Question

equation of vector in plane​

Answer 1

Answer:

Here, we use our knowledge of the dot product to find the equation of a plane in R3 (3D space). Firstly, a normal vector to the plane is any vector that starts at a point in the plane and has a direction that is orthogonal (perpendicular) to the surface of the plane. For example, k = (0,0,1) is a normal vector to the xy plane (the plane containing the x and y axes).

Any three distinct points define our plane, or alternatively, a single point in the plane and a given normal vector to the plane. We'll define it like this first.

All we need is a way of checking whether a given point, with position vector r lies in the plane or not.

Let a be the position vector of the given point in the plane, and

let n be the given normal vector to the plane.

Then r is in the plane if and only if (iff)

(r - a) . n = 0

Explanation: r - a is the vector from A to R, and any vector that lies completely in the plane (like this one does) must be orthogonal to n which implies their dot product must equal 0.

Using 3 points instead

Let a, b, c be the position vectors of 3 points in the plane. Then all we have to do to find the equation of the plane is construct a normal vector - then we can use this and any of the 3 points to find the equation as before.

The two vectors,

b - a (A to B)

c - a (A to C)

both lie in the plane, so if we take their cross product we find our normal vector:

n = (b - a) x (c - a)

So as before our equation for the plane consists of all vectors, r satisfying

(r - a) . n = 0

=>(r - a) . ((b - a) x (c - a)) = 0

Answer 2

Answer:

The Vector Equation of a Plane. ... For example, k = (0,0,1) is a normal vector to the xy plane (the plane containing the x and y axes). Any three distinct points define our plane, or alternatively, a single point in the plane and a given normal vector to the plane.

Step-by-step explanation:

The Vector Equation of a Plane

Here, we use our knowledge of the dot product to find the equation of a plane in R3 (3D space). Firstly, a normal vector to the plane is any vector that starts at a point in the plane and has a direction that is orthogonal (perpendicular) to the surface of the plane. For example, k = (0,0,1) is a normal vector to the xy plane (the plane containing the x and y axes).

Any three distinct points define our plane, or alternatively, a single point in the plane and a given normal vector to the plane. We'll define it like this first.

All we need is a way of checking whether a given point, with position vector r lies in the plane or not.

Let a be the position vector of the given point in the plane, and

let n be the given normal vector to the plane.

Then r is in the plane if and only if (iff)

(r - a) . n = 0

Explanation: r - a is the vector from A to R, and any vector that lies completely in the plane (like this one does) must be orthogonal to n which implies their dot product must equal 0.

Using 3 points instead

Let a, b, c be the position vectors of 3 points in the plane. Then all we have to do to find the equation of the plane is construct a normal vector - then we can use this and any of the 3 points to find the equation as before.

The two vectors,

b - a (A to B)

c - a (A to C)

both lie in the plane, so if we take their cross product we find our normal vector:

n = (b - a) x (c - a)

So as before our equation for the plane consists of all vectors, r satisfying

(r - a) . n = 0

=>(r - a) . ((b - a) x (c - a)) = 0