The line is perpendicular to the plane 2x–3y+z–w=7 and passes the origin. Find parametric form vector formThis is my solution:The line perpendicular to the plane implies being parallel to the normal vector. n=[2,−3,1,−1]vector form: [x,y,z,w,v]=[0,0,0,0]+t[2,−3,1,−1]parametric form: x=2t,y=−3t,z=t,w=−t,v=0right?

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Ashlynn Stephens · Accepted Answer

Since the normal to the plane is             n      →        =  (  2  ,  −  3  ,  1  ,  −  1  ) therefore the parametric equation of the perpendicular line in parametric form is given by  (  0  ,  0  ,  0  ,  0  )  +  t  (  2  ,  −  3  ,  1  ,  −  1  )

Gauge Roach · Accepted Answer

I think we are in the space             R        4   with elements   (  x  ,  y  ,  z  ,  w  ). Hence the normal vector is   n  =  (  2  ,  −  3  ,  1  ,  −  1  ), therefore the line is given by  {  t  ⋅  (  2  ,  −  3  ,  1  ,  −  1  )  :  t  ∈      R    }  .

The line is perpendicular to the plane 2x–3y+z–w=7 and passes the origin. Find parametric form vector form

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