Question

A plane is defined by the equations x-4y+2z=16. Find two vectors parallel to the plane.

Answer 1

You need 3 points.
Probably the easiest ones would be the intercepts
(0,0,8)
(0,-4,0) and
(16,0,0)
using the first two, a direction vector would be (0,4,8)
using the last two, a direction vector would be (16,4,0)
we could shorten these to (0,1,2) and (4,1,0)

check:
a normal to our plane is (1,-4,2)
taking the dot product of that normal with each of my two vectors yields zero, as expected.

Answer 2

If i want to find the x-, y- and z- intercepts do i write a new vector equation because we previously found the intercepts but the x-intercept =16 y-intercept = -4 and the z-intercept = 8

i tried writing a vector equation using the one vector as my point and the scalar equation as the direction vector but something seems to be wrong as i do not get the above values!