Final answer:
The distance of the plane 2x + 3y + 4z = 4 from the origin is approximately 1.170 units.
Step-by-step explanation:
To find the distance of the plane 2x + 3y + 4z = 4 from the origin, we can use the formula for the distance between a point and a plane.
The formula is given by |Ax + By + Cz + D| / √(A^2 + B^2 + C^2),
where (A, B, C) are the coefficients of the plane equation and (x, y, z) are the coordinates of the point.
In this case, the plane equation is 2x + 3y + 4z = 4.
The coefficients are A = 2, B = 3, C = 4, and D = -4.
The origin has coordinates (0, 0, 0).
Plugging these values into the formula, we get
|2(0) + 3(0) + 4(0) + (-4)| / √(2^2 + 3^2 + 4^2)
= 4 / √(4 + 9 + 16)
=4 / √29.
=1.170
So the distance of the plane 2x + 3y + 4z = 4 from the origin is approximately 1.170 units.