Find the point on the plane 4x+3y +z =1 that is closest to (1,-1,1).?

Question

I don't understand this question that was given in my lecture today, thanks in advance, all help is appreciated!

Answer 1

Use the formula for the distance from a point to a plane.

For the point (h, k, p) and the plane ax + by + cz + d = 0 the distance formula is:

Distance = | ah + bk + cp + d | / √(a² + b² + c²)

In the question at hand we have:
Point (1, -1, 1) and
Plane 4x + 3y + z - 1 = 0.

Distance = | 4*1 + 3*(-1) + 1*1 - 1 | / √(4² + 3² + 1²)

Distance = 1 / √26

Answer 2

The equation defines a plane, which is a set of points. You need to find the point within that set that has the smallest distance from x=1,y=-1,z=1

You need to find the direction perpendicular to that plane. That will give you a set of lines in three dimensions. Then you need to find the line that goes through the point you have. Then, finally, you need to find the point that is on both the plane and the line.

Since this seems to be your homework, I'll leave the details to you.