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Question

# Find a unit vector perpendicular to each of the vector and , where and .

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Solution

## 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.

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