Vector Algebra

To solve the problem, we need to find the magnitude of the vector .

1. First, compute the cross product :

Let and .

The cross product is calculated as follows:

This determinant expansion gives us:

\mathbf{a} \times \mathbf{b} = (\hat{\mathbf{i}}(1(0) - (-2)(1)) - \hat{\mathbf{j}}(2(0) - (-2)(1)) + \hat{\mathbf{k}}(2(1) - 1(1)) = 2\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + \hat{\mathbf{k}}.

1. Now, given that , this indicates that is in the same direction as , i.e., where is a constant.
2. The condition leads to:

We know:

|\mathbf{c} - \mathbf{a}|^2 = (k-1)^2 (4 + 1 + 4).\

So:

(4 + 1 + 4)(k-1)^2 = 8.\

Simplifying gives:

9(k-1)^2 = 8 \Rightarrow (k-1)^2 = \frac{8}{9}\

Thus, .

1. The angle between and is .

The magnitude of the cross product is:

|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}| = |\mathbf{a} \times \mathbf{b}||\mathbf{c}|\sin30^\circ.\

Calculate :

|\mathbf{a} \times \mathbf{b}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3.\

1. The magnitude of , where

.

Therefore:

|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}| = 3 \cdot \frac{3}{\sqrt{3}} \cdot \frac{1}{2} = \frac{3}{2}.\

Thus, the magnitude of is .

To solve the given expression, we need to evaluate: .

We will examine each term individually:

1. Let's take an arbitrary vector .
2. The first term is .
   • The dot product, .
   • The cross product, .
   • Therefore, .
3. The second term is .
   • The dot product, .
   • The cross product, .
   • Therefore, .
4. The third term is .
   • The dot product, .
   • The cross product, .
   • Therefore, .
5. Adding all the terms:
   • .
   • Adding them, all the terms cancel each other out, thus the sum is .

Hence, the result of the given expression is: 0.