How to find two orthogonal vectors that are both orthogonal to a third vector?

For example,


"Find two orthogonal vectors, v and w,


which are both orthogonal to a vector x = (1, 1, 1)." --


so, v and w are orthogonal, and both of these vectors are orthogonal to x








Thanks.|||there is not a unic answer ,





first find a vecor ortogonal to first one


set one component zero, ( 1,1,0 )


change the other 2 component ( 1,1,0)


set one of them negative ( -1 ,1,0 )


it is an ortogonal vector to first one.





use them to find 3th vector


1 1 1


-1 1 0


e1 e2 e3


= e3 -e2 -e1 +e3 = -e1 -e2 +2e3


3th vector = ( -1, -1, 2)|||The question is testing your understanding of cross and dot products.





if v and w are orthogonal, then v . w = 0, where . is the dot product.





If x is perpendicular to both v and w, then x = v cross w is one way to get it.





to find the first two vectors, say v = (a, b, c), then if v is perpendicular to x which is given,





v . x = (a*1 + b*1 + c*1) = a + b + c = 0, which is too many unknowns (3) and you only have one equation. Instead try constructing a vector with only one unknown and figure out what value that has to be to make it orthogonal to x.





i.e. v = (1, 1, c), then v. x = (1*1 + 1*1 + c*1) = 2 + c = 0, so if c = -2, then v = (1,1,-2) is orthogonal to x. Great. Check to see if this works:





v . x = (1*1 + 1*1 + 1*(-2)) = 2 - 2 = 0, indeed.





You could have made v = any numbers you wanted with one unknown, it would just change the value of 'c' here.





to find the last vector w, you can just do w = v cross x, which is guaranteed by definition of the cross product to be orthogonal to both.





You can check to see all vectors are orthonal to each other by computed v . w, v. x, and x . w in the end.





If you are finding it strange to understand why, say v . w = 0 means the vectors are orthogonal, then you can recall that





v . w = |v| |w| cos(angle between them), but if v . w = 0, then





0 = |v| |w| cos(angle between them), and |v| and |w| are not zero, so you only have one unknown, the angle, call it theta.





0 = cos(theta) =%26gt; cos^-1 (0) = 90 degrees, i.e. orthogonal.