# Scalar (or Dot) Product of Two Vectors

We have already studied about the addition and subtraction of vectors. Vectors can be multiplied in two ways, scalar or dot product where the result is a scalar and vector or cross product where is the result is a vector. In this article, we will look at the scalar or dot product of two vectors.

## Definition

The scalar or dot product of two non-zero vectors $$\vec{a}$$ and $$\vec{b}$$, denoted by $$\vec{a}$$.$$\vec{b}$$ is

$$\vec{a}$$.$$\vec{b}$$ = |$$\vec{a}$$| |$$\vec{b}$$|Â $$\cos \theta$$

whereÂ $$\theta$$ is the angle betweenÂ $$\vec{a}$$ and $$\vec{b}$$ and 0Â â‰¤Â $$\theta$$Â â‰¤Â $$\pi$$ as shown in the figure below.

It is important to note that if eitherÂ $$\vec{a}$$ =Â $$\vec{0}$$ orÂ $$\vec{b}$$ =Â $$\vec{0}$$, thenÂ $$\theta$$ is not defined, and in this case

$$\vec{a}$$.$$\vec{b}$$ = 0

## Important Observations

• $$\vec{a}$$.$$\vec{b}$$ is a real number
• If $$\vec{a}$$ and $$\vec{b}$$ are two non-zero vectors, thenÂ $$\vec{a}$$.$$\vec{b}$$ = 0 if and only ifÂ $$\vec{a}$$ and $$\vec{b}$$ are perpendicular to each other.

$$\vec{a}$$.$$\vec{b}$$ = 0Â $$\Longleftrightarrow$$Â $$\vec{a}$$Â âŠ¥ $$\vec{b}$$

• IfÂ $$\theta$$ = 0, thenÂ $$\vec{a}$$.$$\vec{b}$$ = |$$\vec{a}$$| |$$\vec{b}$$|. In particular,

$$\vec{a}$$.$$\vec{a}$$ = $$\vec{|a|}^2$$, as in this caseÂ $$\theta$$ is 0.

• IfÂ $$\theta$$ = $$\pi$$, thenÂ $$\vec{a}$$.$$\vec{b}$$ = – |$$\vec{a}$$| |$$\vec{b}$$|.Â In particular,

$$\vec{a}$$.$$\vec{(- a)}$$ = – $$\vec{|a|}^2$$, as in this caseÂ $$\theta$$ is $$\pi$$.

• Following from observations 2 and 3, for mutually perpendicular vectorsÂ $$\hat{i}$$,Â $$\hat{j}$$, andÂ $$\hat{k}$$, we have

$$\hat{i}$$.$$\hat{i}$$ =Â $$\hat{j}$$.$$\hat{j}$$ =Â $$\hat{k}$$.$$\hat{k}$$ = 1

And, $$\hat{i}$$.$$\hat{j}$$ =Â $$\hat{j}$$.$$\hat{k}$$ =Â $$\hat{k}$$.$$\hat{i}$$ = 0

• The angle between two non-zero vectors is given by,

$$\cos \theta$$ =Â $$\frac{\vec{a}.\vec{b}}{|\vec{a}| |\vec{b}|}$$

Or, $$\theta$$ =Â  cos-1Â [$$\frac{\vec{a}.\vec{b}}{|\vec{a}| |\vec{b}|}$$]

• The scalar product is commutative, i.e.

$$\vec{a}$$.$$\vec{b}$$ =Â $$\vec{b}$$.$$\vec{a}$$

## Two Important Properties of Dot Product

### Property 1: Distributivity of a scalar or dot product over addition

IfÂ $$\vec{a}$$,Â $$\vec{b}$$ and $$\vec{c}$$ are any three vectors, then

$$\vec{a}$$.($$\vec{b}$$ +Â $$\vec{c}$$) =Â $$\vec{a}$$.$$\vec{b}$$ +Â $$\vec{a}$$.$$\vec{c}$$

### Property 2:

IfÂ $$\vec{a}$$ and $$\vec{b}$$ are any two vectors andÂ $$\lambda$$ is a scalar, then

($$\lambda$$$$\vec{a}$$).$$\vec{b}$$ =Â $$\lambda$$($$\vec{a}$$.$$\vec{b}$$) =Â $$\vec{a}$$.($$\lambda$$$$\vec{b}$$)

If twoÂ vectorsÂ $$\vec{a}$$ and $$\vec{b}$$ are given in the component form as a1$$\vec{i}$$ + a2$$\vec{j}$$ + a3$$\vec{k}$$ andÂ b1$$\vec{i}$$ + b2$$\vec{j}$$ + b3$$\vec{k}$$, then their scalar or dot product is as follows:

$$\vec{a}$$.$$\vec{b}$$ = (a1$$\vec{i}$$ + a2$$\vec{j}$$ + a3$$\vec{k}$$).(b1$$\vec{i}$$ + b2$$\vec{j}$$ + b3$$\vec{k}$$)

Or, $$\vec{a}$$.$$\vec{b}$$ =Â a1$$\vec{i}$$(b1$$\vec{i}$$ + b2$$\vec{j}$$ + b3$$\vec{k}$$) +Â a2$$\vec{j}$$(b1$$\vec{i}$$ + b2$$\vec{j}$$ + b3$$\vec{k}$$) +Â a3$$\vec{k}$$(b1$$\vec{i}$$ + b2$$\vec{j}$$ + b3$$\vec{k}$$)

On solving further, and using the properties 1 and 2 above, we get

$$\vec{a}$$.$$\vec{b}$$ =Â a1b1($$\vec{i}$$.$$\vec{i}$$) + a1b2($$\vec{i}$$.$$\vec{j}$$)Â + a1b3($$\vec{i}$$.$$\vec{k}$$)Â + a2b1($$\vec{j}$$.$$\vec{i}$$)Â + a2b2($$\vec{j}$$.$$\vec{j}$$)Â + a2b3($$\vec{j}$$.$$\vec{k}$$)Â + a3b1($$\vec{k}$$.$$\vec{i}$$)Â + a3b2($$\vec{k}$$.$$\vec{j}$$)Â + a3b3($$\vec{k}$$.$$\vec{k}$$)

Using Observation 5 mentioned above, we get

$$\vec{a}$$.$$\vec{b}$$ =Â a1b1 +Â a2b2 +Â a3b3

## Solved Problems for You

Question: Find the angle between two vectors $$\vec{a}$$ and $$\vec{b}$$ with magnitudes $$\sqrt{3}$$ and 2 ,
respectively havingÂ $$\vec{a}$$.$$\vec{b}$$ = $$\sqrt{6}$$.

Solution: By definition of the scalar or dot product of vectors, we know that

$$\vec{a}$$.$$\vec{b}$$ = |$$\vec{a}$$| |$$\vec{b}$$|Â $$\cos \theta$$

whereÂ $$\theta$$ is the angle betweenÂ $$\vec{a}$$ and $$\vec{b}$$. In this question, we have,

|$$\vec{a}$$| =Â $$\sqrt{3}$$,Â |$$\vec{b}$$| = 2, andÂ $$\vec{a}$$.$$\vec{b}$$ = $$\sqrt{6}$$

Replacing these values in the formula, we get

$$\vec{a}$$.$$\vec{b}$$ = |$$\vec{a}$$| |$$\vec{b}$$|Â $$\cos \theta$$
$$\Longleftrightarrow$$Â $$\sqrt{6}$$ =Â $$\sqrt{3}$$ x 2 xÂ $$\cos \theta$$
$$\Longleftrightarrow$$Â $$\sqrt{3}$$ xÂ $$\sqrt{2}$$ =Â $$\sqrt{3}$$ x $$\sqrt{2}$$ xÂ $$\sqrt{2}$$ x $$\cos \theta$$

Cancelling the common terms on both sides, we get

1 =Â $$\sqrt{2}$$ xÂ $$\cos \theta$$
Or, $$\cos \theta$$ =Â $$\frac{1}{ \sqrt{2}}$$
Therefore, $$\theta$$ =Â $$\frac{\pi}{4}$$
Hence, the angle between theÂ vectors $$\vec{a}$$ and $$\vec{b}$$ isÂ $$\frac{\pi}{4}$$.

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