Product of Two Vectors

If ā and b̄ are any two vectors, then the scalar product of these vectors is

ā · b̄ = |ā| |b̄| cos θ = ab cos θ, where θ is the angle between ā and b̄.

Two important properties of scalar product :

Property:  (Distributivity of scalar product over addition) 
Let `vec a,  vec b` and `vec c` be any three vectors , then `vec a (vec b + vec c) = vec a . vec b + vec a. vec c`

Property:  Let `vec a` and `vec b` be any two vectors, and l be any scalar. Then
`(lambda vec a). vec b = (lambda vec a).vec b = lambda (vec a . vec b) = vec a . (lambda vec b)`
If two vectors `vec a`  and  `vec b` are given in component form as

`a_1 hat i + a_2hat j + a_3 hat k`  and  `b_1 hat i + b_2hat j + b_3 hat k .`,  then their scalar product is given as
`vec a . vec b` = `(a_1hat i + a_2 hat j + a_3 hat k) . (b_1 hat i + b_2 hat j + b_3 hat k)`
= `a_1hat i . (b_1 hat i + b_2 hat j + b_3 hat k) + a_2 hat j . (b_1 hat i + b_2 hat j + b_3 hat k) + a_3 hat k . (b_1 hat i + b_2 hat j + b_3 hat k)`
= `a_1 b_1 (hat i . hat i) + a_1b_2 (hat i . hat j) +a_1b_3 (hat i . hat k)+a_2b_1 (hat j . hat i) + a_2 b_2 (hat j . hat j) + a_2b_3 (hat j . hat k) + a_3b_1 (hat k . hat i) + a_3b_2 (hat k . hat j) + a_3b_3 (hat k . hat k)`   (Using the above Properties 1 and 2)
= `a_1b_1 + a_2b_2 + a_3b_3 `                            (Using Observation 5)

Thus `vec a . vec b = a_1b_1 + a_2b_2 + a_3b_3`

Observations: 
1) `vec a xx vec b` is a vector. 

2) Let `vec a` and `vec b` be two  nonzero vectors. Then `vec a xx vec b = vec 0` if and only if `vec a` and `vec b` are parallel (or collinear) to each other, i.e.,
`vec a xx vec b = vec 0 <=>  vec a||vec b`
In particular , `vec a xx vec b = vec 0` and  `vec a xx (-vec a) = vec 0`, since in the first situation, θ = 0 and in the second one, θ = π, making the value of sinθ to be 0.

3) If  , θ = `π/2` then `vec a xx vec b = |vec a||vec b|.`

4)  In view of the Observations 2 and 3, for mutually perpendicular unit vectors `hat i , hat j` and` hatk` fig.

`hat i xx hat i = hat j xx hat j = hat k xx hat k = vec 0`
`hat i xx hat j = hat k ,  hat j xx hat k = hat i , hat k xx hat i = hat j`

5) In terms of vector product, the angle between two vectors `vec a` and `vec b` may be given as
`sin theta = (|vec a xx vec b|)/(|vec a||vec b|)`

6)  It is always true that the vector product is not commutative, as `vec a xx vec b = - vec b xx vec a.`
Indeed `vec a xx vec b = |vec a||vec b|  sin theta   hat n_1`, where `vec b, vec a   "and"   hat n_1` form a right handed system, i.e., θ is traversed from `vec b   "to"  vec a`  in following fig. 

While , `vec b xx vec a = |vec a||vec b|   sin theta   hat n_1` , where `vec b, vec a  "and"   hat n_1` form a right handed system i.e. θ is traversed from `vec b " to"   vec a`,
Fig.

Thus, if we assume `vec a` and `vec b` to lie  in the plane of the paper, then `hat n` and `hat n_1` both will be perpendicular to the plane of the paper. But, `hat n`  being directed above the paper while `hat n_1` directed below the paper i.e. `hat n_1 = -hat n.`
Hence `vec a xx vec b = |vec a||vec b|  sin theta  hat n`
=` - |vec a||vec b|  = sin theta  hatn_1` 
= -`vec b xx vec a =`

7) In view of the Observations 4 and 6, we have 
`hat j xx hat i = - hat k , hat k xx hat j = - hat i` and  `hat i xx hat k = -hat j.`

 8) If `vec a` and `vec b` represent the adjacent sides of a triangle then its area is given as `1/2 |vec a xx vec b|`.
By definition of the area of a triangle, we have from fig. 

Area of triangle ABC =  `1/2` AB .CD
But AB = `|vec b|` (as given ), and  CD = `|vec a|   sin θ. `

Thus,  Area of triangle ABC =  `1/2 |vec b||vec a| sin theta = 1/2 | vec a xx vec b|`. 

9)  If `vec a  "and"   vec b` represent the adjacent sides of a  parallelogram, then its area is given by |vec a xx vec b|.From the following Fig. we have 

\[\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\]

Special Cases

• Perpendicular → \[\overline{\mathrm{a}}\cdot\overline{\mathrm{b}}=0\]

• Parallel → \[\mathbf{\overline{a}}\cdot\mathbf{\overline{b}}=\mathbf{ab}\]

Projection

• Scalar= \[\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\]

• Vector \[=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^{2}}\cdot\mathbf{b}\]