text: a)  Not sure how you 'show' this... it's kind of trivial.  Anything parallel to the x-axis is of the form y = constant, z = constant - and here the constants have to be b and c to go through the point, as there's no x dependence.\n\nb)  |a+b|^2 = |a|^2 + a.b + |b|^2\nSo for it to balance, a.b = 0; that is, a and b are orthogonal.\n\nc)  Let the vector be (x,y,z).  Then:\n\n(1,1,0).(x,y,z) = 0  => x + y = 0\n(0,1,1).(x,y,z) = 0  => y + z = 0\n\nSo x = z = -y, and the vector is parallel to (1,-1,1).  Normalising this to unit length gives:\n\n(1 / sqrt(3)) (1, -1, 1).