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

Marco Hudson

Marco Hudson

Open question

2022-08-14

The line is perpendicular to the plane 2x–3y+z–w=7 and passes the origin. Find parametric form vector form
This 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=0
right?

Answer & Explanation

Ashlynn Stephens

Ashlynn Stephens

Beginner2022-08-15Added 25 answers

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

Gauge Roach

Beginner2022-08-16Added 3 answers

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 } .

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