# Vector Geometry – Explanation & Examples

Modeling is important in all branches of mathematics, including **vector geometry**. This is:

*â€œThe study of geometric representations of vectors, namely the representation as directed linesegments or arrows.” *

In this topic, we will discuss the following aspects of vector geometry:

- What is a Vector in Geometry?
- Vector Definition in Geometry

## Â

## What is a Vector in Geometry?

Quantities with both magnitude and direction are known as vectors. We can use a graph to represent vectors visually. For example, a vector connecting two points, A and B, is called:

**AB**

A vector in standard position will have the origin as its starting point.

In component or column form, vectors are written in an ordered pair (x, y). A vector written in this form begins at the origin and ends at the ordered pair’s point.

The **negative **of a given **vector** is found by reversing the direction of the vector. In this case, its magnitude (or length) is the same as that of the original vector.

For example, the vector:

**BA** = –**AB**

is the negative of the vector **AB,** and:

||**BA**|| = ||-**AB**|| = ||**AB**||

### Vector Definition in Geometry

Given two points, P and Q, the arrow from P to Q will have both length and direction.

Let us assume that P and Q be two arbitrary points in space **R**^{3}. The line segment from P to Q is denoted as **PQ. **In geometry, this is known as the **vector** from P to Q.

This vector will have magnitude and direction. Point P is called the **tail **(or the initial point) of the vector **PQ, **while the point Q is called the **tip **(or the head or the terminal point) of the vector **PQ**. Its length is denoted as ||**PQ**||.

Suppose the point P has coordinates (x_{1}, y_{1}), and the point Q has coordinates (x_{2}, y_{2}) in the plane **R**^{2}. Then the length of the vector **PQ** is defined by the steps given below.

**Step 1:** First, subtract the first component of point P from the first component of point Q. Then, find the square of the resulting difference.

**Step 2:** Similarly, subtract the second component of point P from the second component of the point Q and square that number.

**Step 3:** Then, add the two squares together.

**Step 4:** Finally, take the square root of the number you found in step three. This scalar number will be the length of the vector.

- Note that the length or magnitude of a vector is a scalar quantity.

*Examples*

Now, let’s try a few examples to practice working with vector geometry.

*Example 1*

Given two points, O at the origin, (0,0), and A with the coordinates (3,2), determine the length ||OA||.

__Solution:__

Comparing components, we get

(x_{1}, y_{1}) = (0,0) and (x_{2}, y_{2}) = (3, 2).

Using the above-mentioned formula for finding the length of two a vector, we have:

â€–**OA**â€– = âˆš((x_2-x_1 )^2+(y_2 – y_1 )^2 ).

Now, substituting the above values gives us:

=âˆš((0-3)^2+(0-2)^2 ).

Simplifying, we get:

â€–**OA**â€– = âˆš((-3)^2+(-2)^2 )

Additional simplification gives us:

â€–**OA**â€–Â = âˆš(9+4)

â€–**OA**â€–Â = âˆš13.

Thus, ||OA|| = âˆš13 is the vector’s length connecting the two points.

Note that if the initial point or the tail of a vector is at the origin, i.e., if the vector is written in standard form, then the magnitude of **OA** is:

||**A**|| = âˆš(x^{2} + y^{2})

Where **A** = **OA** = (x, y).

*Example 2*

Given two points, A = (3, 2) and B = (2, 2), determine the vector AB length.

__Solution__*:*

Comparing components, we get

(x_{1}, y_{1}) = (3, 2) and (x_{2}, y_{2}) = (2, 2).

Using the above-mentioned formula for finding the length of two a vector, we have

â€–**AB**â€– = âˆš((x_2-x_1 )^2+(y_2 – y_1 )^2 )

*Â *Now, substituting the above values gives us:

â€–**AB**â€–Â =Â âˆš((3-2)^2+(2-2)^2 )

Simplifying, we get:

â€–**AB**â€–Â =Â âˆš((1)^2+(0)^2 )

â€–**AB**â€–Â =Â âˆš1+ 0

â€–**AB**â€–Â = 1

Thus, ||**AB**|| = 1 is the vector’s length connecting the two points.

*Example 3*

In the figure below, two vectors **AB** = 2**a** and **BC** = 3**b,** are given. Point D is the midpoint of BE and E is the midpoint of BC.

Using the information given above, determine the following vectors in terms of **a** and **b:**

**(a)****BC****(b)****BD****(c)****AD**

__Solution:__

- (a) Using the triangle law of vector addition, we have

**BC** = **BA** + **AC**

We note the relationship between **BA** and the vector of known length, **AB**:

= (-**AB**) + **AC**

Substituting the known values of** AB **and** AC **gives us**:**

= -2**a** + 3**b**.

Thus, **BC** = -2**a** + 3**b** is the length of the vector.

- (b) Since E is given as the midpoint of BC, and D is the midpoint of BE, then D one-fourth of the distance BC:

**BD** = Â¼ **BC**

Substituting the value of** BC **found in part (a) gives us:

= Â¼ (-2**a** + 3**b**)

= -1/2 **a** + 3/4 **b**.

Thus, **BD** = -1/2 **a** + 3/4 **b** is the length of the vector.

- (c) Using triangle law of vector addition, we have:

**AD** = **AB** + **BD**

Substituting the known values of** AB **and** AC **gives us:

= 2**a** + (-1/2 **a** + 3/4 **b**)

This can be simplified:

= (2**a** + -1/2 **a**) + 3/4 **b**

= 3/2 **a** + 3/4 **b**.

Thus, **AD** = 3/2 **a** + 3/4 **b** is the length of the vector.

*Example 4*

Let A = (2, 0, -4) and B = (2, 1, -2) be two points in the space **R**^{3}. If 2**a** – ||**b**||**b** = 3/2 (**a** â€“ 2**x**), where **a** = **OA**, **b** = **OB** and **x** = **OX** and X = (x_{1}, x_{2}, x_{3}), find the values of x_{1}, x_{2} and x_{3}.

__Solution__:

We are given the relationship:

2**a** – ||**b**||**b** = 3/2 (**a** â€“ 2**x**),

Where **a** = (2, 0, -4), **b** = (2, 1, -2) and **x** = (x_{1}, x_{2}, x_{3}).

Firstly we will evaluate the norm (magnitude) ||**b**|| by using previous definitions and the procedure for finding the magnitude of a vector whose starting point is at the origin. This gives us:

Â Â Â ||**b**|| = âˆš(2^{2} + 1^{2} + (-2)^{2})

= âˆš(4 + 1 + 4)

= âˆš9

=3

Therefore, ||**b**|| = 3. Now we have:

Â ||**b**|| **b **= 3 (2, 1, -2)

= (3âˆ™2, 3âˆ™1, 3âˆ™(-2))

= (6, 3, -6)

Similarly, we have:

2**a **= 2 (2, 0, -4) = (4, 0, -8).

Therefore, the left-hand side will become

Â Â Â Â 2**a** – ||**b**||**b** = (4, 0, -8) – (6, 3, -6)

= (4-6, 0-3 , -8-(-6))

= (-2, -3, -8+6)

= (-2, -3, -2).

The right-hand side will become:

Â 3/2 (**a** â€“ 2**x**) = 3/2 ((4, 0, -8)â€“ 2(x_{1}, x_{2}, x_{3}))

= 3/2 ((4, 0, -8) â€“ (2x_{1}, 2x_{2}, 2x_{3}))

= 3/2 (4-2x_{1}, 2x_{2}, -8-2x_{3}).

Now substituting these values into the left- and right-hand sides of the equation, we get:

2**a** – ||**b**||**b** = 3/2 (**a** â€“ 2**x**)

(-2, -3, -2) = 3/2 (4-2x_{1}, 2x_{2}, -8-2x_{3})

2 (-2, -3, -2) = 3 (4-2x_{1}, 2x_{2}, -8-2x_{3})

(-4, -6, -4) = (3(4-2x_{1}), 3(2x_{2}), 3(-8-2x_{3})).

Comparing the x, y, and z components give us:

-4 = 3(4-2x_{1})

-4= 12 – 6x_{1}

6x_{1} =12 â€“ 4

6x_{1}= 8

x_{1} = 8/6

Therefore, x_{1}= 4/3.

-6 = 3(2x_{2})

-6= 6x_{2}

x_{2} = -6/6

Therefore, x_{2}= -1

-4 = 3(-8-2x_{3})

-4= -24 -6x_{3}Â

6x_{3} = -24 + 4

6x_{3}= -20

x_{3} = -20/6

Therefore, x_{3}= -10/3.

Thus:

x_{1} = 4/3, x_{2} = -1, and x_{3} = -10/3.

*Practice Questions*

- Given two points, V = (2, 5, 1) and C = (3, -2, 1), determine the length of the vector
**VC**. - Given two points, G = (5, 5) and H = (4, -10), determine ||
**OG**|| and ||**OH**||^{2}, where O = (0, 0) is the origin. - In the triangle ABC, CB = 4CX, XA = 5XY, and Z is the midpoint of AB. Let
**CX**=**m**and**BZ**=**n**.

**Â **

** **

**(a)**Determine the length of**XB**and**XA**in terms of**m**and**n****(b)**Show that**CZ**= (4**m**+**n**)**(c)**Evaluate â€–**CY**â€–/â€–**CZ**â€– and (area ACY)/(area ACZ)

- Let ABCDEF be a regular hexagon, and let
**AB**=**n**.

**(a)**Explain why the vector**ED**=**n**.**(b)**If**BC**=**m**and**CD**=**p,**find**AC**and**AD****(c)**Find**FD**.

- Given a straight line ABC, let
**BC**= 3**AB**. If**OA**=**a**and**AB**=**b,**express**OC**in terms of**a**and**b**.

**Â **__Answers__

- ||
**VC**|| = 5âˆš2 - ||
**OG**|| = 5âˆš2, ||**OH**||Â² = 116 - In the given triangle,

**(a)****XB**= 3**m**and**XA**=3**m**+ 2**n****(b) CZ=CB+BZ=4m+n****(c)**The values are: â€–**CY**â€–/â€–**CZ**â€– = 2/5, and (area ACY)/(area ACZ) = 2/5

- Â

4. Let ABCDEF be a regular hexagon and **AB** = **n**.

**(a)**It is opposite**AB**.**(b)**If**BC**=**m**and**CD**=**p**. Then**AC**=**n**+**m**and**AD**=**n**+**m**+**p****(c)****FD**= -(**m**+**n**) (i.e, negative of the vector**AC**).

**OC**=**a**+4**b**