Analytic Geometry / Vectors - Find point with min distance to plane

[B.Cantarelli]

Question:

Find the point in the plane x+2y-3z=1 with minimum distance to point (1,1,1).

Attempt at resolution:

Well, techinically, I already have the means to solve this.

I can find the distance between the point and the plane to be 1/sqrt(14) and then I can either solve for the sphere (x-1)^2+(y-1)^2+(z-1)^2=1/14 and the plane x+2y-3z=1, or I can use Calculus (Lagrange's method) to get L=x^2+y^2+z^2-2x-2y-2z+3+l(x+2y-3z-1) and take all the partial derivatives.

Both yield (15/14, 8/7, 11/14), but I'm quite sure I'm missing some way more efficient method for doing this using vectors or something.

 

[Mark44]

Question:

Find the point in the plane x+2y-3z=1 with minimum distance to point (1,1,1).

Attempt at resolution:

Well, techinically, I already have the means to solve this.

I can find the distance between the point and the plane to be 1/sqrt(14) and then I can either solve for the sphere (x-1)^2+(y-1)^2+(z-1)^2=1/14 and the plane x+2y-3z=1, or I can use Calculus (Lagrange's method) to get L=x^2+y^2+z^2-2x-2y-2z+3+l(x+2y-3z-1) and take all the partial derivatives.

Both yield (15/14, 8/7, 11/14), but I'm quite sure I'm missing some way more efficient method for doing this using vectors or something.

There is an alternate approach that doesn't require calculus. Instead, it uses right triangle trig and vectors. From your plane, you can note by inspection that <1, 2, -3> is a normal to the plane. Find a point in the plane - any point will do - and form a displacement vector from that point to the point (1, 1, 1). Next, find the projection of the vector onto the vector <1, 2, -3>. When you have that vector, its magnitude will be the minimum distance. Knowing the distance, you can quickly find the point on the plane that is that distance away from the point (1, 1, 1).

 

[B.Cantarelli]

There is an alternate approach that doesn't require calculus. Instead, it uses right triangle trig and vectors. From your plane, you can note by inspection that <1, 2, -3> is a normal to the plane. Find a point in the plane - any point will do - and form a displacement vector from that point to the point (1, 1, 1). Next, find the projection of the vector onto the vector <1, 2, -3>. When you have that vector, its magnitude will be the minimum distance. Knowing the distance, you can quickly find the point on the plane that is that distance away from the point (1, 1, 1).

Thank you very much for the response. The trouble I was having was exactly with the "Knowing the distance, you can quickly find the point on the plane that is that distance away from the point (1, 1, 1)" part. Particularly with the "quickly". I got the ray 1/sqrt(14) by doing exactly what you described, but was failing to see that the point I wanted consisted of (1,1,1)+(1/sqrt(14))(1,2,-3)/||(1,2,-3)||. Instead of realizing I just had to "walk" from (1,1,1) in the direction of the normilized norm times the ray, I was thinking in terms of solving for tangent point of the sphere and the plane. Took me quite a while to realize what you meant by quickly. Again, thank you.

 

[HallsofIvy]

There is an alternate approach that doesn't require calculus. Instead, it uses right triangle trig and vectors. From your plane, you can note by inspection that <1, 2, -3> is a normal to the plane. Find a point in the plane - any point will do -

I would do this just a little differently. Instead of chooseing a point in the plane, construct the line through the given point (1, 1, 1) in the direction of the normal vector: x= t+ 1, y= 2t+ 1, z= -3t+ 1.
Now find the point where that line intersects the plane by replacing x, y, and z in the equation of the plane with those expressions: (t+ 1)+ 2(2t+ 1)- 3(-3t+ 1)= 1. Solve that linear equation for t, then use that in the parametric equations of the line to find the point.

and form a displacement vector from that point to the point (1, 1, 1). Next, find the projection of the vector onto the vector <1, 2, -3>. When you have that vector, its magnitude will be the minimum distance. Knowing the distance, you can quickly find the point on the plane that is that distance away from the point (1, 1, 1).