# Thread: Finding a non-zero vector that...

1. ## Finding a non-zero vector that...

I need help to find a non-zero vector x in R^3 that belongs both to span {y; u} and to span {v;w} where y = (1; 0; 0); u = (0; 0; 1), v = (1; 1; 1) and w = (2; 3;-1).
Thank you.

2. I'll get you started:

In order for x to be a linear combination of y and u, it's second component should be zero. So x=(a,0,c) for some real numbers a and c. Now express x as a linear combination of v and w, and solve for a and c.

3. Let 'a' be the required vector.
you need to satisfy box[a,y,u]=0 and box[a,v,w]=0; where box[a,y,u] is the scalar triple product of a, y and u.
Since span of two given vectors is a plane, 'a' lies on the intersection of two planes hence 'a' is the vector along the line of intersection of the two planes.

4. Thank you both for your reply....just one short question abhishekkgp,because i need to clarify something.Why do I need to satisfy box[a,y,u]=0 and box[a,v,w]=0?because "a" is at the intersection of the two planes generated by the spans?

5. Originally Posted by AkilMAI
Thank you both for your reply....just one short question abhishekkgp,because i need to clarify something.Why do I need to satisfy box[a,y,u]=0 and box[a,v,w]=0?because "a" is at the intersection of the two planes generated by the spans?
Because 'a' lies in the plane spanned by y and u. when three vectors lie in a plane then their box product(or scalar triple product as you may call it) is zero. this is because the volume of the parallelogram formed by these three vectors is zero. now do you get it?

6. yes......[A,B,C]=det(ABC)...so det(a,y,u)=>y=0....and det(a,v,w)=-4x+y+z=0.....y=0,so x=1 and z=4