Question

Let vector a=icap acosthita + jcap asinthita,be any vector.another vector b which is normal to a is

Answer 1

Answer:

b = bsin∅ i  - bcos∅ j or  bcos∅ j - bsin∅ i

Explanation:

here is a trick dear

if we draw a graph of a vector tan it will be in first quadrant so its perpendicular will be in second or fourth quadrant

if we take it to be in second quadrant then bcos∅ j will be positive because in second quadrant j is positive and similarly bsin∅ i will be negative because i is negative

so in second quadrant a vectors' perpendicular = bcos∅ j - bsin∅ i

if we take it to be in fourth quadrant then bcos∅ j will be negative because in fourth quadrant j is negative and similarly bsin∅ i will be positive because i is positive.

so in fourth quadrant a vectors' perpendicular = bsin∅ i - bcos∅ j

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