If vector `vecv=(1,sqrt(3))` and vector `vecu=(3,-2) ` find the value of `|3vecv-vecu|`

Options

A. 5.4
B. 6
C. 7
D. 7.2

Explanation

To solve the problem, we need to find the value of \(|3\vec{v} - \vec{u}|\) where \(\vec{v} = (1, \sqrt{3})\) and \(\vec{u} = (3, -2)\).

### Step-by-Step Solution:

1. **Calculate \(3\vec{v}\)**:
\[ 3\vec{v} = 3 \cdot (1, \sqrt{3}) = (3 \cdot 1, 3 \cdot \sqrt{3}) = (3, 3\sqrt{3}) \]

**Hint**: Remember to multiply each component of the vector by the scalar.

2. **Subtract \(\vec{u}\) from \(3\vec{v}\)**:
\[ 3\vec{v} - \vec{u} = (3, 3\sqrt{3}) - (3, -2) = (3 - 3, 3\sqrt{3} - (-2)) = (0, 3\sqrt{3} + 2) \]

**Hint**: When subtracting vectors, subtract their corresponding components.

3. **Find the magnitude \(|3\vec{v} - \vec{u}|\)**:
\[ |3\vec{v} - \vec{u}| = |(0, 3\sqrt{3} + 2)| = \sqrt{0^2 + (3\sqrt{3} + 2)^2} \]

**Hint**: The magnitude of a vector \((x, y)\) is calculated using the formula \(\sqrt{x^2 + y^2}\).

4. **Calculate \((3\sqrt{3} + 2)^2\)**:
\[ (3\sqrt{3} + 2)^2 = (3\sqrt{3})^2 + 2 \cdot 3\sqrt{3} \cdot 2 + 2^2 = 27 + 12\sqrt{3} + 4 = 31 + 12\sqrt{3} \]

**Hint**: Use the expansion formula \((a + b)^2 = a^2 + 2ab + b^2\).

5. **Final magnitude calculation**:
\[ |3\vec{v} - \vec{u}| = \sqrt{31 + 12\sqrt{3}} \]

**Hint**: Remember that the square root of a sum is not the sum of the square roots.

### Conclusion:
The value of \(|3\vec{v} - \vec{u}|\) is \(\sqrt{31 + 12\sqrt{3}}\).