# Lines and Planes

### Two-Space and Three-Space

## LINES IN TWO-SPACE AND THREE-SPACE

Any line in two-space (a space that has only two dimensions i.e. **x** and **y**) can be defined using its slope (m) and y-intercept (b). The slope of the lines represents the direction and the y-intercept represents the exact position in two-space.

The slope of the line is not defined in three space (a space that has three dimensions i.e. **x**, **y** and **z**), so line must be defined in other ways...

A line can be defined by an equation in **Standard form **(Ax+By+C=0), which is also known as the **Scalar Equation, **or in **slope-y-intercept form** (y=mx+b).

Do you know that a vector can also be used to define a line?

Scalar Equation **-x+2y-10=0**

A **direction vector** parallel to line is **m **= [2, 1]

A **position vector **that has its tip on the line is **r0** = [2, 6]

Another position vector **r** = [6, 8] can be drawn to form a closed triangle.

The third side of the triangle is vector **s**

Therefore, the closed triangle can be shown using the vector sum.

**r = r0 +s [vector r = vector r0 + vector s]**

**or**

**s = tm [ vector s = t x vector m] **(since s is parallel to m)

Therefore, **r = r0 + tm**

which can also be written as [x, y] = [x0, y0]+t[m1, m2] in vector sum.

Therefore x = x0+tm1

&

y = y0 +tm2

This form of equation is known as **Parametric Equation**

**Vector Equation...**

**Scalar Equation**

A **normal vector **to a line '**l**' is vector '**n**', and vector n is perpendicular to the line

Fox Example:

If line '**l**' is **A**x+**B**y+C = 0, then normal vector **n**=[**A**, **B**]

If a point (x ,y) is given along with a normal vector [A, B], then what will be the Scalar equation?

Lets recall that Scalar equation is in the form **A**x+**B**y+**C**=0

Therefore, substitute the given values **A, B, x and y **and find the unknown **C**

**If, a vector equation is given, then write the equation in terms of parametric equations,then isolate "t" in both of the parametric equations and finally, set both of the t equations equations to each other and bring all terms on one side. The end result will be a scalar equation**

## Equation of Lines in Three-Space

**vector equation**or by

**parametric equation.**However, we cannot use

**scalar equation**to define a line is three space. The reason behind this is that in three space, we will use the scalar equation to defines a

**plane.**

Just for you knowledge, a plane is a two-dimensional flat surface extends infinitely in all directions

**z**.

Therefore, we will add the variable **z** to the two-space vector equation and parametric equation, which will then give us the equation of line in three space.

**Vector equation** in three space will be...

r = r0 + tm or [x, y, z] = [x0, y0, z0]+t[m1, m2, m3]

and the **Parametric equation** will be...

**x** = x0 + tm1

**y** = y0 + tm2

**z** = z0 + tm3

## In order to review Equation of Lines in Two-Space and Three-space in more detail, you can watch the following video...

## Examples for Vector Equation of a Line in Two-Space

## Example 1

**Write a vector equation for a line given a direction vector m = [2, -1] and point P0 (4, 9)**

First of all the position vector contains only ’x’ and ’y’ values.

Therefore, the vector equation for a line will be in two-space.

Given: A point and a position vector

Equation is... [x, y] = [x0, y0]+t[m1, m2]

[m1, m2] = [2, -1] (direction vector is given directly)

[x0, y0] = ???

We have to find the position vector.

if point (4, 9) is one the line, then the position vector [4, 9] has its tip on the line

A vector is also known as the difference between two points

Therefore, [x0, y0] = (4, 9) - (0 , 0)

= [4, 9]

**Therefore [x, y] = [4, 9]+t[2, -1]**

**and we can also conclude that a given point on the line can also be used as a position vector**

## Example 2

**Determine if point P(-2, -10) is on the line [x, y] = [2, -3]+t[4, 7]**

If point (-2, -10) is on the line, then the position vector [-2, -10] has its tip on the line. Therefore, the only one "t" value exits that make the equation true.

[x, y] = [2, -3]+t[4, 7]

[-2, -10] = [2, -3]+t[4, 7]

Now, Evaluate the x-coordinates and the y-coordinates. (**Hint: parametric equations can be used to evaluate the coordinates)**

**x-coordinate:**

-2 = 2+4t

-2-2 = 4t

-4 = 4t

t = -1

**y-coordinate:**

-10 = -3+7t

-10+3 = 7t

-7 = 7t

t = -1

**Since the t-values are equal, the point (-2, -10) lies on the line**

## Examples for Parametric and Scalar Equation of a Line in Two-Space

## Example 1

**Write a vector equation and a scalar equation of the line defined by the following set of parametric equations.**

**x = 2-5t**

**y= 3+6t**

From parametric equation, we can choose r0 = [2, 3] and m = [-5, 6]

Therefore, the **vector equation** is...

**[x, y] = [2, 3]+t[-5, 6]**

As told earlier, we can isolate "t" in both of the parametric equations.

x = 2-5t

x-2 = -5t

t = (-x+2)/5

y = 3+6t

y-3 = 6t

t = (y-3)/6

Equate both equations...

(-x+2)/5 = (y-3)/6

Therefore** scalar equation** is...

**6x+5y-27=0**

## Example 2

**Determine if the following lines are parallel to each other**

**Line one = [x, y] = [1, 3]+t[-3, 4]**

**Line two = [x, y] = [2, -5]+[-6, 8]**

The direction vectors of parallel lines are scalar multiples of one another

Direction vector for line one = [-3, 4]

Direction vector for line two = [-6, 8]

If the lines are parallel, therefore [-3, 4] = k[-6, 8]

-3 = -6k

-3/-6 = k

k = 1/2

4 = 8k

4/8 = k

k = 1/2

**Both of the k values are equal, therefore both of the lines are parallel to each other**

## Examples for Vector and Parametric Equation of a Line in Three-Space

## Example 1

**A line passes through points A(2, 3, 5) and B(3, 6, 2)**

**Write a vector and parametric equation for the line**

**For vector equation...**

Find a direction vector first.

Direct vector is:

m = B - A

m = [3, 6, 2] - [2, 3, 5]

m = [1, 3, -3]

And as described earlier, we can use any point on the line as a position vector

Possible position vector is [2, 3, 5]

**Therefore, the vector equation for the line is...**

**[x, y] = [2, 3, 5]+t[1, 3, -3]**

**For parametric equation...**

Using the vector equation found above, we can conclude that...

**x = 2+t**

**y = 3+3t**

**z = 5-3t**

## In conclusion (for a line in two and three space)...

For lines in two-space, a perpendicular vector and a point can be used to define a line

## EQUATION OF PLANES

In three space, a plane can be defines by a vector equation, parametric equation or a scalar equation.

## Equations of Planes

Starting from the origin. vector r0 goes to a known point on the plane, then to two vectos in the plane that are scalar multiples of the direction vectors a and b.

The resultant of these three vectors is the vector r

**Therefore, a plane in three-space can be defined with a vector equation.**

**r = r0 + ta + sb or [x, y, z] = [x0, y0, z0]+t[a1, a2, a3]+s[b1, b2, b3]**

r is the position vector for any point on the plane

r0 is a position vector for a known point on the plane

a and b are non-parallel direction vectors parallel to the planes

t and a are scalars

**Parametric equations can also be used to define a plane**

Parametric equations of a plane in three-space are:

**x = x0 + ta1 + sb1**

**y = y0 + ta2 + sb2**

**z = z0 + ta3 + sb3**

These equations are very similar in format to Parametric equations of a line as explained previously.

**What about Scalar equation?**

The scalar equation is form of **Ax+By+Cz+D = 0 **can also be used.

A, B and C represents the** normal vector [A, B, C]**

and

**D** can be found using a **given point on the line**

**Additionally, **

**X-int**of a plane can be found by setting y=z=0

**Y-int **of a plane can be found by setting x=z=0

and

**Z-int** of a plane can be found by setting x=y=0

To find if a point lies on the plane (given a scalar equation), plug in the given point in the equation and check if LHS = RHS

To find if a point lies on the plane (given a vector or parametric equation), then there exists a single set of t- and s-values that satisfy the equations.

**To find if two planes are parallel, their normals must be parallel.**

**Normals can also be found by using cross product of two directional vectors of the plane.**

**If the normals are scalar multiples of each other, then the both planes are parallel to each other.**

## In order to review Equation of Planes in more detail, you can watch the following video.

## Examples of Equations of Planes

## Example 1

**Determine the x-, y- and z- intercept of the plane defined by the scalar equation 2x+4y-6z-24 = 0**

**X-int:**

2x+4(0)-6(0)-24 = 0

2x-24 = 0

2x = 24

x = 24/2

x = 2

**Therefore, the x-int is (2, 0, 0)**

**Y-int:**

2(0)+4y-6(0)-24 = 0

4y-24 = 0

4y = 24

y = 24/4

y = 6

**Therefore, the y-int is (0, 6, 0)**

**Z-int:**

2(0)+4(0)-6z-24 = 0

-6z-24 = 0

-6z = 24

z = 24/-6

z = -4

**Therefore, the z-int is (0, 0, -4)**

## Example 2

**Determine if the point A(1,3,7) lies on the plane defined by the scalar equation 3x-2y+z = 4**

3(1)-2(3)+7 = 4

3-6+7 = 4

-3+7 = 4

4 = 4

**LHS = RHS, Therefore the point A lies on the plane**

## Example 3

**Consider the plane with direction vectors a = [2, 3, 5] and b = [1, 4, 6] through a point P0 (3, 7, 1)**

**Write the vector and parametric equations for the plane**

[x, y, x] = [x0, y0, z0]+t[a1, a2, a3]+s[b1, b2, b3]

[x0, y0, z0] = [3, 7, 1], a position vector for a point on the plane

t[a1, a2, a3] = [2, 3, 5], a direction vector

s[b1, b2, b3] = [1, 4, 6], also a direction vector

**Therefore, the vector equation will be...**

**[x, y, x] = [3, 7, 1]+t[2, 3, 5]+s[1, 4, 6]**

**The parametric equation will be...**

**x = 3+2t+s**

**y = 7+3t+4s**

**z = 1+5t+6s**

## Example 4

**Determine if the planes in each pair are parallel.**

**Plane one: [x, y, z] = [10, 3, 6]+t[2, 3, 5]+s[1, 2, 3]**

**and**

**Planetwo: [x, y, z] = [1, 5, 9]+t[4, 6, 10]+s[2, 4, 6]**

First of all, find the normals of the both planes by taking cross product of directional vectors.

The Normal of plane one is...

[-1, -1, 1]

and

The Normal of plane two is...

[-4, -4, 4]

Therefore, [-4, -4, 1] = k[-1, -1, 1]

-4 = -k; k = 4

-4 = -k; k = 4

4= k ; k = 4

**The normal of plane two is the scalar multiple of the normal of plane one.**

**Therefore, both of the planes are parallel to each other**

## PROPERTIES OF PLANES

## Equations of properties of planes

**Ax + By +Cz +D = 0**, where **n = [A, B, C]** is a normal vector to the plane

**Scalar equation**

The scalar equation is form of **Ax+By+Cz+D = 0 **is used.

A, B and C represents the** normal vector [A, B, C]**

and

**D** can be found using a **given point on the line**

**Scalar Equation...**

## In order to review the properties of plane in more detail, you can watch the following video...

## Example 1

**Consider the plane that has a normal vector n = [2, 1, 3] and contains the point P0(4, 5, 7).**

**Find the scalar equation of the plane.**

Scalar equation is int he form Ax+By+Cz+D=0

Since A, B and C are the values of the normal vector,

[A, B, C] = [2, 1, 3]

Therefore, substitute the given normal vector and the point into the Scalar equation.

Plug in normal vector: 2x+y+3z+D = 0

Plug in the given point to find D:

2(4)+5+3(7)+D = 0

34+D = 0

D = -34

**Therefore, the scalar equation is 2x+y+3z-34 = 0**

## Example 2

**Is vector a = [5, 4, -23] parallel to the plane 3x+2y+z-34 = 0 ?**

If dot product of vector a and the normal vector of the plane is equal to zero, then vector a is parallel to the plane

First of all find the normal vector.

We know that [A, B, C] = normal vector.

In the given plane, A = 3, B = 2 and C = 1

[A, B, C] = [3, 2, 1]

Therefore, [A, B, C]*[5, 4, -23] = 0

[3, 2, 1]*[5, 4, -23] = 0

(3*5)+(2*4)+(1*-23)=0

0 = 0

**Therefore, vector a is parallel to the given plane.**

## Example 3

**Find the scalar equation of the plane given below:**

**[x, y, z] = [2, -5, 4]+t[7, -3, 5]+s[2, -3, -2]**

Since scalar equation is in the form Ax+By+Cz+D = 0 and A, B, C represents the normal vector, we need to find the normal vector first.

**Normal vector is any vector perpendicular to the plane.**

Therefore, cross product can of two given vectors, t and s can be used to find the normal vector.

[7, -3, 5]x[2, -3, -2]

The cross product comes out to be [21, 24, -15], which can we further simplified to [7, 8, -5].

Therefore, n = [7, 8, -5] = [A, B, C]

From our previous lesson, we know that vector [2, -5, 4] also represents the point (2, -5, 4)

Now we have normal vector and a point, we can plug in the value of normal vector and the point into scalar equation to find the value of D.

Ax+By+Cz+D = 0

7x+8y-5z+D = 0

7(2)+8(-5)-5(4)+D = 0

-46+D = 0

D = 46

**Therefore, the scalar equation is 7x+8y-5z+46 = 0**

## INTERSECTION OF PLANES

## Intersection of two planes in three-space

**You can verify these situations algebraically by solving linear system of equations.**

**Please follow the following examples for better understanding of intersection of two planes in three-space.**

**In the above example,**

Multiply one of the equation by a number which will help us eliminate one of the variable from both of the equations.

After subtracting/ adding both of the equations, a third equation is formed.

In the third equation, you can set one of the variable from two left over variables to be equal to** t.**

**"t"** will help us find the parametric equations of the line.

Now solve the third equation in terms of **non t** variable.

Now, you will have two variables **in terms of t.**

Plug in both of them in the either of the equations of plane to find the third variable.

Now you will have all of the variables as parametric equations which can be re-written in form of vector equation of the line.

## Intersection of three planes in three-space

**A system of three planes is consistent if it has one or more solutions.**

There are three possible geometric models of three planes which are consistent.

**A system of three planes is inconsistent if it has no solution.**

There are three possible geometric models of three planes which are inconsistent.

**In order to check if the normals are parallel, each of the given equation of the planes must be scalar multiple of each other.**

**To check if the normals are coplanar, triple scalar product can be used. If the triple scalar product is equal to zero, then the normals are coplanar.**

Two more equations can be drived using by subtracting/ adding first two equations and last two equations.

Now, solve in terms of parametric equations and re-write them in terms of vector equation of a line.

**Please follow the following examples for better understanding of intersection of three planes in three-space.**