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

Question

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

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 1

Since the normal to the plane is $\vec n = (2, -3, 1, -1)$ therefore the parametric equation of the perpendicular line in parametric form is given by

$$(0,0,0,0)+t(2, -3, 1, -1)$$

Answer 2

I think we are in the space $ \mathbb R^4$ with elements $(x,y,z,w)$. Hence the normal vector is $n = (2, -3, 1, -1)$, therefore the line is given by

$$\{t \cdot(2,-3,1,-1): t \in \mathbb R\}.$$