The angle between vector A and (A - B) is
(a) tan^-1((-B/2)/(A-B*sqrt(3)/2))
(b) tan^-1(A/(0.7B))
(c

Q: Why did we use $60^\circ$ instead of $120^\circ$ as shown in the diagram?

The mathematical structure of the correct option (c) implies an angle of $60^\circ$. If the angle were $120^\circ$, the denominator would be $2A + B$. This is a classic case where analyzing the options helps correct a misleading diagram!

Q: How do we find the angle of a resultant vector?

Once you have the x and y components of the resultant vector, the angle it makes with the x-axis is simply $\tan^{-1}(y/x)$.

Q: Why did we reverse vector B?

The problem asks for the angle with $\mathbf{A} - \mathbf{B}$. Geometrically, subtracting a vector is identical to adding its exact opposite, hence we reverse $\mathbf{B}$ to get $-\mathbf{B}$.

Q: Does it matter which axis we place vector A on?

Not at all! Placing $\mathbf{A}$ on the positive x-axis just makes the math much cleaner because its y-component becomes zero. The relative angle will remain the same regardless of orientation.

Q: What if the magnitude of B is greater than 2A?

If $B > 2A$, the denominator $2A - B$ becomes negative. This simply means the resultant vector points into the second or third quadrant relative to $\mathbf{A}$, and the angle $\beta$ would be obtuse.

Question

The angle between vector $\mathbf{A}$ and $(\mathbf{A} - \mathbf{B})$ is

Options:

(A) $	an^{-1}\left(\frac{-B/2}{A-B\sqrt{3}/2}ight)$

(B) $	an^{-1}\left(\frac{A}{0.7B}ight)$

(C) $	an^{-1}\left(\frac{\sqrt{3}B}{2A-B}ight)$

(D) $	an^{-1}\left(\frac{B\cos	heta}{A-B\sin	heta}ight)$

Sigma Percentile · Accepted Answer

This problem is a fantastic example of why you must always trust the mathematical structure of the options over a potentially misleading diagram. Let's dive into the vector geometry and see how the math reveals the true physical setup.

### The Diagram Trap

At first glance, the diagram provided in the question might lead you to believe that the angle between vectors $\mathbf{A}$ and $\mathbf{B}$ is $120^\circ$. However, if we were to calculate the angle of the resultant $\mathbf{A} - \mathbf{B}$ assuming an angle of $120^\circ$, the denominator of our final expression would be $2A + B$.

Looking closely at the correct option (c), we see the denominator is $2A - B$. This specific mathematical signature tells us unequivocally that the true angle between the vectors $\mathbf{A}$ and $\mathbf{B}$ is actually $60^\circ$. Let's prove this by setting up a clean coordinate system.

### Setting Up the Coordinate System

To make our calculations as elegant as possible, let's align vector $\mathbf{A}$ along the positive x-axis.

$$\vec{A} = A\hat{i}$$

Now, knowing the true angle is $60^\circ$, we can express vector $\mathbf{B}$ in terms of its rectangular components:

$$\vec{B} = B\cos 60^\circ \hat{i} + B\sin 60^\circ \hat{j}$$

Substituting the standard trigonometric values, we get:

$$\vec{B} = \frac{B}{2}\hat{i} + \frac{\sqrt{3}B}{2}\hat{j}$$

### The Vector Subtraction

The question asks for the angle between $\mathbf{A}$ and the resultant vector $\mathbf{A} - \mathbf{B}$. Geometrically, subtracting a vector is identical to adding its exact opposite. So, we first find $-\mathbf{B}$ by simply reversing its signs:

$$-\vec{B} = -\frac{B}{2}\hat{i} - \frac{\sqrt{3}B}{2}\hat{j}$$

Now, we add $\vec{A}$ and $-\vec{B}$ to find our resultant vector $\vec{R}$:

$$\vec{R} = \vec{A} - \vec{B} = \left(A - \frac{B}{2}ight)\hat{i} - \frac{\sqrt{3}B}{2}\hat{j}$$

### Calculating the Final Angle

We now have the exact x and y components of our resultant vector $\mathbf{A} - \mathbf{B}$. The angle $\beta$ that this resultant makes with the x-axis (which is where we placed vector $\mathbf{A}$) is found using the tangent ratio:

$$	an\beta = \frac{|y|}{|x|} = \frac{\frac{\sqrt{3}B}{2}}{A - \frac{B}{2}}$$

To clean up this complex fraction, we multiply both the numerator and the denominator by 2:

$$	an\beta = \frac{\sqrt{3}B}{2A - B}$$

Taking the inverse tangent of both sides yields our final answer:

**$$\beta = 	an^{-1}\left(\frac{\sqrt{3}B}{2A - B}ight)$$**

This perfectly matches option (c)! This problem beautifully illustrates that while diagrams are helpful guides, the rigorous application of vector algebra will always lead you to the undeniable truth.

Animated Solution for Physics - Kinematics: The angle between vector $A$ and $(A - B)$ is

Select Answer:

Visualized Solution

\text{Analyzing the Vectors}

\text{The True Angle Between } \mathbf{A} \text{ and } \mathbf{B}

\text{Finding } -\mathbf{B}

\text{Components of } -\mathbf{B}

\text{The Resultant Vector } \mathbf{A} - \mathbf{B}

\text{Calculating the Angle } \beta

\text{Reflecting on the Result}

The Sigma Insight: Vector Addition, Subtraction, and Resolution

The Diagram Trap

Setting Up the Coordinate System

The Vector Subtraction

Calculating the Final Angle

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Animated Solution for Physics - Kinematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle between vector A and (A−B) is