Find the equation of the plane which has intercepts of (4,0,0) and (0,−2,0) and no z-intercept.

Given two points (4,0,0) and (0,-2,0) are two intercepts of plane Π which has no z-intercept.

First step: find direction vector of the two points:
V = <4-0, 0-(-2), 0-0> = <4,2,0>

Since there is no z-intercept, Π must be orthogonal to the z-axis, hence the unit normal vector is Z=<0,0,1>.

To find the normal vector of the plane Π, we take the cross-product of V & Z, i.e.
Π = 
 i  j  k
4 2 0
0 0 1
= <2*1, -4*1, 0>
= <2,-4,0>
The corresponding equation of the plane through (4,0,0) is then
2(x-4)-4y+0(z-0)=0 => 2x-4y=8
Answer: The equation of the plane is 2x-4y=8