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Cross product of vectors

B.Sudhakar
31/12/2016 0 0

Cross product of two vectors ( a x b ) always yeilds another vector 'c' which will be perpendicular to vectors " a & b".

| a x b | = | a | | b | sin θ ñ -------------------------------------------------( A )      

(1)  θ : angle between vectors "a & b"

(2)  ñ : unit vector perpendicular to " a & b"

(3) | a | & | b | are the magnitude of vectors "a & b"

(4) | a x b | is the magnitude of vector 'c' ie: ( a x b )

Ex: find cross product of vectors       a = ( 2i + 4j + 3k )  &  b =( 3i + 2j + 5k )

Just use determinant form as below to find (a x b ) to get vector 'c'

                                              |     i         j        k    |                                                                      

                                              |     2       4       3    |

                                              |     3        2      5    |

= i {( 4 x 5 ) - ( 3 x 2 ) } - j { ( 2 x 5 ) - ( 3 x 3 ) } + k { ( 2 x 2 ) - ( 4 x 3 ) }

= i { 20 - 6 } - j { 10 - 9 } + k { 4 - 12 }

= ( 14 i - j - 8 k )  this is the cross product of ( a x b ) which is vector  'c' perpendicular to vectors "a & b"

‍∴ ( 2 i + 4 j + 3 k ) x ( 3 i + 2 j + 5 k ) = ( 14 i - j - 8 k )

Unlike dot product, cross product is zero if vectors "a & b" are parallel to each other as you can see from the equation ( A ),    ( a x b )= | a | | b | sin 0 n

Ie: | a x b |= | a | | b | ( 0 ) n = 0     ( as sin 0 = 0 )

Angle θ between vectors "a & b" is given by  θ = arcsin | a x b | / | a | | b | n

Let us find the angle between vectors "a & b"

θ = arcsin | 14 i - j -8 k | / | 2 i + 4 j + 3 k |  | 3 i + 2 j + 5 k |

=arcsin [ { √14² + (-1)² + (-8)²} /  { (√2² + 4² + 3² )  ( √3² + 2² + 5² ) } ]

= arcsin { √196 + 1 + 64 ) / ( √4 + 16 + 9 )  ( √9 + 4 + 25 ) }

= arcsin (16.16 ) / ( 5.38 ) ( 6.16 )

= arcsin ( 16.16 ) / ( 33.14 )

= arcsin 0.487

=29.14º

 

 

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