Question:

Consider a system of linear equation in x,y and z

a1x + b1y + c1z = 0,
a2x + b2y + c2z = 0,
a3x + b3y + c3z = 0;

and the vectors a' = (a1, b1, c1), b' = (a2, b2, c2), and c' = (a3, b3, c3). (we position all vectors so that their starting points are at the origin)

Prove that this system has a nontrivial solution ((x,y,z) ≠ (0,0,0)) if and only if the vectors a', b', and c' lie in the same plane (Hint: for the vector n = (x,y,z), we have 0 = a1x + b1y + c1z = a'.n, and similarly for b' and c'.)

Any answers or explanations would be much appreciated.

2. ## tratata

u r in trinity college???? vlad's course.... thisis freakin impossible!

3. Originally Posted by jackiemoon
Question:

Consider a system of linear equation in x,y and z

a1x + b1y + c1z = 0,
a2x + b2y + c2z = 0,
a3x + b3y + c3z = 0;

and the vectors a' = (a1, b1, c1), b' = (a2, b2, c2), and c' = (a3, b3, c3). (we position all vectors so that their starting points are at the origin)

Prove that this system has a nontrivial solution ((x,y,z) ≠ (0,0,0)) if and only if the vectors a', b', and c' lie in the same plane (Hint: for the vector n = (x,y,z), we have 0 = a1x + b1y + c1z = a'.n, and similarly for b' and c'.)

Any answers or explanations would be much appreciated.
Contrary to ooga-booga, it's pretty close to trivial. (x, y, z) lies in $R^3$, a 3 dimensional space. If all vectors a', b', and c', the "row vectors" for this matrix, lie in a plane, then there exist non-zero vectors perpendicular to that plane. The equations above are just the dot product of (x, y, z) with each of the vectors a', b', and c' and any of the vectors perpendicular to the plane will have 0 dot product with those.

4. Thanks Hallsofivy - it makes perfect sense now.

Ooga booga. Who is Vlad - I have never heard of this guy and certainly will not be attending his lecture this evening at 5-6pm in the Chemistry building. You must have me confused with someone else!

Originally Posted by ooga-booga
u r in trinity college???? vlad's course.... thisis freakin impossible!