JEE PYQ: Vector Algebra - Question ID 211e8f6e499d (JEE Main 2021)

To find the angle between two vectors, we use the dot-product formula: $\cos\phi = \frac{\vec{P} \cdot \vec{Q}}{|\vec{P}|\,|\vec{Q}|}.$ Here, the two vectors are $\vec{P} = \vec{A}$ and $\vec{Q} = \vec{A} - \vec{B}$. We are given (from the diagram) that the angle between $\vec{A}$ and $\vec{B}$ is $\theta = 60^\circ$.

1. Express the dot product
We need $\vec{A} \cdot (\vec{A} - \vec{B})$. Expand it: $\vec{A} \cdot (\vec{A} - \vec{B}) = \vec{A} \cdot \vec{A} - \vec{A} \cdot \vec{B} = A^2 - A\,B\cos60^\circ = A^2 - A\,B\cdot\frac12 = A^2 - \frac{A\,B}{2}.$

2. Compute the magnitudes
$\;|\vec{A}| = A$, and $|\vec{A} - \vec{B}|^2 = A^2 + B^2 - 2\,A\,B\cos60^\circ = A^2 + B^2 - A\,B.$ Hence $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - A\,B}.$

3. Write the cosine of the desired angle
$\cos\phi = \frac{\vec{A}\cdot(\vec{A}-\vec{B})}{|\vec{A}|\,|\vec{A}-\vec{B}|} = \frac{A^2 - \tfrac{A\,B}{2}}{A\,\sqrt{A^2 + B^2 - A\,B}} = \frac{A - \tfrac{B}{2}}{\sqrt{A^2 + B^2 - A\,B}}.$

4. Find $\tan\phi$
We use $\tan\phi = \frac{\sin\phi}{\cos\phi}$. From $\sin^2\phi = 1 - \cos^2\phi$ we get $\sin\phi = \sqrt{1 - \frac{\bigl(A - \tfrac{B}{2}\bigr)^2}{A^2 + B^2 - A\,B}} = \frac{\sqrt{3}\,B/2}{\sqrt{A^2 + B^2 - A\,B}}.$ Therefore $\tan\phi = \frac{\sin\phi}{\cos\phi} = \frac{\sqrt{3}\,B/2}{A - B/2} = \frac{\sqrt{3}\,B}{2A - B}.$ Taking the inverse tangent gives $\phi = \tan^{-1}\!\Bigl(\frac{\sqrt{3}\,B}{2A - B}\Bigr).$

5. Match with the given options
The expression we derived is exactly option C.

1. Misidentifying the angle between $\vec{A}$ and $\vec{B}$: The diagram shows $60^\circ$, not $90^\circ$.
2. Forgetting to take the square root when computing $|\vec{A}-\vec{B}|$.
3. Sign errors in the dot product: $\vec{A}\cdot\vec{B} = A\,B\cos60^\circ$, not $-A\,B\cos60^\circ$.
4. Confusing $\tan^{-1}$ with $\cos^{-1}$: The question asks for the angle in terms of $\tan^{-1}$.

Tip: Always draw the vectors, label the known angle, and use the dot-product formula systematically.