Cross Product (Component Form)

Given,

$$\vec a=2\hat i-\hat j+2\hat k$$ and $$\vec b=\hat i+2\hat j-2\hat k$$

First compute

$$\vec a\times\vec b$$

$$= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -1 & 2\\ 1 & 2 & -2 \end{vmatrix}$$

$$=\hat i(2-4)-\hat j(-4-2)+\hat k(4+1)$$

$$=-2\hat i+6\hat j+5\hat k$$

Now,

$$\vec a+2\vec b$$

$$=(2,-1,2)+2(1,2,-2)$$

$$=(4,3,-2)$$

Let

$$\vec p=\vec a\times\vec b=(-2,6,5)$$

and

$$\vec q=\vec a+2\vec b=(4,3,-2)$$

Using

$$|\vec p\times\vec q|^2=|\vec p|^2|\vec q|^2-(\vec p\cdot\vec q)^2,$$

first compute:

$$|\vec p|^2=(-2)^2+6^2+5^2$$

$$=4+36+25$$

$$=65$$

Also,

$$|\vec q|^2=4^2+3^2+(-2)^2$$

$$=16+9+4$$

$$=29$$

Now,

$$\vec p\cdot\vec q=(-2)(4)+6(3)+5(-2)$$

$$=-8+18-10$$

$$=0$$

Therefore,

$$|\vec p\times\vec q|^2=65\times29$$

$$=1885$$

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