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Question: Let \[\vec A = \hat iA\cos \theta + \hat jA\sin \theta \] be any vector. Another vector \[\vec B\], ...

Let A⃗=i^Acos⁡θ+j^Asin⁡θ\vec A = \hat iA\cos \theta + \hat jA\sin \theta be any vector. Another vector B⃗\vec B, which is normal to A⃗\vec A can be represented as
(A) i^Bcos⁡θ−j^Bsin⁡θ\hat iB\cos \theta - \hat jB\sin \theta
(B) i^Bcos⁡θ+j^Bsin⁡θ\hat iB\cos \theta + \hat jB\sin \theta
(C) i^Bsin⁡θ−j^Bcos⁡θ\hat iB\sin \theta - \hat jB\cos \theta
(D) i^Bsin⁡θ+j^Bcos⁡θ\hat iB\sin \theta + \hat jB\cos \theta

Explanation

Solution

When two vectors are normal to each other their dot product is equal to zero. Investigate which value of the vector B⃗\vec B will allow the dot product to zero.
Formula used: In this solution we will be using the following formulae;
A⃗⋅B⃗=AxBx+AyBy\vec A \cdot \vec B = {A_x}{B_x} + {A_y}{B_y} where A⃗⋅B⃗\vec A \cdot \vec B signifies the dot product between two vectors A⃗\vec A and B⃗\vec B, and Ax{A_x} and Ay{A_y} are the x component and y component of the vector A⃗\vec A respectively, and similarly, Bx{B_x} and By{B_y} are the x component and y component of the vector B⃗\vec B respectively

Complete Step-by-Step Solution:
We are to find the value of the vector B⃗\vec B for which it must be normal to the vector A⃗\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 B⃗\vec B.
Generally, vector B⃗\vec B will be given in two dimension as B⃗=i^Bx+j^By\vec B = \hat i{B_x} + \hat j{B_y}
Hence, performing a dot product of vectors A⃗\vec A and B⃗\vec B, generally, we have
A⃗⋅B⃗=AxBx+AyBy\vec A \cdot \vec B = {A_x}{B_x} + {A_y}{B_y} where Ax{A_x} and Ay{A_y} are the x component and y component of the vector A⃗\vec A respectively, and similarly, Bx{B_x} and By{B_y} are the x component and y component of the vector B⃗\vec B respectively.
Hence, inserting known values
hence for normality
Acos⁡θBx+Asin⁡θBy=0A\cos \theta {B_x} + A\sin \theta {B_y} = 0
Now, by observation or trial and error, we can see that if Bx=Bsin⁡θ{B_x} = B\sin \theta and By=−Bcos⁡θ{B_y} = - B\cos \theta , then the equation above will become zero.
Hence, we have that the vector B⃗=i^Bsin⁡θ−j^Bcos⁡θ\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
A⃗⋅B⃗=ABcos⁡α\vec A \cdot \vec B = AB\cos \alpha where α\alpha is the angle between the vectors. Hence, as obvious, when α=90∘\alpha = 90^\circ then, cos⁡α=0\cos \alpha = 0.
Hence dot product is zero when vectors are perpendicular (or normal)