How can I find two perpendicular vectors in R^3 and a value of c?

[Noxide]

my brain is not working i need some major help


Find two nonzero vectors v,w in R^3 and a value of c so that each of v and w is perpendicular to all three vectors (wtf.):

[ 2-c ]
[ 1 ]
[ 0 ]

and...

[ -1 ]
[ -1 ]
[ 1 ]

AND THEN !

[ 1 ]
[ 2 ]
[ 1-c]



/cry

 

[CompuChip]

The only way that you can find a three-dimensional vector which is perpendicular to three other vectors, is if the latter three vectors are not all linearly independent. In other words, at least one of them must be expressable as a linear combination of the other two.

Also, the question doesn't state that v and w must be different, so once you find a vector v, you can take w = r v for any number r :P

 

[HallsofIvy]

So your vectors are <2- c, 1, 0>, <-1, -1, 1>, and <1, 2, 1- c>? And you want to find two vectors perpendicular to all 3?

Okay, first, there does NOT exist a vector, in R³ that is perpendicular to three independent vectors so one of the given vectors must be a linear combination of the other two. Also the "cross product" of two vectors is perpendicular to both so I would start by finding the cross product of two of these.

Take the cross product of the first two vectors to get a vector that is perpendicular to the first two. Then take the dot product of that with the third vector and set it equal to 0. Solve that equation for c. That will give one vector, v, perpendicular to all three and the value of c. Since there are no other conditions on w, just take any multiple of v for w.

 

[Noxide]

WTFPWNED!

Thanks guys. I had no idea there was something called a cross product until now, but it's loleasy!