Lelt two non collinear unit vectors `hata and hatb` form and acute angle. A point P moves so that at any time t the position vector `vec(OP)` (where O is the origin) is given by `hatacost+hatbsint.` When P is farthest from origin O, let M be the length of `vec(OP) and hatu` be the unit vector along `vec(OP)` Then (A) `hatu= (hata+hatb)/(|hata+hatb|) and M=(1+hata.hatb)^(1/2)` (B) `hatu= (hata-hatb)/(|hata-hatb|) and M=(1+hata.hatb)^(1/2)` (C) `hatu= (hata+hatb)/(|hata+hatb|) and M=(1+2hata.hatb)^(1/2)` (D) `hatu= (hata-hatb)/(|hata-hatb|) and M=(1+2hata.hatb)^(1/2)`

Options

A. `,hatu = (hata+hatb)/(|hata + hatb|) and M = (1 + hata.hatb)^(1//2)`
B. `,hatu = (hata-hatb)/(|hata - hatb|) and M = (1 + hata.hatb)^(1//2)`
C. `hatu = (hata+hatb)/(|hata + hatb|) and M = (1 + 2hata.hatb)^(1//2)`
D. `,hatu = (hata-hatb)/(|hata - hatb|) and M = (1 + 2hata.hatb)^(1//2)`

Explanation

To solve the problem step by step, we will analyze the position vector \( \vec{OP} \) given by \( \hat{a} \cos t + \hat{b} \sin t \) and find the maximum distance from the origin \( O \) and the corresponding unit vector \( \hat{u} \).

### Step 1: Find the Magnitude of the Position Vector
The position vector is given by:
\[ \vec{OP} = \hat{a} \cos t + \hat{b} \sin t \]
To find the magnitude \( M \) of \( \vec{OP} \), we calculate:
\[ M = |\vec{OP}| = \sqrt{(\hat{a} \cos t + \hat{b} \sin t) \cdot (\hat{a} \cos t + \hat{b} \sin t)} \]

### Step 2: Expand the Dot Product
Using the properties of dot products:
\[ M^2 = (\hat{a} \cos t) \cdot (\hat{a} \cos t) + (\hat{b} \sin t) \cdot (\hat{b} \sin t) + 2(\hat{a} \cos t) \cdot (\hat{b} \sin t) \]
This simplifies to:
\[ M^2 = \cos^2 t + \sin^2 t + 2 (\hat{a} \cdot \hat{b}) \sin t \cos t \]

### Step 3: Use the Pythagorean Identity
We know from trigonometric identities that:
\[ \cos^2 t + \sin^2 t = 1 \]
Thus, we have:
\[ M^2 = 1 + 2 (\hat{a} \cdot \hat{b}) \sin t \cos t \]

### Step 4: Substitute the Half-Angle Identity
Using the half-angle identity \( \sin 2t = 2 \sin t \cos t \):
\[ M^2 = 1 + (\hat{a} \cdot \hat{b}) \sin 2t \]

### Step 5: Maximize \( M \)
The maximum value of \( \sin 2t \) is 1. Therefore, the maximum value of \( M^2 \) occurs when \( \sin 2t = 1 \):
\[ M^2_{\text{max}} = 1 + (\hat{a} \cdot \hat{b}) \]
Thus, the maximum length \( M \) is:
\[ M = \sqrt{1 + \hat{a} \cdot \hat{b}} \]

### Step 6: Find the Unit Vector \( \hat{u} \)
The unit vector \( \hat{u} \) along \( \vec{OP} \) is given by:
\[ \hat{u} = \frac{\vec{OP}}{|\vec{OP}|} \]
Substituting \( t = \frac{\pi}{4} \) (where \( \sin t = \cos t = \frac{1}{\sqrt{2}} \)):
\[ \vec{OP} = \hat{a} \frac{1}{\sqrt{2}} + \hat{b} \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} (\hat{a} + \hat{b}) \]
Thus,
\[ \hat{u} = \frac{\frac{1}{\sqrt{2}} (\hat{a} + \hat{b})}{M} \]
Substituting \( M = \sqrt{1 + \hat{a} \cdot \hat{b}} \):
\[ \hat{u} = \frac{\hat{a} + \hat{b}}{\sqrt{2(1 + \hat{a} \cdot \hat{b})}} \]

### Conclusion
After evaluating the options, we find:
- The correct unit vector \( \hat{u} \) is:
\[ \hat{u} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|} \]
- The maximum length \( M \) is:
\[ M = \sqrt{1 + \hat{a} \cdot \hat{b}} \]

Thus, the correct answer is:
**(A)** \( \hat{u} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|} \) and \( M = \sqrt{1 + \hat{a} \cdot \hat{b}} \).