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January 19

Scalar triple product and the equation of a plane

I was answering a question on a third semester calculus exam when I realized a connection between the scalar triple product and the equation for a plane. It may be trivial, but I haven't seen it explained in the textbook exposition of the distance formula between a point and a plane. Okay, I crossed two vectors u and v to find a normal vector to a plane parallel to the two vectors. Then to find an equation that defines the plane, given a point P on the plane, I essentially found the dot product of the position vector of the point and normal vector to the plane (P dot (u cross v))--I didn't realize that was I was I doing. Then I was asked to determine whether a second point Q lies on the plane. So I realized that I just had to do same thing (Q dot (u cross v)). But this is just the scalar triple product, which defines the volume of a parallelepiped. What I did seems like can be interpreted as: given two non-zero vectors, the position vector of any point on a plane parallel to the two vectors will produce the same volume when forming a parallelepiped with the two vectors.

This interpretation is not obvious to me and I hope someone can explain it to me. Thank you very much. 69.22.242.15 (talk) 21:17, 19 January 2017 (UTC)[reply]

The scalar triple product is a special case of a determinant. One of the properties of a determinant is that it is zero if, and only if, there is a linear relation among the rows. In the case of interest, u and v are two of the rows, which you've assumed are linearly independent (so they span a plane), a third vector x is in the plane spanned by u and v only if there is a linear relation among x, u, and v. This is the case only when det(x,u,v)=0 (i.e., their scalar triple product is zero). So det(x,u,v)=0 defines the equation of that plane. Sławomir Biały (talk) 21:46, 19 January 2017 (UTC)[reply]

But the scalar triple product, which I understand to be a 3x3 determinant, isn't 0 here (ax+by+cz=-d defines a plane, with <a,b,c> being a cross product). — Preceding unsigned comment added by 69.22.242.15 (talk) 00:26, 20 January 2017 (UTC)[reply]

The volume of the parallelepiped is just the base times the height (the distance from the plane). This height is just the length of the projection of the position vector to the unit normal vector, which is just their dot product. The dot product being zero is the same as the projection to the normal being zero is the same as the position vector lying on the plane is the same as the volume being zero. Hope this helps, too tired to write more clearly.John Z (talk) 11:05, 20 January 2017 (UTC)[reply]

I see it now, using the property P dot (u X v) = (P cross u) dot v. Very cool. 69.22.242.15 (talk) 14:03, 20 January 2017 (UTC)[reply]

I would take this as an easy example of Cavalieri's principle. --Tardis (talk) 03:13, 26 January 2017 (UTC)[reply]