# Distance from plane to origin

• misterau
In summary: The inner product is a useful tool because it allows us to combine vectors in a way that preserves their magnitude. In this case, the result of multiplying x and y is going to be the same as multiplying z and the result of multiplying x and z is going to be the same as multiplying y and z. So the inner product between x and y, between x and z, and between y and z is all going to be the same. This is just a geometrical way of saying that the distance between P and the origin is the sum of the distances between each of the points on the plane.

## Homework Statement

Show that the distance D from the origin of any point (x , y , z) lying on the plane x + y + z= 1 satisfies
D^2 + x^2 + y^2 +(x + y -1)^2 .
By considering partial derivatives, find the point that is closest to the origin. Prove that this distance is genuinely minimal.

## The Attempt at a Solution

I did the first part. Knowing Distance is D^2 = x^2 + y^2 + z^2 then by re-arranging x + y + z =1 to -z = x + y - 1.
z^2 = (-z)^2 = (x + y - 1)^2
D^2 = x^2 + y^2 + (x + y - 1)^2
Not sure how to do the next part. How does considering partial derivatives help me find a point closest to the origin?
Any help would be great.

misterau said:

## Homework Statement

Show that the distance D from the origin of any point (x , y , z) lying on the plane x + y + z= 1 satisfies
D^2 + x^2 + y^2 +(x + y -1)^2 .
Shouldn't this be an equation? Namely, D^2 = x^2 + y^2 +(x + y -1)^2.

misterau said:
By considering partial derivatives, find the point that is closest to the origin. Prove that this distance is genuinely minimal.

## The Attempt at a Solution

I did the first part. Knowing Distance is D^2 = x^2 + y^2 + z^2 then by re-arranging x + y + z =1 to -z = x + y - 1.
z^2 = (-z)^2 = (x + y - 1)^2
D^2 = x^2 + y^2 + (x + y - 1)^2
Not sure how to do the next part. How does considering partial derivatives help me find a point closest to the origin?
Any help would be great.

You have D2 as a function of x and y. You're looking for the smallest value of D2 from the origin to the plane. You can use partial derivatives to find the minimum point in a way analogous to finding the minimum value of a function of one variable (i.e., finding the derivative, and setting it to zero to get the critical points, and then checking with the second derivative to see if you have a local maximum or local minimum).

misterau said:

## Homework Statement

Show that the distance D from the origin of any point (x , y , z) lying on the plane x + y + z= 1 satisfies
D^2 + x^2 + y^2 +(x + y -1)^2 .
By considering partial derivatives, find the point that is closest to the origin. Prove that this distance is genuinely minimal.

## The Attempt at a Solution

I did the first part. Knowing Distance is D^2 = x^2 + y^2 + z^2 then by re-arranging x + y + z =1 to -z = x + y - 1.
z^2 = (-z)^2 = (x + y - 1)^2
D^2 = x^2 + y^2 + (x + y - 1)^2
Not sure how to do the next part. How does considering partial derivatives help me find a point closest to the origin?
Any help would be great.

The distance from a plane P to the origin is equal to a.x + b.y + c.z = d where <a,b,c> is
a normal vector and d is the distance to the origin.

We do this through the notion of a Euclidean Inner Product. Basically what happens here is
that in three dimensional geometry we express the inner product in the form of
<a,b> = |a||b|cos(a,b) where |a| is the length vector of a and |b| is the lebgth vector of
b. cos(a,b) is the cosine of the angle between a and b is denoted in that way.

## What is the meaning of "Distance from plane to origin"?

The distance from plane to origin is the shortest distance between a point on a plane and the origin point (0,0,0) in a three-dimensional Cartesian coordinate system. It is also known as the perpendicular distance since it is measured along a line that is perpendicular to the plane.

## How is the distance from plane to origin calculated?

The distance from plane to origin can be calculated using the formula: d = |Ax + By + Cz + D| / sqrt(A^2 + B^2 + C^2), where A, B, and C are the coefficients of the plane's equation and D is the constant term.

## What is the unit of measurement for the distance from plane to origin?

The unit of measurement for the distance from plane to origin depends on the units used for the coefficients of the plane's equation. For example, if the coefficients are in meters, the distance will be in meters as well.

## Can the distance from plane to origin be negative?

Yes, the distance from plane to origin can be negative. It means that the origin point is on the opposite side of the plane as the point being measured. A positive distance means the origin is on the same side of the plane as the point being measured.

## How is the distance from plane to origin used in real life?

The distance from plane to origin is commonly used in geometry and physics to measure the shortest distance between a point on a plane and the origin point. It is also used in various engineering applications, such as calculating the distance of a plane from a target point or finding the shortest distance between a plane and a curve or surface.