Distributive Property of Cross Product

Let's first simplify the inner cross product vector term, $$\left(\vec{a} + \vec{b}\right) \times \left(\vec{a} \times \vec{b}\right)$$, by distributing it:

$$\left(\vec{a} + \vec{b}\right) \times \left(\vec{a} \times \vec{b}\right) = \vec{a} \times \left(\vec{a} \times \vec{b}\right) + \vec{b} \times \left(\vec{a} \times \vec{b}\right)$$

Using the vector triple product expansion formula $$\vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$$:

1. $$\vec{a} \times \left(\vec{a} \times \vec{b}\right) = (\vec{a} \cdot \vec{b})\vec{a} - |\vec{a}|^2\vec{b}$$
2. $$\vec{b} \times \left(\vec{a} \times \vec{b}\right) = -\vec{b} \times \left(\vec{b} \times \vec{a}\right) = -\left[ (\vec{b} \cdot \vec{a})\vec{b} - |\vec{b}|^2\vec{a} \right] = |\vec{b}|^2\vec{a} - (\vec{a} \cdot \vec{b})\vec{b}$$

Adding these two results together:

$$\left(\vec{a} + \vec{b}\right) \times \left(\vec{a} \times \vec{b}\right) = \left( \vec{a} \cdot \vec{b} + |\vec{b}|^2 \right)\vec{a} - \left( |\vec{a}|^2 + \vec{a} \cdot \vec{b} \right)\vec{b}$$

Now, we compute the final outer cross product with $(\vec{a} - \vec{b})$:

$$\left[ \left( \vec{a} \cdot \vec{b} + |\vec{b}|^2 \right)\vec{a} - \left( |\vec{a}|^2 + \vec{a} \cdot \vec{b} \right)\vec{b} \right] \times (\vec{a} - \vec{b})$$

Since $$\vec{a} \times \vec{a} = 0$$ and $$\vec{b} \times \vec{b} = 0$$, this distributes cleanly into:

$$= -\left( \vec{a} \cdot \vec{b} + |\vec{b}|^2 \right)(\vec{a} \times \vec{b}) - \left( |\vec{a}|^2 + \vec{a} \cdot \vec{b} \right)(\vec{b} \times \vec{a})$$ $$= -\left( \vec{a} \cdot \vec{b} + |\vec{b}|^2 \right)(\vec{a} \times \vec{b}) + \left( |\vec{a}|^2 + \vec{a} \cdot \vec{b} \right)(\vec{a} \times \vec{b})$$ $$= \left( |\vec{a}|^2 - |\vec{b}|^2 \right) (\vec{a} \times \vec{b})$$

Step 2: Determine the value of $$\lambda$$

Now let's compute $$|\vec{a}|^2$$ and $$|\vec{b}|^2$$ from our given vectors:

• $$|\vec{a}|^2 = \lambda^2 + 3^2 + (-1)^2 = \lambda^2 + 10$$
• $$|\vec{b}|^2 = 2^2 + \lambda^2 + 1^2 = \lambda^2 + 5$$

Subtracting them gives a constant:

$$|\vec{a}|^2 - |\vec{b}|^2 = (\lambda^2 + 10) - (\lambda^2 + 5) = 5$$

Substitute this back into our simplified relation:

$$5(\vec{a} \times \vec{b}) = 25\hat{i} - 20\hat{j} - 10\hat{k}$$ $$\vec{a} \times \vec{b} = 5\hat{i} - 4\hat{j} - 2\hat{k}$$

Let's verify this using the cross-product determinant of $$\vec{a}$$ and $$\vec{b}$$:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda & 3 & -1 \\ 2 & \lambda & 1 \end{vmatrix} = (3 + \lambda)\hat{i} - (\lambda + 2)\hat{j} + (\lambda^2 - 6)\hat{k}$$

Comparing components with $$5\hat{i} - 4\hat{j} - 2\hat{k}$$:

• $$3 + \lambda = 5 \implies \lambda = 2$$
• $$-(\lambda + 2) = -4 \implies \lambda = 2$$

Both yield $$\lambda = 2$$.

Step 3: Find the target expression value

We need to find $$\left| \lambda\left(\vec{a} + \vec{b}\right) \times \left(\vec{a} - \vec{b}\right) \right|^2$$.

Expanding the cross product inside the norm:

$$\left(\vec{a} + \vec{b}\right) \times \left(\vec{a} - \vec{b}\right) = \vec{a}\times\vec{a} - \vec{a}\times\vec{b} + \vec{b}\times\vec{a} - \vec{b}\times\vec{b} = -2(\vec{a} \times \vec{b})$$

Substitute this into the expression:

$$\left| -2\lambda (\vec{a} \times \vec{b}) \right|^2 = 4\lambda^2 |\vec{a} \times \vec{b}|^2$$

Using $$\lambda = 2$$ and $$|\vec{a} \times \vec{b}|^2 = 5^2 + (-4)^2 + (-2)^2 = 25 + 16 + 4 = 45$$:

$$\text{Value} = 4(2)^2 \cdot 45 = 16 \cdot 45 = 720$$

Correct Option: (3) 720

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:

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