# 5. Dot Product (aka Scalar Product) in 2 Dimensions

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If we have any 2 vectors **P** and **Q**, the ** dot product** of **P** and **Q** is given by:

P`*`Q=|P| |Q|cosθ

where

|P|and|Q|are the magnitudes ofPandQrespectively, andθ is the angle between the 2 vectors.

The dot product of the vectors **P** and **Q** is also known as **the scalar product** since it always returns a **scalar value**.

The term dot product is used here because of the **•** notation used and because the term "scalar product" is too similar to the term "**scalar multiplication**" that we learned about earlier.

### Example 1

a. Find the dot product of the force vectors ** F _{1} **= 4 N and

**F**= 6 N acting at 40° to each other as in the diagram.

_{2}Answer

The dot product is:

F_{1}`*` F_{2}=

|Fcos θ_{1}| |F_{2}|= (4)(6) cos 40°

= 18.38 N

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b. Find the dot product of the vectors **P** and **Q** if ** |P| **= 7 units and |**Q| **= 5 units and they are acting at right angles to each other.

Answer

The required dot product is:

P `*` Q=

|P| |Q|cos θ= 7 × 5 × cos 90°

= 0

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The second example illustrates an important point about how scalar products can be used to find out if vectors are acting at right angles, as follows.

## Dot Product and Perpendicular Vectors

If 2 vectors act perpendicular to each other, the **dot product **(ie **scalar product**) of the 2 vectors has value **zero**.

This is a useful result when we want to check if 2 vectors are actually acting at right angles.

## Dot Products of Unit Vectors

For the unit vectors **i** (acting in the *x*-direction) and **j** (acting in the *y*-direction), we have the following dot (ie scalar) products (since they are perpendicular to each other):

i `*` j = j `*` i= 0

### Example 2

What is the value of these 2 dot products:

a. **i `*` i **

b. ** j `*` j **

Answer

(a)

i `*` i=

| i | | i |cos 0°= 1 × 1 × 1

= 1

(b)

j • j=

| j | | j |cos 0°= 1 × 1 × 1

= 1

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## Alternative Form of the Dot Product

Recall that vectors can be written using scalar products of unit vectors.

If we have 2 vectors **P** and **Q** defined as:

P =ai+bj

Q=ci +dj,

where

a,b,c,dare constants;

iis the unit vector in thex-direction; and

jis the unit vector in they-direction,

then it can be shown that the **dot product** (**scalar product**) of **P** and **Q** is given by:

P `*` Q=ac+bd

Answer

For convenience, let vectors **P** and** Q** be as shown on the graph. (**P** is horizontal.)

Draw the altitude from vector **P** to the terminal point (*c*, *d*) of **Q**.

Simple trigonometry gives us:

`cos\ θ = c / | Q |`, so

c= |Q| cosθ

and

`sin\ θ = d / | Q |`, so

d= |Q| sinθ

Since **P** is horizontal,

a= |P|

b= 0

So

ac+bd

= a|Q| cosθ+ 0=

| P ||Q| cosθ

Therefore

P `*` Q=| P ||Q| cosθ=ac+bd

This result can be generalized for **P** and **Q** in any orientation.

### Example 3 - Alternative Form of the Dot Product

Find **P • Q** if

P= 6i+ 5jand

Q= 2i− 8j

Answer

Using the alternative way of finding dot product, we have:

P `*` Q

=(6i+ 5j)`*`(2i− 8j)= (6 × 2) + (5 × -8)

= 12 − 40

= −28

Now we see another use for the dot product − finding the angle between vectors.

## Angle Between Two Vectors

We can use the dot product to find the angle between 2 vectors. For the vectors **P** and **Q**, the dot product is given by

P `*` Q=|P| |Q|cos θ

Rearranging this formula we obtain the cosine of the angle between **P** and **Q**:

`cos\ theta=(P * Q)/(|P||Q|)`

To find the angle, we just find the inverse cosine of the expression on the right.

So the angle θ between 2 vectors **P** and **Q** is given by

`theta=arccos((P * Q)/(|P||Q|))`

### Example 4

Find the angle between the vectors **P** = 3 **i** − 5 **j **and **Q** = 4 **i + **6 **j**.

Answer

We aim to find angle POQ. We observe that it is an obtuse angle (between 90° and 180°).

We use the formula we just derived:

`theta=arccos((P*Q)/(|P||Q|))`

Taking the numerator first, we have:

P `*` Q= (3

i− 5j)•(4i +6j)= (3 × 4) + (-5 × 6)

= -18

And now for the denominator (the magnitudes are found using Pythagoras' Theorem):

|P| |Q|^{ }= √(3

^{2}+ (-5)^{2}) × √(4^{2}+ 6^{2})= 42.048

So

θ= arccos(−18 ÷ 42.048)

Therefore the angle between the vectors **P** and **Q** is

θ= 115.3°

This looks a reasonable answer considering the diagram above (which is drawn to scale).

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