`=>` In right handed coordinate system system, when the positive x-axis is rotated counterclockwise into the positive y-axis, a right handed (standard) screw would advance in the direction of the positive z-axis (Fig(i)).

`=>` In a right handed coordinate system, the thumb of the right hand points in the direction of the positive z-axis when the fingers are curled in the direction away from the positive x-axis toward the positive y-axis (Fig(ii)).

`=>` The vector product of two nonzero vectors `vec a` and `vec b` , is denoted by `vec a xx vec b`

and defined as

`color{green}{vec a xx vec b = | vec a | | vec b | sin theta hat n ,}`

`=>`where, `θ` is the angle between `vec a` and `vec b` , ` 0 ≤ θ ≤ π ` and `hat n` is a unit vector perpendicular to both `vec a ` and `vec b` , such that `vec a , vec b` and `hat n` form a right handed system (Fig).

i.e., the right handed system rotated from `vec a` to ` vec b` moves in the direction of `hat n` .

● If either `vec a = vec 0` or `vec b = vec 0` , then θ is not defined and in this case, we define `vec a xx vec b = vec 0`

`"Point to consider :"`

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., `color{red}{vec a xx vec b = vec 0 ⇔ vec a ∥ vec b}`

`=>` In particular, `vec a xx vec a = 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 sin θ 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 `hat k` (Fig 10.24), we have

`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

`color{red}{sin θ = ( | 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 θ hat n` , where `vec a , vec b` and `hat n` form a right handed system,

i.e., ` θ ` is traversed from `vec a` to `vec b` Fig 10.25 (i). While `vec b xx vec a = | vec a | | vec b| sin θ 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(ii).




`=>` 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 θ hat n`

`= - | vec a | | vec b| sin θ hat (n_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 θ = 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 Fig, we have

Area of parallelogram `ABCD = AB * DE.`





But `AB = | vec b |` and `DE = | vec a | sin θ` .

Area of parallelogram `color{orange}{ABCD = | vec b | | vec a | sin θ = | vec a xx vec b |}` .

( Distributivity of vector product over addition): If `vec a , vec b` and `vec c` are any three vectors and λ be a scalar,) then

(i) `vec a xx ( vec b+ vec c ) = vec a xx vec b + vec a xx vec c`

(ii) `λ ( vec a xx vec b ) = ( λ vec a) xx vec b = vec a xx ( λ vec b )`

`=>` Let `vec a ` and `vec b ` be two vectors given in component form as `a_1 hat i + a_2 hat j + a_3 hat k` and

`b_1 hat i + b_2 hat j + b_3 hat k` , respectively . Then their cross product may be given by

`color{red}{vec a xx vec b = | ( hat i , hat j , hat k ), ( a_1 , a_2, a_3 ), ( b_1, b_2 , b_3) |}`


`"Proof"`

`vec a xx vec b = ( a_1 hat i + a_2 hat j + a_3 hat k ) xx ( b_1 hat i + b_2 hat j + b_3 hat k )`

`= a_1 b_1 ( hat i xx hat i) + a_1 b_2 ( hat i xx hat j ) + a_1 b_3 ( hat i xx hat k) + a_2 b_1 ( hat j xx hat i )`

`+ a_2 b_2 ( hat j xx hat j ) + a_2 b_3 ( hat j xx hat k ) `

`+ a_3 b_1 ( hat k xx hat i ) - a_3 b_2 ( hat k xx hat j ) + a_3 b_3 ( hat k xx hat k )`

`= a_1 b_2 ( hat i xx hat j ) - a_1 b_3 ( hat k xx hat i ) -a_2 b_1 ( hat i xx hat j )`

`+ a_2 b_3 ( hat j xx hat k) + a_3 b_1 ( hat k xx hat i ) - a_3 b_2 ( hat j xx hat k )`

(as `hat i xx hat i = hat j xx hat j = hat k xx hat k=0 )` and `hat i xx hat k = - hat k xx hat i , hat j xx hat i = - hat i xx hat j ` and ` hat k xx hat j = - hat j xx hat k` )

`= a_1 b_2 hat k - a_1 b_3 hat j - a_2 b_1 hat k + a_2 b_3 hat i + a_3 b_1 hat j - a_3 b_2 hat i`

(as `hat i xx hat j = hat k , hat j xx hat k = hat i` and ` hat k xx hat i = hat j ` )

`= ( a_2 b_3 - a_3 b_2 ) hat i - ( a_1 b_3 - a_3 b_1 ) hat j + ( a_1 b_2 - a_2 b_1 ) hat k`

`= | ( hat i , hat j, hat k ) ,( a_1, a_2 , a_3 ), (b_1 , b_2 , b_3) |`