Express the following plane in vector form: P 1 ⊆ R 3 with equation 4 x − z = 0The answer is t ( 1 , 0 , 4 ) + s ( 0 , 1 , 0 ). I don't understand how they got (0,1,0) for the direction vector.

Question

Express the following plane in vector form:            P        1    ⊆            R        3   with equation   4  x  −  z  =  0The answer is   t  (  1  ,  0  ,  4  )  +  s  (  0  ,  1  ,  0  ). I don&#039;t understand how they got (0,1,0) for the direction vector.

Brienueentismvh · Accepted Answer

The cartesian equation is z=4xThe parametrization is not unique. Anyway You need two linearly independent points on the plane.One point can be A=(1,0,4). The second one cannot be (2,0,8) or other points you can get from A multiplying by a constant. Otherwise you get a line not a plane.The second point cannot obviously be (0,0,0)So we look for a point where the only nonzero coordinate is y, like (0,1,0) or (0,10,0)Finally the parametric equation of the plane is  (  1  ,  0  ,  4  )  t  +  (  0  ,  1  ,  0  )  s  =  0Note that   (  2  ,  0  ,  8  )  t  +  (  0  ,  10  ,  0  )  s  =  0 would be right.Hope this helps

Lexi Mcneil · Accepted Answer

Let   π  ≡  4  x  −  z  =  0 be your plane. We notice that the point   (  0  ,  0  ,  0  )  ∈  π and the orthogonal vector to the plane is       u    _    =  (  4  ,  0  ,  −  1  )The vector space   dir  (  π  ) has dimension 2 and the vectors that span this space are orthogonal to the vector       u    _  We can take for example the vectors       v    _    =  (  1  ,  0  ,  4  ) and       w    _    =  (  0  ,  1  ,  0  ), infact we have  ⟨      u    _    ,      v    _    ⟩  =  1  ⋅  4  +  0  +  4  ⋅  (  −  1  )  =  0    ⟹     they are orthogonal  ⟨      u    _    ,      w    _    ⟩  =  0  ⋅  4  +  1  ⋅  0  +  0  ⋅  (  −  1  )  =  0    ⟹     they are orthogonalFinally the equation of the plane   π is      {                            x          =          0          +          α                                      y          =          0          +          β                                      z          =          0          +          4          α                          ,  α  ,  β  ∈      R    .

Express the following plane in vector form: cc P_1 sube RR^3 with equation 4x−z=0.

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