Final answer:
The point on the plane 2x + 3y + 4z that is nearest to (2,0,1) is (2,0,1) itself, with a distance of 0.37 units.
Step-by-step explanation:
To find the point on the plane 2x + 3y + 4z that is nearest to (2,0,1), we need to find the equation of the plane first. Given that the normal vector to the plane is (2, 3, 4) and a point on the plane is (2, 0, 1), we can use the equation of a plane to find the equation. The equation of the plane is 2x + 3y + 4z = 2*2 + 3*0 + 4*1 = 8.
Now, we need to find the point on this plane that is nearest to (2,0,1). Since we want to minimize the distance between the point and the plane, we need to find the point that is perpendicular to the plane. The perpendicular distance between a point (x₀, y₀, z₀) and a plane Ax + By + Cz = D is given by: d = |(Ax₀ + By₀ + Cz₀ - D)| / √(A² + B² + C²). In this case, the equation becomes: d = |(2*2 + 3*0 + 4*1 - 8)| / √(2² + 3² + 4²) = 2 / √(29) = 0.37.
Therefore, the point on the plane 2x + 3y + 4z that is nearest to (2,0,1) is (2,0,1) itself, with a distance of 0.37 units.