Mathematics Vector (or cross) product of two vectors for cbse-ncert

### Topic Covered

star Vector (or cross) product of two vectors

### Vector (or cross) product of two vectors

=> 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)).

### Definition

=> 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 |} .

### Properties of vector product

( 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) |

Q 3118067809

Find | vec a xx vec b | , if vec a = 2 hat i + hat j + 3 hat k and vec b = 3 hat i + 5 hat j − 2 hat k
Class 12 Chapter 10 Example 22
Solution:

We have

 | ( hat i , hat j , hat k),(2,1,3),(3 ,5, -2)|

 = hat i (−2 −15) − (−4 − 9) hat j + (10 – 3) hat k = −17 hat i +13 hat j + 7 hat k

Hence  | veca xx vec b| = sqrt( (-17)^2 +(13)^2 +(7)^2) = sqrt( 507)
Q 3118167900

Find a unit vector perpendicular to each of the vectors (vec a + vec b)
and (vec a − vec b), where vec a = hat i + hat j + hat k, vec b = hat i + 2 hat j + 3 hat k .
Class 12 Chapter 10 Example 23
Solution:

We have vec a + vec b = 2 hat i + 3 hat j + 4 hat k and hat a − hat b = − hat j − 2hat k

A vector which is perpendicular to both vec a + vec b and vec a − vec b is given by

(vec a + vec b) xx ( vec a - vec b) = |( hat i , hat j , hat k),( 2,3,4),(0,-1,-2) | = - 2 hat i + 4 hat j - 2 hat k

Now  | vec c| = sqrt( 4 + 16 + 4) = sqrt(24) = 2 sqrt6

Therefore, the required unit vector is

 vec c/(|vec c | ) = (-1)/sqrt6 hat i + 2/sqrt6 hat j - 1/sqrt6 hat k

Note : There are two perpendicular directions to any plane. Thus, another unit vector perpendicular to ab ab veca + vecb and veca - vecb will be (1)/sqrt6 hat i - 2/sqrt6 hat j + 1/sqrt6 hat k But that will be a consequence of (veca + vecb) xx (veca - vecb).
Q 3128167901

Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3)
and C(2, 3, 1) as its vertices.
Class 12 Chapter 10 Example 24
Solution:

We have vec (AB) = hat j + 2 hat k and vec(AC) = hat i + 2 hat j . The area of the given triangle
is 1/2 | vec( AB) xx vec ( AC) |.

Now, vec( AB) xx vec ( AC) = |( hat i , hat j , hat k),( 0,1,2),(1,2,0) | = -4 hat i + 2 hat j - hat k

Therefore  | vec( AB) xx vec ( AC) | = sqrt( 16 + 4 +1) = sqrt(21)

Thus, the required area is  1/2 sqrt (21)
Q 3138167902

Find the area of a parallelogram whose adjacent sides are given
by the vectors vec a = 3 hat i + hat j + 4 hat k and vec b = hat i − hat j + hat k
Class 12 Chapter 10 Example 25
Solution:

The area of a parallelogram with vec a and vec b as its adjacent sides is given
by | vec a xx vec b | .

Now  vec a xx vec b = | ( hat i , hat j , hat k) ,( 3 , 1,4) ,( 1 , -1 ,1) | = 5 hat i + hat j - 4 hat k

Therefore  | vec a xx vec b| = sqrt ( 25 + 1 + 16) = sqrt(42)

and hence, the required area is  sqrt(42)