq'n of a plane that is perpendicular to the plane with normal vector [3,1,2] and passes through points A(2,-6,-1) and B(1,2,-4).

I think that the normal vector can be the direction vector of this new plane. But then, in order to find the scalar eq'n I need a normal vector of this new plane. How do I find this?
 
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Strictly speaking a plane doesn't have a "direction vector". What is true is that the vector [3,1,2] is in the plane you want. You also know that the vector from A(2,6,-1) to B(1,2,-4) (which is, of course, [1-2,2-6,-4-(-1)]= [-1, -4, -3] is in the plane. Do you know how to find a vector that is perpendicular to both [3,1,2] and [1,4,-3]?
 
I understand what you just found. But no, I do not know how to find a vector's that perpendicular to both those vectors.
 
eme_girl,
do u know the direction of a vector that is a cross product of two vectors?

-- AI
 
TenaliRaman's point: the cross product of two vectors is always perpendicular to both.

The cross product of [a1,a2,a3] and [b1,b2,b3] is the vector [a2b3-a3b2,a3b1-a1b3,a1b2-a2b1].
 
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