Question

If \(|\vec{a}|=2\) and \(|\vec{b}|=3\) and the angle between \(\)\(\vec{a}\) and \(\vec{b}\) is 120°, then the length of the vector \(|\frac{1\vec{a}}{2}-\frac{1\vec{b}}{3}|^2\) is

• A. 3
• B. \(\frac{1}{6}\)
• C. 2
• D. 1

Answer 1

The Correct Option is A

Given: 
\[ |\vec{a}| = 2,\quad |\vec{b}| = 3,\quad \theta = 120^\circ \] We are asked to find: \[ \left|\frac{1}{2}\vec{a} - \frac{1}{3}\vec{b}\right|^2 \] Let: \[ \vec{u} = \frac{1}{2}\vec{a} - \frac{1}{3}\vec{b} \] Then: \[ |\vec{u}|^2 = \left(\frac{1}{2}\vec{a} - \frac{1}{3}\vec{b}\right) \cdot \left(\frac{1}{2}\vec{a} - \frac{1}{3}\vec{b}\right) \] Use the identity: \[ |\vec{u}|^2 = \left(\frac{1}{2}\right)^2|\vec{a}|^2 + \left(\frac{1}{3}\right)^2|\vec{b}|^2 - 2\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)\vec{a} \cdot \vec{b} \] Step 1: Compute dot product
\[ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = 2 \cdot 3 \cdot \cos(120^\circ) = 6 \cdot \left(-\frac{1}{2}\right) = -3 \] Step 2: Plug in values
\[ |\vec{u}|^2 = \frac{1}{4}(4) + \frac{1}{9}(9) - 2 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot (-3) \] \[ = 1 + 1 + 1 = 3 \] ✅ Correct answer: 3

Answer 2

We are given:

• \(|\vec{a}| = 2\)
• \(|\vec{b}| = 3\)
• The angle between \(\vec{a}\) and \(\vec{b}\) is \(120^\circ\)

We need to find:

\[ \left| \frac{1}{2} \vec{a} - \frac{1}{3} \vec{b} \right|^2 \]

Let:

\[ \vec{u} = \frac{1}{2} \vec{a}, \quad \vec{v} = \frac{1}{3} \vec{b} \]

Then:

\[ |\vec{u} - \vec{v}|^2 = \vec{u} \cdot \vec{u} - 2 \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{v} \]

Now, calculate each term:

• \(\vec{u} \cdot \vec{u} = \left( \frac{1}{2} \right)^2 |\vec{a}|^2 = \frac{1}{4} \cdot 4 = 1\)
• \(\vec{v} \cdot \vec{v} = \left( \frac{1}{3} \right)^2 |\vec{b}|^2 = \frac{1}{9} \cdot 9 = 1\)
• \(\vec{u} \cdot \vec{v} = \frac{1}{2} \cdot \frac{1}{3} \cdot |\vec{a}| |\vec{b}| \cos(120^\circ) = \frac{1}{6} \cdot 2 \cdot 3 \cdot \cos(120^\circ) = 1 \cdot \left(-\frac{1}{2}\right) = -\frac{1}{2}\)

Now, plug in these values:

\[ |\vec{u} - \vec{v}|^2 = 1 - 2\left(-\frac{1}{2}\right) + 1 = 1 + 1 + 1 = 3 \]

Therefore, the value of \(\left| \frac{1}{2} \vec{a} - \frac{1}{3} \vec{b} \right|^2\) is \(\boxed{3}\).