# Proving: All Points on Ax + By + Cz = 0 Lie in Plane Perpend. to Ai + Bj + Ck

• MozAngeles
In summary, the position vector r points from the origin to an arbitrary point in space. We need to use the fact that the dot product of two vectors is zero to prove that all points satisfying the given equation lie in a plane perpendicular to a given vector and passing through the origin. We can do this by considering a point P on the given vector and a point Q on the plane, and using the dot product to show that the two vectors are perpendicular.
MozAngeles

## Homework Statement

The vector r = xi + yj + zk, called the position vector, points from the origin (0, 0, 0) to an arbitrary point in space with coordinates (x, y, z).

Use what you know about vectors to prove the following: All points (x, y, z) that satisfy the equation Ax + By + Cz = 0, where A, B and C are constants, lie in a plane that passes through the origin and that is perpendicular to the vector Ai + Bj + Ck.

A∙B=0 ⇒A⊥B

## The Attempt at a Solution

I have no idea how to begin this proof. I know that A∙B is necessary and sufficient condition for two vectors to be perpendicular to each other. I do not know what else i would put down...

Hi MozAngeles!

The vector Ai + Bj + Ck ends at a point …

start by calling that point P, and calling a point on the plane Q.

## 1. How do you prove that all points on the equation Ax + By + Cz = 0 lie in a plane perpendicular to Ai + Bj + Ck?

To prove this, we must show that the dot product between the normal vector of the plane and any vector on the plane is equal to 0. This can be done by substituting the values of x, y, and z from the given equation into the equation for the dot product and simplifying to show that it equals 0. This demonstrates that all points on the given equation lie in a plane perpendicular to the normal vector.

## 2. What is the significance of the equation Ax + By + Cz = 0 in proving that all points lie in a plane perpendicular to Ai + Bj + Ck?

The equation Ax + By + Cz = 0 represents a specific type of plane known as a perpendicular or orthogonal plane. This means that the plane is at a 90 degree angle to the vector Ai + Bj + Ck. By proving that all points on this equation lie in this type of plane, we can also prove that they are perpendicular to the given vector.

## 3. Can this proof be used for any values of A, B, and C?

Yes, this proof can be used for any values of A, B, and C as long as they are not all equal to 0. This is because a plane cannot be defined by an equation with all zero coefficients.

## 4. What other methods can be used to prove that all points on Ax + By + Cz = 0 lie in a plane perpendicular to Ai + Bj + Ck?

Another method that can be used is to show that the cross product between the normal vector of the plane and any two vectors on the plane is equal to 0. This method may be more intuitive for some individuals as it involves visualizing the vectors and their relationship to the plane.

## 5. Is this proof applicable to higher dimensions?

Yes, this proof can be extended to higher dimensions by using the same concepts and equations. In higher dimensions, the equation for a plane will have more variables, but the same principles of showing that the dot product or cross product between the normal vector and any vector on the plane is equal to 0 still apply.

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