# Planes

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## Planes

As a line is defined as needing a vector to the line and a vector parallel to the line, so a plane similarly needs a **vector to the plane** and then **two vectors in the plane** (these two vectors should not be parallel).

It is then possible to get to any point in the plane by firstly getting to the plane and then moving around the plane using multiples of the two vectors in the plane.

**This gives the vector equation of a plane as:**

[u]r[/u] = [u]a[/u] + λ[u]b[/u] + μ[u]c[/u]

Where [u]a[/u] is a vector to the plane, and [u]b[/u] and [u]c[/u] are vectors in the plane.

**Example:**

To find the vector equation of the plane that contains the three points, (1, 5, 3), (-2, 3, 5), and (2, 1, 0):

1. Choose the vector to one point as the vector to the plane. For example,

2. Find two vectors in the plane by finding vectors that join the known points.

**For example:**

3. Put these together to make the equation of the plane.

Where [u]r[/u] is the vector to any point on the plane.

If a point lies on the plane then there will be values for λ and μ that are true for the x, y and z coordinates.

**Example:**

Show that the point (10, -3, -10) lies on the plane

**Match the x, y and z coordinates to get:**

1 - 3λ + μ = 10

5 - 2λ - 4λ = -3

3 + 2λ - 3λ = -10

Solve the first two equations to get λ = -2 and μ = 3.

Check that these work in the third equation: 3 - 4 - 9 = -10 (Yes!)

**Therefore: (10, -3, -10) lies on the plane.**

The vector equation of a plane is good, but it requires three pieces of information, and it is possible to define a plane with just two.

As before we need to know a point in the plane, but rather than use two vectors in the plane we can instead use **the normal - the vector at right angles to the plane.**

**To find an alternative equation for the plane we need:**

- A point in the plane (with position vector [u]a[/u]).
- The normal to the plane (the vector [u]n[/u]).

Using the fact that the scalar product between perpendicular vectors is 0, we get:

([u]r[/u] - [u]a[/u]).[u]n[/u] = 0

Where [u]r[/u] = the vector to any point on the plane.

**This can be rewritten as:**

[u]r[/u].[u]n[/u]= [u]a[/u].[u]n[/u]

This means that the scalar product of a point on the plane and the normal has the same value 'wherever the point is on the plane'. (If you know the normal and one point you know the equation of the plane.)

**Example:**

**Using the formula, [u]r[/u].[u]n[/u] = [u]a[/u].[u]n[/u] we get:**

Therefore the equation of the plane is:

This means that the position vector to any general point (x, y, z) on the plane has 1 as its scalar product with the normal. (Check that (2, 1, 2), (1, 0, 0), (-7, -4, -10), all lie on the plane, and find z if (4, -9, z) is also on the plane.) *(Answer: z = -12)*

**This turns the vector equation into a cartesian equation as follows:**

**This is the cartesian equation of the plane.**

**Finding the cartesian equation from the vector equation of a plane**

**Earlier we used the equation:**

Which came from the plane containing the three points (1, 5, 3), (-2, 3, 5), and (2, 1, 0).

In order to write this in cartesian form we need to once more use the scalar product to find the normal.

are vectors in the plane, and are perpendicular to the normal.

If the normal vector is

then:

**This gives us two equations, but unfortunately three unknowns!**

-3a − 2b + 2c = 0

a − 4b − 3c = 0

Eliminating 'a' we get:

-14b − 7c = 0

At this point it is useful to remember that **the normal is a vector and so can have any length (and hence any y-value for instance).** This allows us to choose a suitable value for b to represent the y-value in the normal vector.

If we let b = 1, then -14 − 7c = 0.

This gives us c = -2.

If b = 1, and c = -2, then a = -2 (found by substituting b = 1, c = -2 into the equation above).

Once we have the normal vector we can use one point, say (1, 5, 3), and the formula [u]r[/u].[u]n[/u] = [u]a[/u].[u]n[/u] to get:

**To get the cartesian equation:**

**-2x + y − 2z = -3.**

(This could be rewritten as 2x - y + 2z = 3 by multiplying by ?1 and is a neater result (here b = -1).)

(It is worth checking that all the points do satisfy this equation.)

To find where a line meets a plane we need to find a point on the line that satisfies the equation for the plane. To do this we:

- Write down the coordinates of a general point on the line.
- Use these coordinates in the formula for the equation of the plane.
- Solve the equation and find the coordinates.

This is easiest to illustrate with an **example:**

Find where the line

meets the plane x + 2y − 2z = 12.

A general point on the line has coordinates (2 - 2λ, 4λ, -1 − λ).

Therefore if the line is to meet the plane:

(2 - 2λ) + 2(4λ) − 2(-1 - λ) = 12

8λ = 8

λ = 1.

**The distance between a point and a plane.**

**Therefore the line meets the plane at (0, 4, -2).**

**This method for finding where a line meets a plane is used to find the distance of a point from a plane. Using the fact that the shortest distance from the point to the plane is at right angles to the plane, the line that joins the point to the plane has the normal to the plane as its direction vector.**

**This means the line is in the form:**

[u]r[/u] = [u]a[/u] + λ[u]n[/u]

where [u]a[/u] = position vector to the point, and [u]n[/u] = normal to the plane.

- Find the coordinates of the point where this line meets the plane.
- Find the vector joining these two points together.
- Use Pythagoras' Theorem to find the distance between them.

**Example:**

Find the distance the point (7, 4, -3) is from the plane x + 2y - 2z = 12.

The line joining (7, 4, -3) to the plane x + 2y - 2z = 12 is:

**This meets the plane when:
**

(7 + λ) + 2(4 + 2λ) − 2(-3 - 2λ) = 12

9λ + 21 = 12

λ = -1

Therefore the line meets the plane at (6, 2, -1)

The vector joining (7, 4, -3) to (6, 2, -1) is

The length of this vector is √(1+4+4) = 3 units.

(7, 4, -3) is 3 units from the plane x + 2y − 2z = 12.

**Note:** The distance of the origin from the plane is the value of **d** in the general equation **ax + by + cz = d**, (where

is the normal vector with a length of 1 unit.)

**Example:**

Find the distance the plane x + 2y − 2z = 12 is from the origin. The length of the normal vector is √(1+4+4) = 3 units. Therefore, divide both sides of the equation by 3 to get a normal vector length 1, and a distance from the origin of 12/3 = 4 units.

The **angle between two planes** is the same as the **angle between the normals** to the planes.

Therefore use the scalar product on the normals, (choosing the acute angle as a sensible final answer).

To find the **angle between a line and a plane**, find the angle between the direction of the line and the normal, and then subtract this from 90. (This is the same as using this variation of the scalar product.)

Where [u]a[/u] = direction of the line, and [u]n[/u] = normal vector.

**Note:** the reason we can use the sin function in the 'scalar product equation' is because cos θ = sin (90 − θ).

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