.
The other tutor retold the general theory, but the meaning of the problem and the meaning of the request
is to point (to find) three concrete linearly independent vectors in perpendicular to vector V.
It was not done in the post by the other tutor, so I will do it right now.
The first vector in , perpendicular to vector v, is x = (1,0,-4,0).
Indeed, you can check it on your own that the scalar product of vectors v and x is equal to zero
(v,x) = 4*1 + (-9)*0 + 1*(-4) + 9*0 = 4 + 0 - 4 + 0 = 0.
The second vector in , perpendicular to vector v, is y = (0,-1,-9,0).
Indeed, you can check it on your own that the scalar product of vectors v and y is equal to zero
(v,y) = 4*0 + (-9)*(-1) + 1*(-9) + 9*0 = 0 + 9 - 9 + 0 = 0.
Finally, the third vector in , perpendicular to vector v, is z = (0,0,-9,1).
Indeed, you can check it on your own that the scalar product of vectors v and z is equal to zero
(v,z) = 4*0 + (-9)*0 + 1*(-9) + 9*1 = 0 + 0 - 9 + 9 = 0.
Next, it is OBVIOUS that the three vectors
x = (1, 0, -4, 0),
y = (0,-1, -9, 0) and
z = (0, 0, -9, 1)
are linearly independent (due to construction of their components).
Thus we constructed (guessed, based on intuition) three linearly independent vectors in perpendicular to the given vector v.
Hence, these three vectors x, y and z form a basis in the orthogonal complement to vector V in .
It is what has to be done.
Solved, explained and completed.