Find a unit vector perpendicular to each of the vector and , where and .
The given vectors are a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ − 2 k ^
a → + b → = 4 i ^ + 4 j ^ and a → − b → = 2 i ^ + 4 j ^ .
The cross product of two vectors ( a 1 i ^ + a 2 j ^ + a 3 k ^ ) and ( b 1 i ^ + b 2 j ^ + b 3 k ^ ) is given by,
a → × b = → | i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 | (1)
A unit vector perpendicular to each vector can be found by their cross product.
( a → + b → ) × ( a → − b → ) = | i ^ j ^ k ^ 4 4 0 2 0 4 | = i ^ ( 16 ) − j ^ ( 16 ) + k ^ ( − 8 ) (2)
| ( a → + b → ) ( a → − b → ) | = ( 16 ) 2 + ( − 16 ) 2 + ( − 8 ) 2 = 2 2 × 8 2 + 2 2 × 8 2 + 8 2 = 8 2 2 + 2 2 + 1 = 8 9 (3)
Further simplify the equations,
| ( a → + b → ) ( a → − b → ) | = 8 × 3 = 24
We need to find unit vector perpendicular to both ( a → + b → ) and ( a → − b → ) . So, divide (2) and (3),
± i ^ ( 16 ) − j ^ ( 16 ) + k ^ ( − 8 ) 24 = ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^
Thus, the unit vector perpendicular to each vectors a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ − 2 k ^
can be found by their cross product.
The given vectors are
The cross product of two vectors
A unit vector perpendicular to each vector can be found by their cross product.
Further simplify the equations,
We need to find unit vector perpendicular to both
Thus, the unit vector perpendicular to each vectors
can be found by their cross product.
and
,
where
and
.