Unit Vector

Let's expand the expression for $$\vec{v}$$ using the distributive property of the cross product:

$$\vec{v} = (3\vec{a} - 2\vec{b}) \times (2\vec{a} + 4\vec{b})$$ 

$$\vec{v} = (3\vec{a} \times 2\vec{a}) + (3\vec{a} \times 4\vec{b}) - (2\vec{b} \times 2\vec{a}) - (2\vec{b} \times 4\vec{b})$$ 

$$\vec{v} = 6(\vec{a} \times \vec{a}) + 12(\vec{a} \times \vec{b}) - 4(\vec{b} \times \vec{a}) - 8(\vec{b} \times \vec{b})$$

Now, apply the fundamental rules of cross products:

1. The cross product of any vector with itself is zero: $$\vec{a} \times \vec{a} = \vec{0}$$ and $$\vec{b} \times \vec{b} = \vec{0}$$.
2. The cross product is anti-commutative: $$\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$$.

Substituting these back into our expansion:

$$\vec{v} = \vec{0} + 12(\vec{a} \times \vec{b}) - 4[-(\vec{a} \times \vec{b})] - \vec{0}$$ 

$$\vec{v} = 12(\vec{a} \times \vec{b}) + 4(\vec{a} \times \vec{b})$$ 

$$\vec{v} = 16(\vec{a} \times \vec{b})$$

We need to find the magnitude of $$\vec{v}$$. Taking the magnitude on both sides gives:

$$|\vec{v}| = |16(\vec{a} \times \vec{b})| = 16|\vec{a} \times \vec{b}|$$

Recall the geometric definition for the magnitude of a cross product:

$$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta$$

We are given:

• $$\vec{a}$$ and $$\vec{b}$$ are unit vectors, so $$|\vec{a}| = 1$$ and $$|\vec{b}| = 1$$.
• The angle $$\theta = \frac{\pi}{3}$$.

Substitute these values:

$$|\vec{a} \times \vec{b}| = (1)(1)\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$

Now, substitute this back into our expression for $$|\vec{v}|$$:

$$|\vec{v}| = 16 \left(\frac{\sqrt{3}}{2}\right) = 8\sqrt{3}$$

Finally, square the magnitude to get the required value:

$$|\vec{v}|^2 = (8\sqrt{3})^2 = 64 \times 3 = 192$$