Keeping one vector constant, if direction of other to be added in the first vector is changed continuously,tip of the resultant vector describes a circle. In the following figure vector ra is kept constant. Whenvector rb added to ra changes its direction, the tip of the resultant vector =+rrrr ab describes circle ofradius b with its center at the tip of vector ar. Maximum angle between vector ar and the resultant=+rrrr ab is

1. Visualising the triangle

Placing vectors tip-to-tail forms a triangle O-A-R:

• $OA = a$ (length of $\vec r_a$)
• $AR = b$ (length of $\vec r_b$)
• $OR = |\vec R|$ (length of the resultant)

The angle we care about is $\varphi = \angle AOR$ – the angle at O between $\vec r_a$ and $\vec R$. It is the angle opposite the fixed side $AR = b$.

2. Using the cosine rule at the required angle

With $b$ known and $a$ fixed, only the third side $OR$ can change while $R$ sweeps the circle.

From the cosine rule on triangle O-A-R at angle $\varphi$:

$$b^{2} = a^{2} + OR^{2} - 2a\,OR\,\cos\varphi$$

Solve for $\cos\varphi$:

$$\cos\varphi = \frac{a^{2} + OR^{2} - b^{2}}{2a\,OR}$$

Since $\varphi$ gets larger when $\cos\varphi$ gets smaller, we need the minimum value of $\cos\varphi$ over the allowed range of $OR$.

3. Allowed range of $OR$

Because $OR$ is the third side of a triangle with sides $a$ and $b$:

$$|a-b| \le OR \le a+b$$

4. Extremising $\cos\varphi$

Treat $f(x)=\frac{a^{2}+x^{2}-b^{2}}{2ax},\quad x\equiv OR$

Differentiate and set $f'(x)=0$:

$$\frac{\mathrm d}{\mathrm dx}f(x)=\frac{1}{2a}-\frac{a^{2}-b^{2}}{2ax^{2}}=0 \;\Longrightarrow\; x^{2}=a^{2}-b^{2}$$

So an interior extremum exists only if $b\le a$:

• When $b\le a$ : $OR_{\text{opt}}=\sqrt{a^{2}-b^{2}}$ lies inside the allowed interval.
• When $b>a$ : the interior root is impossible; the minimum $\cos\varphi$ occurs at the endpoint $OR=|a-b|=b-a$.

5. Evaluate the maximum angle

Case 1: $b \le a$

Insert $OR=\sqrt{a^{2}-b^{2}}$ into $\cos\varphi$:

$$\cos\varphi_{\text{min}}=\frac{a^{2}-b^{2}}{a\sqrt{a^{2}-b^{2}}}=\frac{\sqrt{a^{2}-b^{2}}}{a}$$

Hence

$$\boxed{\varphi_{\max}=\arccos\!\left(\frac{\sqrt{a^{2}-b^{2}}}{a}\right)} \;\;\text{or}\;\;\boxed{\sin\varphi_{\max}=\frac{b}{a}}$$

Case 2: $b \ge a$

At $OR=b-a$, the calculation gives $\cos\varphi_{\min}=-1$ so $\varphi_{\max}=180^{\circ}$. The resultant can point exactly opposite to $\vec r_a$.

Most JEE questions implicitly assume $b<a$ to keep the answer neat, therefore you usually quote

$$\boxed{\varphi_{\max}=\sin^{-1}\!\left(\dfrac{b}{a}\right)}\quad (\text{valid when } b\le a).$$

What’s going on here?

Imagine you stand at point O and hold a stick of fixed length a. That stick points to point A – this is your constant vector $\vec r_a$.

Now you get a second stick of length b (vector $\vec r_b$). One end is always tied to point A, but you are free to swing its other end in any direction so that the tip traces a perfect circle of radius b around A.

Every time you add the two sticks tip-to-tail, you get a new combined stick (the resultant $\vec R = \vec r_a + \vec r_b$) that runs from O to some point R on that circle.

The question:
“How far can the new stick lean away from the old fixed stick?”
In other words, what is the biggest angle between $\vec r_a$ and the resultant $\vec R$ while you keep swinging $\vec r_b$ all around?

Step-by-step JEE-style solution

Let $|\vec r_a|=a$ and $|\vec r_b|=b$.

1. Form triangle $OAR$ with sides $OA=a,\; AR=b,\; OR=|\vec R|$

2. Angle between $\vec r_a$ and $\vec R$ is $\varphi = \angle AOR$.

3. Apply cosine rule at $\angle AOR$:

$$b^{2}=a^{2}+OR^{2}-2a\,OR\cos\varphi\tag{1}$$

4. Solve for $\cos\varphi$:

$$\cos\varphi=\frac{a^{2}+OR^{2}-b^{2}}{2a\,OR}\tag{2}$$

5. The variable side $OR$ can vary in the interval $|a-b|\le OR\le a+b$.

6. Minimise $\cos\varphi$ (to maximise $\varphi$). Differentiate (2):

$$\frac{\mathrm d}{\mathrm d(OR)}\cos\varphi=\frac{1}{2a}-\frac{a^{2}-b^{2}}{2a\,OR^{2}}=0 \implies OR^{2}=a^{2}-b^{2}$$

7. Validity: this root exists only if $b\le a$.

8. Substituting $OR=\sqrt{a^{2}-b^{2}}$ into (2):

$$\cos\varphi_{\min}=\frac{a^{2}+(a^{2}-b^{2})-b^{2}}{2a\sqrt{a^{2}-b^{2}}}=\frac{\sqrt{a^{2}-b^{2}}}{a}$$

9. Therefore

$$\boxed{\sin\varphi_{\max}=\frac{b}{a}},\quad\text{so }\boxed{\varphi_{\max}=\sin^{-1}\!\left(\dfrac{b}{a}\right)}\;(b\le a)$$

10. If $b>a$, minimum $OR=b-a$ gives $\cos\varphi_{\min}=-1$ hence $\varphi_{\max}=180^{\circ}$.

Hence, under the usual condition $b<a$ (most exam problems), the sought maximum angle is $\boxed{\sin^{-1}(b/a)}$.