How to find a vector that is perpendicular to every vector in a linear subspace?

AI Thread Summary
To find a vector perpendicular to every vector in the linear subspace defined by the equation x - y + z = 0, one must identify the normal vector of the subspace. The normal vector can be derived directly from the coefficients of the equation, which gives the vector <1, -1, 1>. This vector is already of length √3, so to obtain a unit vector, it needs to be normalized. The unit vector perpendicular to the subspace is then <1/√3, -1/√3, 1/√3>. Understanding the geometric interpretation of the subspace as a plane in R3 aids in visualizing the perpendicular relationship.
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Homework Statement


Hi, i don't know if you can help me but i am currently studying for my finals and i have come across a question which i am very confused about. i have looked it up in books but there seems to be no answer there. the question is Write down a vector of length 1 that is perpendicular to every vector in the linear subspace of r3 described by x-y+z=0. If you could help me i would be very greatful! thank you.
I know that to find a normal vector which is perpendicular to another vector you use the cross product but i do not see how this will benefit me in this question as i am not given any vectors!
 
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A first step would be to find out which subspace we're talking about, i.e. write down what form the vectors of this subspace take.
 
The subspace is {<x, y, z> in R3 | x - y + z = 0}. What sort of a geometric object is this subspace?
 

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