At this point, you might have an understanding of vectors, the importance of vectors, and why we use them in our daily life? Let's have a small recap, a vector is a line that has a direction and magnitude. Both properties define a vector. It means that changing one property changes the whole vector. Few applications of vectors are calculation of velocity, newton's law of fluid mechanics, GPS, aerodynamics, and many more.

Let's say you have two vectors, your teacher asks you to combine both vectors. Now, what would you do? Add them? Adding them doesn't mean combine, for that we use a fundamental way which is known as the dot product. Another name of the dot product is the scalar product. So, what is a dot product? Dot product tells us the quantity of one vector from the second vector's direction perspective. In easy words, it tells how much one vector is in amount but in the direction of the second vector. Furthermore, it also combines two vectors, it joins the tail of one vector to the tail of the second vector. In addition, you also find how much your first vector deviates from the second vector.

The scalar product or dot product of two vectors \vec { u } and \vec { v } is equal to:

\vec { u } . \vec { v } = \left | \vec { u } \right | . \left | \vec { v } \right | . \cos { \alpha }

Where,

\left | \vec { u } \right | = \sqrt { { { x }_{ 1 } }^{ 2 } + { { y }_{ 1 } }^{ 2 } + { { z }_{ 1 } }^{ 2 } }

\left | \vec { v } \right | = \sqrt { { { x }_{ 2 } }^{ 2 } + { { y }_{ 2 } }^{ 2 } + { { z }_{ 2 } }^{ 2 } }

In some text books, dot product formula might be written as:

\vec { u } . \vec { v } = \sqrt { { { x }_{ 1 } }^{ 2 } + { { y }_{ 1 } }^{ 2 } + { { z }_{ 1 } }^{ 2 } } . \sqrt { { { x }_{ 2 } }^{ 2 } + { { y }_{ 2 } }^{ 2 } + { { z }_{ 2 } }^{ 2 } } . \cos { \alpha }

Both are same formulas and will gave you the same result. The only difference between both formulas is that the first formula, the magnitude of both vectors \vec { u } & \vec { v } are written in magnitude form. You need to calculate the magnitude of both vectors separately and then use it in that formula. On the other hand, the second formula has magnitude formula infused. In simple words, you don't need extra effort, just replace all the variables with their values and that is it!

One of the keen interest points of the dot product is the angle between both vectors. Many researchers are only concern with the degree of deviation. The angle part of the dot product is defined by the \cos { \alpha }. If both vectors are perpendicular to each other, the dot product will be equal to zero. No matter how big the magnitude of the vector is, if both vectors are perpendicular to each other, it will always result in zero. This is because \cos { \alpha } is zero at { 90 }^{ \circ }. However, if both vectors are opposite to each other, it will always result in a negative number because \cos { { 180 }^{ \circ } } is equal to -1. Last but not least, if the angle between both vectors is zero then the dot product will be maximum and why is it? Because \cos { { 0 }^{ \circ } } is equal to +1.

Example

\vec { u } = (3, 0) \qquad \vec { v } = (5, 5) \qquad \widehat{ uv } = { 45 }^{ \circ }

\vec { u } . \vec { v } = \sqrt { { 3 }^{ 2 } + { 0 }^{ 2 } } . \sqrt { { 5 }^{ 2 } + { 5 }^{ 2 } } . \cos { { 45 }^{ \circ } }

\vec { u } . \vec { v } = 3 . 5 . \sqrt { 2 } . \frac { \sqrt { 2 } }{ 2 } = 15

 

It can also be expressed as:

\vec { u } . \vec { v } = { u }_{ 1 } . { v }_{ 1 } + { u }_{ 2 } . { v }_{ 2 }

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Example

\vec { u } = (3, 0) \qquad \vec { v } = (5, 5)

\vec { u } . \vec { v } = 3. 5 + 0 . 5 = 15

 

Magnitude of a Vector

\left | \vec { u } \right | = \sqrt { \vec { u } . \vec { u } } = \sqrt { { u }_{ 1 } . { u }_{ 1 } + { u }_{ 2 } . { u }_{ 2 } } = \sqrt { { u }_{ 1 }^{ 2 } + { u }_{ 2 }^{ 2 } }

\vec { u } = (3, 0) \qquad \vec { v } = (5, 5)

\left | \vec { u } \right | = \sqrt { \vec { u } . \vec { u } } = \sqrt { 3 . 3 + 0 . 0 } = 3

\left | \vec { v } \right | = \sqrt { \vec { v } . \vec { v } } = \sqrt { 5 . 5 + 5 . 5 } = 5 \sqrt { 2 }

 

Properties of Dot Product

Commutative

\vec { u } . \vec { v } = \vec { v } . \vec { u }

Associative

k . (\vec { u } . \vec { v } ) = (k . \vec { u }). \vec { v }

Distributive

\vec { u } . (\vec { v } + \vec { w }) = \vec { u } . \vec { v } + \vec { u } . \vec { w }

The dot product of a non-zero vector is always positive

\vec { u } \neq 0 \Rightarrow \vec { u } . \vec { u } > 0

Example

Calculate the scalar product of the following vectors:

1. \vec { u } = (3, 4) and \vec { v } = (-8, 6)

\vec { u } . \vec { u } = [3 . (-8)] + (4 . 6) = 0

 

2. \vec { u } = (5, 6) and \vec { v } = (-1, 4)

\vec { u } . \vec { u } = [5 . (-1)] + (6 . 4) = 19

3. \vec { u } = (3, 5) and \vec { v } = (-1, 6)

\vec { u } . \vec { u } = [3 . (-1)] + (5 . 6) = 27

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If B = \left \{ \vec { u } , \vec { v } \right \}  is a basis of vectors in the plane, such that \left | \vec { u } \right | = \left | \vec { v } \right | = 2 and \cos { (\vec { u }, \vec { v }) } = \frac { 1 }{ 2 } and:

\vec { x } = 3 \vec { u } + 2 \vec { v } and \vec { y } = \vec { u } + 2 \vec { v }

Calculate \vec { x } . \vec { y }.

\vec { x } . \vec { y } = (3\vec { u } + 2 \vec { v }) . (\vec { u } + 2 \vec { v })

3 {( \vec { u } )}^{ 2 } + 6 \vec { u } . \vec { v } + 2 \vec { v } . \vec { u } + 4 { (\vec { v }) }^{ 2 }

The scalar product is commutative.

3 {( \vec { u } )}^{ 2 } + 8 \vec { u } . \vec { v } + 4 { (\vec { v }) }^{ 2 }

\cos { (\vec { u }, \vec { u }) } = \cos { (\vec { v }, \vec { v }) } = 1

= 3 . 2 . 2 . 1 + 8 . 2 . 2 . \frac { 1 }{ 2 } + 4 . 2 . 2 . 1 = 12 + 16 + 16 = 44

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Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.