Vectors - Equation Of Plane

Slide 1: Vectors - Equation of Plane

In three-dimensional space, a plane can be defined using a vector and a point on the plane.
The equation of a plane is given by ax + by + cz = d, where a, b, and c are the components of the normal vector to the plane, and d is a constant.
For example, the equation of a plane with normal vector n = <a, b, c> and a point P(x₁, y₁, z₁) on the plane is ax + by + cz = ax₁ + by₁ + cz₁.
The normal vector represents the direction perpendicular to the plane, and determines the orientation of the plane.
The equation of a plane can also be written using dot product as n · (r - r₀) = 0, where n is the normal vector, r is the position vector of any point on the plane, and r₀ is the position vector of a point on the plane.

Slide 2: Vector Equation of a Plane

A plane can also be represented using a vector equation.
The vector equation of a plane passing through a point A(x₀, y₀, z₀) and parallel to two non-collinear vectors u and v is r = A + su + tv, where r is the position vector of any point on the plane, s and t are scalar parameters.
This equation represents an infinite number of points lying on the plane.
To find the scalar parameters s and t, the given equation can be equated with the coordinates of a random point P(x, y, z) on the plane, resulting in a system of linear equations.
Solving the system of equations will give the values of s and t, which will determine the position of the point P(x, y, z) on the plane.

Slide 3: Example 1

Find the equation of the plane passing through the point P(1, 2, -3) and with a normal vector n = <2, -1, 3>.
The equation of the plane is ax + by + cz = d, where a, b, c are the components of the normal vector n and d can be found by substituting the coordinates of point P.
Plugging in the values, we get 2x - y + 3z = 2(1) - 1(2) + 3(-3) = 0.
Therefore, the equation of the plane passing through P(1, 2, -3) with normal vector n = <2, -1, 3> is 2x - y + 3z = 0.

Slide 4: Example 2

Find the equation of the plane passing through the points A(1, 2, -3), B(2, -1, 4), and C(-1, 3, 1).
To find the equation of the plane, we need a point on the plane and a normal vector.
Let AB and AC be two vectors lying on the plane, which can be found using the coordinates of points A, B, and C.
The normal vector n can be found as the cross product of AB and AC.
Once we have the normal vector, we can use one of the given points, such as A, to find the equation of the plane.
Solving this system of equations, we obtain x - 2y + z = -3 as the equation of the plane passing through A(1, 2, -3), B(2, -1, 4), and C(-1, 3, 1).

Slide 5: Intersection of Two Planes

When two planes intersect, their intersection line is either a line or a plane of the same dimension.
To find the intersection, we need to find the common point or the equation of the intersection line or plane.
If the normal vectors of both planes are parallel, they are either coincident or parallel, and there is no intersection.
If the normal vectors of both planes are not parallel, the planes intersect at a line.
The common point can be found by solving the system of equations formed by the equations of the two planes.
The direction of the intersection line can be found by taking the cross product of the normal vectors of the two planes.

Slide 6: Example 3

Find the intersection of the planes P₁: 2x + y - z = 5 and P₂: x - y + 2z = 3.
To find the common point or the equation of the intersection line, we need to solve the system of equations formed by the two planes.
Adding the two equations, we get 3x + z = 8.
Solving for z, we find z = 8 - 3x.
Substituting this value for z in the equations of P₁ and P₂, we can find the values of x and y.
The intersection line can be found by taking the cross product of the normal vectors n₁ and n₂ of the two planes.

Slide 7: Equations of Planes in Different Forms

The equation of a plane ax + by + cz = d can be written in different forms to highlight specific characteristics.
If the normal vector n = <a, b, c> is divided by a scalar k, the equation becomes (a/k)x + (b/k)y + (c/k)z = d/k, which represents the same plane.
The equation can also be written in the form r · n = d, where r is the position vector of any point on the plane and n is the normal vector.
Another form of the equation is n · (r - r₀) = 0, where r₀ is the position vector of a known point on the plane.
These forms can be useful in different scenarios, such as finding the equation of a plane passing through a certain point, or when working with vectors.

Slide 8: Example 4

Write the equation of the plane passing through the point P(1, 2, 3) and with the normal vector n = <2, -1, 3> in different forms.
The equation of the plane in the standard form is 2x - y + 3z = 2(1) - 1(2) + 3(3) = 8.
In the form r · n = d, we have <x, y, z> · <2, -1, 3> = 8.
Simplifying, we get 2x - y + 3z = 8.
Using the form n · (r - r₀) = 0, we have <2, -1, 3> · <x - 1, y - 2, z - 3> = 0.
Expanding and simplifying, we obtain 2x - y + 3z = 8.

Slide 9: Equation of a Plane in Intercept Form

The equation of a plane can also be written in the intercept form, where x/a + y/b + z/c = 1.
The intercepts a, b, and c are the distances from the plane to the coordinate axes, divided by the magnitude of the normal vector.
To find the equation of a plane in intercept form, the intercepts can be found by substituting known points on the plane into the equation and solving for the intercepts.
This form of the equation is helpful in finding the intercepts and understanding the behavior of the plane with respect to the coordinate axes.

Slide 10: Examples 5

Find the equation of the plane in intercept form for the following conditions:
- P₁(6, 0, 0), P₂(0, 4, 0), P₃(0, 0, 3)
- P₁(0, 3, 0), P₂(1, 0, -1), P₃(0, 0, 4)
For the first set of conditions, the intercepts a, b, and c can be found by substituting the coordinates of the given points into the equation x/a + y/b + z/c = 1.
Simplifying, we obtain x/6 + y/4 + z/3 = 1.
For the second set of conditions, substituting the coordinates and simplifying, we get x + 3y + z/4 = 1 as the equation of the plane in intercept form.

Slide 11: Line of Intersection of Two Planes

When two planes are not parallel, they intersect to form a line.
To find the line of intersection, we need the direction vector of the line as well as a point on the line.
The direction vector can be found by taking the cross product of the normal vectors of the two planes.
The point on the line can be found by solving the system of equations formed by the equations of the two planes.

Slide 12: Example 6

Find the line of intersection of the planes P₁: 2x + y - z = 3 and P₂: 3x - 2y + z = 4.
The direction vector of the line can be found by taking the cross product of the normal vectors of P₁ and P₂.
The normal vector of P₁ is n₁ = <2, 1, -1> and the normal vector of P₂ is n₂ = <3, -2, 1>.
Taking the cross product, we get n₁ × n₂ = <1, -1, -4> as the direction vector of the line.
To find a point on the line, we solve the system of equations formed by the equations of P₁ and P₂.
Solving, we find x = 1, y = -1, and z = -1 as the coordinates of a point on the line.
Therefore, the line of intersection of the planes P₁: 2x + y - z = 3 and P₂: 3x - 2y + z = 4 is given by the equations:
- x = 1 + t
- y = -1 - t
- z = -1 - 4t

Slide 13: Distance from a Point to a Plane

To find the distance from a point P(x, y, z) to a plane ax + by + cz + d = 0, we can use the formula:
- d(P, Plane) = |ax + by + cz + d| / √(a² + b² + c²)
The numerator represents the magnitude of the signed perpendicular distance from the point to the plane.
The denominator represents the magnitude of the normal vector of the plane.
The distance can also be represented as the absolute value of this signed distance.

Slide 14: Example 7

Find the distance from the point P(2, -1, 3) to the plane 2x - y + 3z = 7.
The equation of the plane is 2x - y + 3z - 7 = 0, where a = 2, b = -1, c = 3, and d = -7.
Substituting the values into the distance formula, we get:
- d(P, Plane) = |2(2) - (-1)(-1) + 3(3) - 7| / √(2² + (-1)² + 3²)
Simplifying, we obtain:
- d(P, Plane) = |4 + 1 + 9 - 7| / √(4 + 1 + 9) = |7| / √14 ≈ 2.65
Therefore, the distance from the point P(2, -1, 3) to the plane 2x - y + 3z = 7 is approximately 2.65 units.

Slide 15: Angle between Two Planes

The angle between two planes can be found using their normal vectors.
Let n₁ and n₂ be the normal vectors of the planes.
The angle θ between the planes can be found using the dot product formula:
- cos(θ) = (n₁ · n₂) / (|n₁| |n₂|)
- θ = cos⁻¹((n₁ · n₂) / (|n₁| |n₂|))

Slide 16: Example 8

Find the angle between the planes P₁: 2x - y + 3z = 0 and P₂: 3x + 2y + z = 7.
The normal vector of P₁ is n₁ = <2, -1, 3> and the normal vector of P₂ is n₂ = <3, 2, 1>.
Calculating the dot product and the magnitudes:
- n₁ · n₂ = (2)(3) + (-1)(2) + (3)(1) = 6 - 2 + 3 = 7
- |n₁| = √(2² + (-1)² + 3²) = √(4 + 1 + 9) = √14
- |n₂| = √(3² + 2² + 1²) = √(9 + 4 + 1) = √14
Substituting the values into the formula, we get:
- θ = cos⁻¹(7 / (√14)(√14))
Simplifying, we find:
- θ = cos⁻¹(7 / 14) ≈ 45°

Slide 17: Parallel and Perpendicular Planes

Two planes are parallel if their normal vectors are parallel.
Two planes are perpendicular if their normal vectors are orthogonal (i.e. their dot product is zero).
The dot product of two normal vectors can be used to determine whether two planes are parallel or perpendicular.

Slide 18: Example 9

Determine whether the planes P₁: 3x - 2y + z = 0 and P₂: 6x - 4y + 2z = 1 are parallel or perpendicular.
The normal vector of P₁ is n₁ = <3, -2, 1> and the normal vector of P₂ is n₂ = <6, -4, 2>.
Calculating the dot product of n₁ and n₂:
- n₁ · n₂ = (3)(6) + (-2)(-4) + (1)(2) = 18 + 8 + 2 = 28
Since the dot product is not zero, the planes are not perpendicular.
To determine if they are parallel, we check if the ratios of their components are equal:
- 3/6 = -2/-4 = 1/2
Since the ratios are equal, the planes are parallel.

Slide 19: Condition for Coincident Planes

Two planes are coincident if they have the same normal vector and pass through the same point.
If the normal vectors are not the same or if they pass through different points, the planes are not coincident.

Slide 20: Examples 10

Determine whether the planes P₁: 2x - y + 3z = 1 and P₂: 4x - 2y + 6z = 2 are coincident, parallel, or neither.
The normal vector of P₁ is n₁ = <2, -1, 3> and the normal vector of P₂ is n₂ = <4, -2, 6>.
The planes have the same normal vector, but they do not pass through the same point.
Therefore, the planes are neither coincident nor parallel.

Slide 21: Equation of a Line in 3D

In 3D, a line can be represented by a parametric equation using two points on the line.
The parametric equation of a line passing through point A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is:
- x = x₁ + t(x₂ - x₁)
- y = y₁ + t(y₂ - y₁)
- z = z₁ + t(z₂ - z₁)
Here, t is a scalar parameter that determines different points on the line.
By varying t, we can obtain various points on the line.
The direction vector of the line is given by <x₂ - x₁, y₂ - y₁, z₂ - z₁>.

Slide 22: Example 11

Find the parametric equations for the line passing through the points A(1, 2, -3) and B(4, -1, 2).
Using the given points, we have:
- **x = 1 + t(4 - 1