0Cross 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
