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Let \[\vec A = \hat iA\cos \theta + \hat jA\sin \theta \] be any vector. Another vector \[\vec B\], which is normal to \[\vec A\] can be represented as
(A) \[\hat iB\cos \theta - \hat jB\sin \theta \]
(B) \[\hat iB\cos \theta + \hat jB\sin \theta \]
(C) \[\hat iB\sin \theta - \hat jB\cos \theta \]
(D) \[\hat iB\sin \theta + \hat jB\cos \theta \]

seo-qna
Last updated date: 20th Jun 2024
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Answer
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Hint: When two vectors are normal to each other their dot product is equal to zero. Investigate which value of the vector \[\vec B\] will allow the dot product to zero.
Formula used: In this solution we will be using the following formulae;
\[\vec A \cdot \vec B = {A_x}{B_x} + {A_y}{B_y}\] where \[\vec A \cdot \vec B\] signifies the dot product between two vectors \[\vec A\] and \[\vec B\], and \[{A_x}\] and \[{A_y}\] are the x component and y component of the vector \[\vec A\] respectively, and similarly, \[{B_x}\] and \[{B_y}\] are the x component and y component of the vector \[\vec B\] respectively

Complete Step-by-Step Solution:
We are to find the value of the vector \[\vec B\] for which it must be normal to the vector \[\vec A\]. From prior knowledge, we know that the dot product of two vectors which are normal (or perpendicular) to each other is always equal to zero. Hence, we can use such property to find the vector \[\vec B\].
Generally, vector \[\vec B\] will be given in two dimension as \[\vec B = \hat i{B_x} + \hat j{B_y}\]
Hence, performing a dot product of vectors \[\vec A\] and \[\vec B\], generally, we have
\[\vec A \cdot \vec B = {A_x}{B_x} + {A_y}{B_y}\] where \[{A_x}\] and \[{A_y}\] are the x component and y component of the vector \[\vec A\] respectively, and similarly, \[{B_x}\] and \[{B_y}\] are the x component and y component of the vector \[\vec B\] respectively.
Hence, inserting known values
 hence for normality
\[A\cos \theta {B_x} + A\sin \theta {B_y} = 0\]
Now, by observation or trial and error, we can see that if \[{B_x} = B\sin \theta \] and \[{B_y} = - B\cos \theta \], then the equation above will become zero.
Hence, we have that the vector \[\vec B = \hat iB\sin \theta - \hat jB\cos \theta \]

Hence, the correct option is C

Note: For clarity, the fact that the dot product of two vector which are normal to each other is zero is evident from the definition of dot product which can be given as
\[\vec A \cdot \vec B = AB\cos \alpha \] where \[\alpha \] is the angle between the vectors. Hence, as obvious, when \[\alpha = 90^\circ \] then, \[\cos \alpha = 0\].
Hence dot product is zero when vectors are perpendicular (or normal)