Let two non-collinear vectors `veca` and `vecb` inclined at an angle `(2pi)/(3)` be such that `|veca|=3` and `vec|b|=2`. If a point P moves so that at any time t its position vector `vec(OP)` (where O is the origin) is given as `vec(OP) = (t+1/t)veca+(t-1/t)vecb` then least distance of P from the origin is

Options

A. `sqrt(2sqrt(133)-10)`
B. `sqrt(2sqrt(133)+10)`
C. `sqrt(5+sqrt(133))`
D. none of these

Explanation

To solve the problem step by step, we will analyze the given information and use the properties of vectors and their dot products.

### Step 1: Understand the given vectors and their properties
We have two vectors \(\vec{a}\) and \(\vec{b}\) with magnitudes:
\[ |\vec{a}| = 3 \quad \text{and} \quad |\vec{b}| = 2 \]
The angle between these vectors is:
\[ \theta = \frac{2\pi}{3} \]

### Step 2: Write the position vector of point \(P\)
The position vector \(\vec{OP}\) is given by:
\[ \vec{OP} = \left(t + \frac{1}{t}\right) \vec{a} + \left(t - \frac{1}{t}\right) \vec{b} \]

### Step 3: Calculate the square of the magnitude of \(\vec{OP}\)
To find the least distance from the origin, we need to compute \(|\vec{OP}|^2\):
\[ |\vec{OP}|^2 = \left(\left(t + \frac{1}{t}\right) \vec{a} + \left(t - \frac{1}{t}\right) \vec{b}\right) \cdot \left(\left(t + \frac{1}{t}\right) \vec{a} + \left(t - \frac{1}{t}\right) \vec{b}\right) \]

### Step 4: Expand the dot product
Using the distributive property of the dot product:
\[ |\vec{OP}|^2 = \left(t + \frac{1}{t}\right)^2 |\vec{a}|^2 + \left(t - \frac{1}{t}\right)^2 |\vec{b}|^2 + 2\left(t + \frac{1}{t}\right)\left(t - \frac{1}{t}\right) \vec{a} \cdot \vec{b} \]

### Step 5: Substitute the magnitudes and compute
Substituting the magnitudes:
\[ |\vec{OP}|^2 = \left(t + \frac{1}{t}\right)^2 \cdot 9 + \left(t - \frac{1}{t}\right)^2 \cdot 4 + 2\left(t + \frac{1}{t}\right)\left(t - \frac{1}{t}\right) \cdot |\vec{a}||\vec{b}| \cos\theta \]
Given \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\), we have:
\[ |\vec{OP}|^2 = 9\left(t + \frac{1}{t}\right)^2 + 4\left(t - \frac{1}{t}\right)^2 - 6\left(t^2 - \frac{1}{t^2}\right) \]

### Step 6: Simplify the expression
Calculating each term:
1. \(\left(t + \frac{1}{t}\right)^2 = t^2 + 2 + \frac{1}{t^2}\)
2. \(\left(t - \frac{1}{t}\right)^2 = t^2 - 2 + \frac{1}{t^2}\)

Substituting back:
\[ |\vec{OP}|^2 = 9\left(t^2 + 2 + \frac{1}{t^2}\right) + 4\left(t^2 - 2 + \frac{1}{t^2}\right) - 6\left(t^2 - \frac{1}{t^2}\right) \]
\[ = 9t^2 + 18 + \frac{9}{t^2} + 4t^2 - 8 + \frac{4}{t^2} - 6t^2 + 6\frac{1}{t^2} \]
\[ = (9t^2 + 4t^2 - 6t^2) + (18 - 8) + \left(\frac{9}{t^2} + \frac{4}{t^2} + 6\frac{1}{t^2}\right) \]
\[ = 7t^2 + 10 + \frac{19}{t^2} \]

### Step 7: Find the minimum value
To find the minimum value of \(7t^2 + \frac{19}{t^2}\), we can use the AM-GM inequality:
\[ \frac{7t^2 + \frac{19}{t^2}}{2} \geq \sqrt{7t^2 \cdot \frac{19}{t^2}} = \sqrt{133} \]
Thus,
\[ 7t^2 + \frac{19}{t^2} \geq 2\sqrt{133} \]

Adding 10 to both sides gives:
\[ 7t^2 + \frac{19}{t^2} + 10 \geq 2\sqrt{133} + 10 \]

### Step 8: Conclusion
The least distance of point \(P\) from the origin is:
\[ |\vec{OP}| \geq \sqrt{2\sqrt{133} + 10} \]