WBJEE 2024 Physics & Chemistry Paper

To determine the direction of the vector expression$\overset{\rightarrow}{B}-B \sin \; \theta \hat{a}_{\perp}$, let's break it down step-by-step:1. Understanding the given vectors: 2. Projection of $\overset{\rightarrow}{B}$ :The term $B \sin \; \theta$ is equivalent to the magnitude of the component of $\overset{\rightarrow}{B}$ that is perpendicular to $\overset{\rightarrow}{A}$. Thus, multiplying it with $\hat{a}_{\perp}$ gives a vector component of $\overset{\rightarrow}{B}$ perpendicular to $\overset{\rightarrow}{A}$, which can be written as:$B \sin \; \theta \hat{a}_{\perp}$ 3. Subtracting the perpendicular component from $\overset{\rightarrow}{B}$ :When we subtract this perpendicular component from $\overset{\rightarrow}{B}$, we get:$\overset{\rightarrow}{B}-B \sin \; \theta \hat{a}_{\perp}$This operation essentially removes the part of $\overset{\rightarrow}{B}$ that is perpendicular to $\overset{\rightarrow}{A}$, leaving only the component of $\overset{\rightarrow}{B}$ that is parallel to $\overset{\rightarrow}{A}$. 4. Conclusion:The resulting vector will be pointing in the direction along $\overset{\rightarrow}{A}$ because the perpendicular component has been removed. Thus, the direction of$\overset{\rightarrow}{B}-B \sin \; \theta \hat{a}_{\perp}$ is along $\overset{\rightarrow}{A}$Therefore, the correct answer is:Option C: along $\overset{\rightarrow}{A}$.