Question

Let \( \theta \) be the angle between two vectors \( \vec{A} \) and \( \vec{B} \). If \( \hat{a}_\perp \) is the unit vector perpendicular to \( \vec{A} \), then the direction of \( \vec{B} - \vec{B} \sin \theta \hat{a}_\perp \) is:

• A. along \( \vec{B} \)
• B. perpendicular to \( \vec{B} \)
• C. along \( \vec{A} \)
• D. perpendicular to \( \vec{A} \)

Hint:

When subtracting a vector’s perpendicular component from the original vector, the resultant vector is aligned with the direction of the vector from which the perpendicular component is subtracted.

Answer 1

The Correct Option is A

Step 1: Let \(\vec{A}\) and \(\vec{B}\) be two vectors with an angle \(\theta\) between them.

Step 2: The term \(\vec{B} - \vec{B} \sin \theta \hat{a}_\perp\) can be interpreted as the component of \(\vec{B}\) along the direction of \(\vec{A}\). This is because \(\hat{a}_\perp\) is the unit vector perpendicular to \(\vec{A}\), and \(\vec{B} \sin \theta \hat{a}_\perp\) represents the perpendicular component of \(\vec{B}\) to \(\vec{A}\).

Step 3: The remaining vector \(\vec{B} - \vec{B} \sin \theta \hat{a}_\perp\) is thus aligned with the direction of \(\vec{A}\). Therefore, the correct direction is along \(\vec{A}\).

Answer 2

Direction of Vector $\vec{B} - B \sin\theta \hat{a}_{\perp}$

Let $\vec{A}$ and $\vec{B}$ be two vectors with an angle $\theta$ between them. Let $\hat{a}_{\perp}$ be the unit vector perpendicular to $\vec{A}$ in the plane containing $\vec{A}$ and $\vec{B}$. We want to find the direction of the vector $\vec{B} - B \sin\theta \hat{a}_{\perp}$.

Resolution of Vector $\vec{B}$

We can resolve vector $\vec{B}$ into two orthogonal components with respect to the direction of $\vec{A}$:

Component parallel to $\vec{A}$ ($\vec{B}_{\parallel}$): The magnitude of this component is $B \cos\theta$, and its direction is along the unit vector $\hat{a}_{A}$ in the direction of $\vec{A}$. So, $\vec{B}_{\parallel} = B \cos\theta \hat{a}_{A}$.

Component perpendicular to $\vec{A}$ ($\vec{B}_{\perp}$): The magnitude of this component is $B \sin\theta$, and its direction is along the unit vector $\hat{a}_{\perp}$. So, $\vec{B}_{\perp} = B \sin\theta \hat{a}_{\perp}$.

Therefore, the vector $\vec{B}$ can be expressed as the sum of these two components:

$\vec{B} = \vec{B}_{\parallel} + \vec{B}_{\perp} = B \cos\theta \hat{a}_{A} + B \sin\theta \hat{a}_{\perp}$

Evaluating $\vec{B} - B \sin\theta \hat{a}_{\perp}$

Now, let's substitute the expression for $\vec{B}$ into the given vector expression:

$\vec{B} - B \sin\theta \hat{a}_{\perp} = (B \cos\theta \hat{a}_{A} + B \sin\theta \hat{a}_{\perp}) - B \sin\theta \hat{a}_{\perp}$

$\vec{B} - B \sin\theta \hat{a}_{\perp} = B \cos\theta \hat{a}_{A} + (B \sin\theta - B \sin\theta) \hat{a}_{\perp}$

$\vec{B} - B \sin\theta \hat{a}_{\perp} = B \cos\theta \hat{a}_{A} + 0 \hat{a}_{\perp}$

$\vec{B} - B \sin\theta \hat{a}_{\perp} = B \cos\theta \hat{a}_{A}$

Direction of the Resultant Vector

The resulting vector is $B \cos\theta \hat{a}_{A}$. This vector has a magnitude of $|B \cos\theta|$ and its direction is determined by the unit vector $\hat{a}_{A}$, which is the direction of vector $\vec{A}$.

Therefore, the direction of $\vec{B} - B \sin\theta \hat{a}_{\perp}$ is along $\vec{A}$.

Final Answer: (C) along $\vec{A}$