Question

The vectors that are perpendicular to (1,1,1) and (1,2,3) lie in a........

Answer 1

I don't understand this question, can someone help me?

Answer 2

You are given two vectors in three dimensional space. By inspection neither is a multiple of the other ie. they are not co-linear. Thus all linear combinations of these two vectors will define a plane in 3D. So what is the name of the class/variety of geometric object, in three dimensions, that is perpendicular to a plane?

Answer 3

It's another plane???

Answer 4

It's a line. If you are familiar with analytic geometry that line is defined as a set of points referenced by all multiples of a vector which is the cross product of any two non co-linear vectors in that plane ( with a possible fixed offset vector from the origin, depending upon whether you consider our vectors to be 'based' or 'free', and other issues ).

In any event the somewhat deeper point is that one can have an entire set of vectors that is orthogonal to another entire set of vectors. Meaning that if you calculate the inner product of any vector from one of the sets with any vector from the other set you will get zero ( the number zero ).

When ( or if ) you solve Ax = 0, you are finding all the vectors x which have an inner product of zero with each row of A. That is because each element of the right hand side zero vector is the inner product of some row of A with x. So the row space of A ( all linear combinations of the rows of A ) is orthogonal to the set of vectors x. Now x = 0 will always give Ax = 0 for any A, but we are really after non-zero vectors that have that quality. ( Mind you if, for a given A, we find that x = 0 is the only solution for Ax = 0 then that is of especial interest. )

In generality the matrix A ( size m x n ) can be viewed as a transform of vectors from a space of dimension n to another space of dimension m. When we talk of a subspace A being orthogonal to a subspace B then both A and B are subspaces of the single same vector space. Prof Strang's lecture on the orthogonality of the subspaces arising in linear algebra is well worth viewing a number of times, and don't be disheartened it takes a few runs at it to get the gist.

Answer 5

Wow.... It's a complex idea, but no impossible to think about. Understand the gist of matrices and their relationchip with vectors (and vectorial spaces) is fundamental... Thank's for your attention hewsmike... Hugs from Brazil!!!!!

Answer 6

A doubt: cross product of non-colinear vector is the same that dot product?

Answer 7

No, not at all. A cross product is an operation on two vectors that produces a third vector, usually denoted C = A x B. This third vector is perpendicular to the first two. The operation of cross product only makes sense in 3D space ( not higher or lower dimensions ), and C is only a non-zero vector if neither A nor B is zero and they are independent ( one is not a multiple of the other ). Nor does cross product commute, as there is a right hand sense to the result ( ie. if you take the standard 3D Cartesian axis set then k = i x j ). Thus A x B = - (B x A), meaning that the vector A x B points in opposite sense to the vector B x A, while they both lie along the same line.

In any case this cross product operation produces a 'normal' vector ( which is orthogonal to all vectors in plane ) if it is generated by taking the basis vectors for the plane - in your example these could be (1,1,1) and (1,2,3).

There's a whole other area of properties for cross products, in combination with taking dot products too, but that's the gist. For this problem we simply are producing a normal vector to a plane.

Answer 8

That's it!!!! Thank's again hewsmike! You're fantastic!!

Hugs from Brazil!!