Lagrange's Identity

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,

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

$$=1885$$

1. We are given the structural vector equation:

$$\vec{x} + \vec{a} \times \vec{x} = \vec{b}$$

2. Take the scalar dot product of both sides of the equation with vector $$\vec{a}$$:

$$\vec{a} \cdot \vec{x} + \vec{a} \cdot (\vec{a} \times \vec{x}) = \vec{a} \cdot \vec{b}$$

Since the vectors $$\vec{a}$$ and $$\vec{b}$$ are given to be orthogonal, $$\vec{a} \cdot \vec{b} = 0$$. Additionally, any scalar triple product with a repeated vector is zero ($$\vec{a} \cdot (\vec{a} \times \vec{x}) = 0$$). This simplifies to:

$$\vec{a} \cdot \vec{x} = 0$$

3. Now, find the squared magnitude of both sides of the original vector equation:

$$|\vec{x} + \vec{a} \times \vec{x}|^2 = |\vec{b}|^2$$

4. Expand the left-hand side using the vector dot product expansion rules:

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

5. Apply vector identities to simplify the terms:

* The scalar triple product $$\vec{x} \cdot (\vec{a} \times \vec{x})$$ contains a repeated vector, so it evaluates to 0.

* Use Lagrange's identity for the cross-product magnitude:

$$|\vec{a} \times \vec{x}|^2 = |\vec{a}|^2|\vec{x}|^2 - (\vec{a} \cdot \vec{x})^2$$

6. Substitute these identity simplifications back into the expanded magnitude equation:

$$|\vec{x}|^2 + \left( |\vec{a}|^2|\vec{x}|^2 - (\vec{a} \cdot \vec{x})^2 \right) + 0 = |\vec{b}|^2$$

7. Substitute the known constant parameters ($$|\vec{a}| = 1$$, $$|\vec{b}| = 2 \implies |\vec{b}|^2 = 4$$, and our derived step $$\vec{a} \cdot \vec{x} = 0$$):

$$|\vec{x}|^2 + \left( (1)^2|\vec{x}|^2 - 0^2 \right) = 4$$

$$|\vec{x}|^2 + |\vec{x}|^2 = 4 \implies 2|\vec{x}|^2 = 4$$

$$|\vec{x}|^2 = 2$$

8. Calculate the final requested numerical value:

$$10|\vec{x}|^2 = 10 \cdot 2 = 20$$