The user wants me to solve a vector algebra problem and verify the answer.

1. Understand the Given Information:

• We have two non-collinear unit vectors, $\widehat a$ and $\widehat b$, meaning $|\widehat a| = 1$ and $|\widehat b| = 1$.
• The angle $\theta$ between them is acute, so $0 < \theta < \pi/2$. This implies $\widehat a \cdot \widehat b = |\widehat a| |\widehat b| \cos \theta = \cos \theta > 0$.
• The position vector of a moving point $P$ is given by $\overrightarrow{OP}(t) = \widehat a \cos t + \widehat b \sin t$.
• We need to find the maximum length of $\overrightarrow{OP}$, denoted by $M$, and the unit vector $\widehat u$ along $\overrightarrow{OP}$ when its length is maximum.

2. Find the Magnitude of $\overrightarrow{OP}$: To find the length of $\overrightarrow{OP}$, we calculate its magnitude. It's easier to work with the square of the magnitude. $|\overrightarrow{OP}|^2 = \overrightarrow{OP} \cdot \overrightarrow{OP} = (\widehat a \cos t + \widehat b \sin t) \cdot (\widehat a \cos t + \widehat b \sin t)$ Expanding the dot product: $|\overrightarrow{OP}|^2 = (\widehat a \cdot \widehat a) \cos^2 t + 2(\widehat a \cdot \widehat b) \cos t \sin t + (\widehat b \cdot \widehat b) \sin^2 t$ Since $\widehat a$ and $\widehat b$ are unit vectors, $\widehat a \cdot \widehat a = |\widehat a|^2 = 1$ and $\widehat b \cdot \widehat b = |\widehat b|^2 = 1$. $|\overrightarrow{OP}|^2 = (1) \cos^2 t + 2(\widehat a \cdot \widehat b) \cos t \sin t + (1) \sin^2 t$ Using the identities $\cos^2 t + \sin^2 t = 1$ and $2 \cos t \sin t = \sin(2t)$: $|\overrightarrow{OP}|^2 = 1 + (\widehat a \cdot \widehat b) \sin(2t)$

3. Maximize the Magnitude: The length of $\overrightarrow{OP}$ will be maximum when $|\overrightarrow{OP}|^2$ is maximum. The expression is $|\overrightarrow{OP}|^2 = 1 + (\widehat a \cdot \widehat b) \sin(2t)$. We know that $\widehat a \cdot \widehat b > 0$ because the angle between the vectors is acute. To maximize this expression, we need to maximize the term $\sin(2t)$. The maximum value of $\sin(2t)$ is 1. Therefore, the maximum value of $|\overrightarrow{OP}|^2$ is: $M^2 = \max(|\overrightarrow{OP}|^2) = 1 + (\widehat a \cdot \widehat b)(1) = 1 + \widehat a \cdot \widehat b$ So, the maximum length $M$ is: $M = \sqrt{1 + \widehat a \cdot \widehat b} = (1 + \widehat a \cdot \widehat b)^{1/2}$

4. Find the Vector $\overrightarrow{OP}$ at Maximum Length: The maximum length occurs when $\sin(2t) = 1$. This happens when $2t = \frac{\pi}{2} + 2k\pi$ for any integer $k$. Let's take the simplest case, $t = \frac{\pi}{4}$. For $t = \frac{\pi}{4}$, we have $\cos t = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin t = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. Substitute these values back into the expression for $\overrightarrow{OP}$: $\overrightarrow{OP}_{max} = \widehat a \cos\left(\frac{\pi}{4}\right) + \widehat b \sin\left(\frac{\pi}{4}\right) = \widehat a \left(\frac{1}{\sqrt{2}}\right) + \widehat b \left(\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}(\widehat a + \widehat b)$

5. Find the Unit Vector $\widehat u$: The unit vector $\widehat u$ is the vector $\overrightarrow{OP}_{max}$ divided by its magnitude $M$. $\widehat u = \frac{\overrightarrow{OP}_{max}}{|\overrightarrow{OP}_{max}|} = \frac{\frac{1}{\sqrt{2}}(\widehat a + \widehat b)}{M}$ Alternatively, and more simply, $\widehat u$ is the unit vector in the direction of $\overrightarrow{OP}_{max}$. The direction of $\overrightarrow{OP}_{max}$ is given by the vector $(\widehat a + \widehat b)$. So, the unit vector is: $\widehat u = \frac{\widehat a + \widehat b}{|\widehat a + \widehat b|}$

6. Compare with Options: We have found:

• $M = (1 + \widehat a \cdot \widehat b)^{1/2}$
• $\widehat u = \frac{\widehat a + \widehat b}{|\widehat a + \widehat b|}$

Let's check the given options: A: $\widehat u = {{\widehat a + \widehat b} \over {\left| {\widehat a + \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + \widehat a.\,\widehat b} \right)^{1/2}}$ - This matches our result. B: $\widehat u = {{\widehat a - \widehat b} \over {\left| {\widehat a - \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + \widehat a.\,\widehat b} \right)^{1/2}}$ - The unit vector is incorrect. C: $\widehat u = {{\widehat a + \widehat b} \over {\left| {\widehat a + \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + 2\widehat a.\,\widehat b} \right)^{1/2}}$ - The magnitude M is incorrect. D: $\widehat u = {{\widehat a - \widehat b} \over {\left| {\widehat a - \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + 2\widehat a.\,\widehat b} \right)^{1/2}}$ - Both are incorrect.

Thus, option A is the correct answer.