find the point on the plane 5x+4y+z=12 that is nearest to (2,0,1)

To find the point on the plane that is nearest to a given point, we need to find the perpendicular distance between the given point and the plane. Since the plane is defined by an equation in the form ax + by + cz = d, we can use the formula for distance between a point and a plane:

Distance = |ax + by + cz - d| / sqrt(a^2 + b^2 + c^2)

In this case, the plane is defined by the equation 5x + 4y + z = 12 and the given point is (2,0,1). We can substitute these values into the formula to find the distance.

Distance = |5(2) + 4(0) + (1) - 12| / sqrt(5^2 + 4^2 + 1^2)
= |10 + 1 - 12| / sqrt(25 + 16 + 1)
= |-1| / sqrt(42)
= 1 / sqrt(42)

Now that we have the distance, we need to find the coordinates of the point on the plane that is nearest to (2,0,1). To do this, we need to use the equation of the plane and solve for the variables x, y, and z.

5x + 4y + z = 12

We can rearrange this equation to solve for z:
z = 12 - 5x - 4y

Substitute the values of x and y from the given point (2,0,1) into the equation to find the value of z.

z = 12 - 5(2) - 4(0)
= 12 - 10
= 2

Therefore, the point on the plane 5x + 4y + z = 12 that is nearest to (2,0,1) is (2,0,2).

just use the usual distance formula. The distance between (a,b,c) and Ax+By+Cz+D=0 is

|Aa+Bb+Cc+D|/√(A^2+B^2+C^2)