Notes On Properties of Cross Product - CBSE Class 12 Maths

Property 1: p   x  q   is also a vector.

The cross product of p   and  q   is given by:

p   x q   = | p  | | q  |  sin ϴ n ^

Hence, the cross product of two vectors p and q, and the unit vector, n, are collinear.


Property 2: p   x  q   = 0   ⇔ p     ||   q   .

Let p   and q   be two non-zero vectors.

If p    q  , ϴ = 0o

p   x q   = | p  | | q  |  sin 0  n ^   =  0  

(sin 0o = 0)


Observation 1: p   x p   =  0  

Since ϴ = 0o


Observation 2: p   x ( - p  ) =  0  

Since ϴ = 180o


Observation 3:

 i ^ x i ^ =  0  

 j ^ x j ^ 0  

 k ^ x k ^ 0  

If p   Á q  , ϴ = 90o


Property 3: Given p   Á q  ,

p   x q   =  | p  | | q  |  sin 90 n ^ =  | p  | | q   n ^  

⇒  p    x   q   =  | p  | | q  | | n ^ |  = | p  | | q  | 

(sin 90o = 1 and n ^ is a unit vector)


Property 4: Angle ϴ between two vectors p   and q   is given by:

sin ϴ = | p _ x q _ | | p _ | | q _ |

p   x q   = | p  | | q  | sin ϴ n ^ ...(1)

 q   x p   =  | p  | | q  | sin ϴ n ^ , ...(2)

n ^ = - n ^

 q   x p   = - | p  | | q  | sin ϴ n ^ ...(3)

From (1) and (3):

p   x q   = q    x p  


Property 5: The cross product of vectors is not commutative, since p   x q   = q    x p  .

For mutually perpendicular vectors i ^ and j ^ :

 i ^ x j ^ = | i ^ | | j ^ | sin 90 n ^

 i ^ x j ^ = n ^ (since  | i ^ | = | j ^ | = sin 90 = 1)

⇒  i ^    x j ^ =  k ^ (since n ^  ⊥ i ^    x j ^ )

Since cross product is not commutative:

 j ^ x i ^ = -  k ^  

Similarly:

 j ^ x  k ^ i ^ and k ^ x  j ^ = i ^

k ^ x i ^ =  j ^ and i ^ x  k ^ = -  j ^  

Property 6: If p   and q   represent the adjacent sides of a triangle, then the area of the triangle = ½ p   x  q  |.

In ∆OPQ, let QR Á OP.

Area of ∆OPQ = ½ Base x Height

⇒ Area of ∆OPQ = ½ (OP). ( QR) ...(1)

In right-angled ∆OQR:

sin ϴ = QR OQ

⇒ QR = OQ sin ϴ

⇒ Area of ∆OPQ = ½ (OP) (OQ) sin ϴ

OP = | p  | and OQ = | q  |

⇒ Area of ∆OPQ = ½ | p  | | q  | sin ϴ

Since |  p   x  q  | = | p  | | q  | sin ϴ

Area of ∆OPQ = ½ | p   x q  |

The cross product of p   and q   is given by: p   x q   = | p   | | q  | sin ϴ n ^


Property 7: If p   and q   represent adjacent sides of a parallelogram, the area of the parallelogram = | p  q  |.

In parallelogram OPRQ, let QS Á OP.

Area of parallelogram OPRQ = Base x Height

⇒ Area of parallelogram OPRQ = OP .QS ...(1)

In right-angled ∆OQS:

sin ϴ = QS OQ

⇒ QS = OQ sin ϴ

⇒ Area of parallelogram OPRQ = OP . OQ sin ϴ

OP = | p  | and OQ = | q  |

⇒ Area of parallelogram OPRQ =  | p  | | q  |  sin ϴ

Since | p  q  | = | p  | | q  | sin ϴ

Area of parallelogram OPRQ = | p  q  |


Property 7: Distributive property of cross product.

For any three vectors  p   , q   and r    :

p   x ( q    + r  ) =   p   x   q   +   p   x r  


Property 8: For any two vectors p   and q   and a scalar λ :

  λ ( p  q  ) = (λ  p  ) x  q   = p   x (λ q  )

  a   =  a1 i ^ + a2  j ^ + a3  k ^    

  b   =  b1 i ^ + b2  j ^ + b3  k ^   

 a   x b   = ( a1 i ^ + a2  j ^ + a3  k ^   ) x (b1 i ^ + b2  j ^ + b3  k ^ )

 a   x b   =  a1 i ^   x (b1 i ^ + b2  j ^ + b3  k ^ ) +   a2  j ^   x (b1 i ^ + b2  j ^ + b3  k ^ ) + a3  k ^   x (b1 i ^ + b2  j ^ + b3  k ^ )   

= a1b1 ( i ^   x  i ^ ) + a1b2 ( i ^   x  j ^ ) + a1b3 ( i ^   x  k ^ ) + a2b1 ( j ^   x  i ^ ) + a2b2 ( j ^   x  j ^ ) + a2b3 ( j ^   x  k ^ ) + a3b1 ( k ^   x  i ^ ) + a3b2 ( k ^   x  j ^ ) + a3b3 ( k ^   x  k ^ )

We know:

  i ^   x  i ^   =  j ^   x  j ^    =   k ^   x  k ^    = 0  

i ^ x j ^ k ^ and j ^ x i ^ = k ^

j ^ x k ^ = i ^ and k ^ x j ^ = i ^

k ^ x i ^ j ^ and i ^ x k ^ = - j ^  

 a   x b   =  a1b2 ( i ^   x  j ^ ) + a1b3 ( i ^   x  k ^ ) + a2b1 ( j ^   x  i ^ + a2b3 ( j ^   x  k ^ ) + a3b1 ( k ^   x  i ^ ) + a3b2 ( k ^   x  j ^     

=   a1b2  k ^    - a1b3  j ^    - a2b1  k ^     + a2b3  i ^    + a3b1  j ^    - a3b2  i ^            

 a   x b   = ( a2b3  - a3b2 i ^    - (a1b3a3b1) j ^    + (a1b2  - a2b1 k ^   

The cross product of two vectors can also be expressed as a determinant.

 a   x b   =  i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3

Summary

Property 1: p   x  q   is also a vector.

The cross product of p   and  q   is given by:

p   x q   = | p  | | q  |  sin ϴ n ^

Hence, the cross product of two vectors p and q, and the unit vector, n, are collinear.


Property 2: p   x  q   = 0   ⇔ p     ||   q   .

Let p   and q   be two non-zero vectors.

If p    q  , ϴ = 0o

p   x q   = | p  | | q  |  sin 0  n ^   =  0  

(sin 0o = 0)


Observation 1: p   x p   =  0  

Since ϴ = 0o


Observation 2: p   x ( - p  ) =  0  

Since ϴ = 180o


Observation 3:

 i ^ x i ^ =  0  

 j ^ x j ^ 0  

 k ^ x k ^ 0  

If p   Á q  , ϴ = 90o


Property 3: Given p   Á q  ,

p   x q   =  | p  | | q  |  sin 90 n ^ =  | p  | | q   n ^  

⇒  p    x   q   =  | p  | | q  | | n ^ |  = | p  | | q  | 

(sin 90o = 1 and n ^ is a unit vector)


Property 4: Angle ϴ between two vectors p   and q   is given by:

sin ϴ = | p _ x q _ | | p _ | | q _ |

p   x q   = | p  | | q  | sin ϴ n ^ ...(1)

 q   x p   =  | p  | | q  | sin ϴ n ^ , ...(2)

n ^ = - n ^

 q   x p   = - | p  | | q  | sin ϴ n ^ ...(3)

From (1) and (3):

p   x q   = q    x p  


Property 5: The cross product of vectors is not commutative, since p   x q   = q    x p  .

For mutually perpendicular vectors i ^ and j ^ :

 i ^ x j ^ = | i ^ | | j ^ | sin 90 n ^

 i ^ x j ^ = n ^ (since  | i ^ | = | j ^ | = sin 90 = 1)

⇒  i ^    x j ^ =  k ^ (since n ^  ⊥ i ^    x j ^ )

Since cross product is not commutative:

 j ^ x i ^ = -  k ^  

Similarly:

 j ^ x  k ^ i ^ and k ^ x  j ^ = i ^

k ^ x i ^ =  j ^ and i ^ x  k ^ = -  j ^  

Property 6: If p   and q   represent the adjacent sides of a triangle, then the area of the triangle = ½ p   x  q  |.

In ∆OPQ, let QR Á OP.

Area of ∆OPQ = ½ Base x Height

⇒ Area of ∆OPQ = ½ (OP). ( QR) ...(1)

In right-angled ∆OQR:

sin ϴ = QR OQ

⇒ QR = OQ sin ϴ

⇒ Area of ∆OPQ = ½ (OP) (OQ) sin ϴ

OP = | p  | and OQ = | q  |

⇒ Area of ∆OPQ = ½ | p  | | q  | sin ϴ

Since |  p   x  q  | = | p  | | q  | sin ϴ

Area of ∆OPQ = ½ | p   x q  |

The cross product of p   and q   is given by: p   x q   = | p   | | q  | sin ϴ n ^


Property 7: If p   and q   represent adjacent sides of a parallelogram, the area of the parallelogram = | p  q  |.

In parallelogram OPRQ, let QS Á OP.

Area of parallelogram OPRQ = Base x Height

⇒ Area of parallelogram OPRQ = OP .QS ...(1)

In right-angled ∆OQS:

sin ϴ = QS OQ

⇒ QS = OQ sin ϴ

⇒ Area of parallelogram OPRQ = OP . OQ sin ϴ

OP = | p  | and OQ = | q  |

⇒ Area of parallelogram OPRQ =  | p  | | q  |  sin ϴ

Since | p  q  | = | p  | | q  | sin ϴ

Area of parallelogram OPRQ = | p  q  |


Property 7: Distributive property of cross product.

For any three vectors  p   , q   and r    :

p   x ( q    + r  ) =   p   x   q   +   p   x r  


Property 8: For any two vectors p   and q   and a scalar λ :

  λ ( p  q  ) = (λ  p  ) x  q   = p   x (λ q  )

  a   =  a1 i ^ + a2  j ^ + a3  k ^    

  b   =  b1 i ^ + b2  j ^ + b3  k ^   

 a   x b   = ( a1 i ^ + a2  j ^ + a3  k ^   ) x (b1 i ^ + b2  j ^ + b3  k ^ )

 a   x b   =  a1 i ^   x (b1 i ^ + b2  j ^ + b3  k ^ ) +   a2  j ^   x (b1 i ^ + b2  j ^ + b3  k ^ ) + a3  k ^   x (b1 i ^ + b2  j ^ + b3  k ^ )   

= a1b1 ( i ^   x  i ^ ) + a1b2 ( i ^   x  j ^ ) + a1b3 ( i ^   x  k ^ ) + a2b1 ( j ^   x  i ^ ) + a2b2 ( j ^   x  j ^ ) + a2b3 ( j ^   x  k ^ ) + a3b1 ( k ^   x  i ^ ) + a3b2 ( k ^   x  j ^ ) + a3b3 ( k ^   x  k ^ )

We know:

  i ^   x  i ^   =  j ^   x  j ^    =   k ^   x  k ^    = 0  

i ^ x j ^ k ^ and j ^ x i ^ = k ^

j ^ x k ^ = i ^ and k ^ x j ^ = i ^

k ^ x i ^ j ^ and i ^ x k ^ = - j ^  

 a   x b   =  a1b2 ( i ^   x  j ^ ) + a1b3 ( i ^   x  k ^ ) + a2b1 ( j ^   x  i ^ + a2b3 ( j ^   x  k ^ ) + a3b1 ( k ^   x  i ^ ) + a3b2 ( k ^   x  j ^     

=   a1b2  k ^    - a1b3  j ^    - a2b1  k ^     + a2b3  i ^    + a3b1  j ^    - a3b2  i ^            

 a   x b   = ( a2b3  - a3b2 i ^    - (a1b3a3b1) j ^    + (a1b2  - a2b1 k ^   

The cross product of two vectors can also be expressed as a determinant.

 a   x b   =  i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3

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