Question

Find a homogeneous linear system of two equations in three unknowns whose solution space consists of those vectors in R^3 that are orthogonal to a=<-3, 2, -1> and b=<0,-2,-2>. What kind of geometric object is the solution space

Answer 1

Well first off lets look at the two vectors we are supposed to be orthogonal to..

Answer 2

Not entirely sure what you are driving at here.

You want a vector or vectors orthogonal to a pair of vectors, the cross? In both directions?

Answer 3

Actually there's a couple of clues given here.. Which would you prefer to take, an algebraic approach or a geometric one?

Answer 4

If you are asking me, I prefer a combo..

Answer 5

Well since the original student left I can explain to you. ;)

Answer 6

His vectors are sitting in some plane, so orthogonal is the cross, up or down according to left or right handed system.

Or have I got it all wrong?

Answer 7

Oh, I see, he wants the entire subspace...

Answer 8

Since we know that the solution is orthogonal to those two vectors, we know that the dot product of any point x,y,z in our solution and each of the vectors must be 0.

So:
\[<-3,2,-1>\cdot <x,y,z> = 0\]and\[<0,-2,-2>\cdot <x,y,z> = 0\]

Now thinking back to matrix multiplication I can write this system as a matrix:
\[\left[ \begin{array}{c c c} -3\ \ \ 2 \ \ \ -1 \\0\ -2\ \ \ -2\end{array} \right]X = 0\]

Answer 9

Then for the 3rd row I just have to use a row of 0's or a row which is a linear combination of the other two, and I think that should be the system it's looking for.

Answer 10

If I just put i,j,k as the top row, that's the cross a x b, with value equal to the vector normal to a and b.

Your method is OK but you can just write the cross directly.

The geometric object is a 1-dim subspace of the 8-dim R3.

Answer 11

I prefer to think that my way is the reason the cross product works ;)

Answer 12

Ah, but it doesn't work, well not really, only in 3D.

Anyway the value you get is the area of the ab (implicit) parallelogram.

So I prefer a wedge b which is an already oriented area (with the same absolute value as the cross ( and it works in any number of dimensions).

not to say the cross doesn't have it's uses though...