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Step-by-Step Solution
Step 1: Understand the Given Vectors
We have two vectors:
$\overrightarrow{a} = \hat{i} - 2\hat{j} + \hat{k}$
$\overrightarrow{b} = \hat{i} - \hat{j} + \hat{k}$
Step 2: Note the Conditions on $\overrightarrow{c}$
We want a vector $\overrightarrow{c}$ such that:
$\overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{b} \times \overrightarrow{a}$
$\overrightarrow{c} \cdot \overrightarrow{a} = 0$
Step 3: Convert the Cross Product Condition into a Parallel Condition
From the given cross product equation,
$\overrightarrow{b} \times \overrightarrow{c} \;=\; \overrightarrow{b} \times \overrightarrow{a}.$
Move terms to one side:
$\overrightarrow{b} \times \overrightarrow{c} \;-\; \overrightarrow{b} \times \overrightarrow{a} \;=\; 0 \quad \Longrightarrow \quad \overrightarrow{b} \times \bigl(\overrightarrow{c} - \overrightarrow{a}\bigr) \;=\; 0.$
The cross product of two vectors is zero if and only if they are parallel. Hence,
$\overrightarrow{c} - \overrightarrow{a} \;\text{is parallel to}\; \overrightarrow{b}, \quad \text{i.e.,} \quad \overrightarrow{c} - \overrightarrow{a} = \lambda \,\overrightarrow{b}.$
So we get:
$\overrightarrow{c} = \overrightarrow{a} + \lambda \,\overrightarrow{b}.$
Step 4: Apply the Dot Product Condition $\overrightarrow{c} \cdot \overrightarrow{a} = 0$
We are also given that $\overrightarrow{c} \cdot \overrightarrow{a} = 0.$ Substitute
$\overrightarrow{c} = \overrightarrow{a} + \lambda \,\overrightarrow{b}$ into this:
$\bigl(\overrightarrow{a} + \lambda \,\overrightarrow{b}\bigr) \cdot \overrightarrow{a} = 0.$
Expand using distributive property of the dot product:
$\overrightarrow{a} \cdot \overrightarrow{a} + \lambda \,\bigl(\overrightarrow{b} \cdot \overrightarrow{a}\bigr) = 0.$
Thus,
$\lvert \overrightarrow{a} \rvert^2 + \lambda \,(\overrightarrow{a} \cdot \overrightarrow{b}) = 0.$
Step 4a: Calculate $\lvert \overrightarrow{a} \rvert^2$ and $\overrightarrow{a} \cdot \overrightarrow{b}$ if needed
$\overrightarrow{a} \cdot \overrightarrow{a} = (\hat{i} - 2\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 1 + 4 + 1 = 6.$ Hence, $\lvert \overrightarrow{a} \rvert^2 = 6.$
$\overrightarrow{a} \cdot \overrightarrow{b} = (\hat{i} - 2\hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = 1 + 2 + 1 = 4.$
Substitute these values:
$6 + \lambda \times 4 = 0 \quad \Longrightarrow \quad \lambda = -\frac{6}{4} = -\frac{3}{2}.$
Step 5: Determine $\overrightarrow{c}$
Now that $\lambda = -\frac{3}{2}$, substitute back into
$\overrightarrow{c} = \overrightarrow{a} + \lambda \,\overrightarrow{b}.$
$\overrightarrow{c} = \bigl(\hat{i} - 2\hat{j} + \hat{k}\bigr) \;+\; \left(-\frac{3}{2}\right)\bigl(\hat{i} - \hat{j} + \hat{k}\bigr).$
Simplify:
$\overrightarrow{c} = \hat{i} - 2\hat{j} + \hat{k} - \frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} - \frac{3}{2}\hat{k} = \left(1 - \frac{3}{2}\right)\hat{i} + \left(-2 + \frac{3}{2}\right)\hat{j} + \left(1 - \frac{3}{2}\right)\hat{k}.$
So,
$\overrightarrow{c} = -\tfrac{1}{2}\,\hat{i} - \tfrac{1}{2}\,\hat{j} - \tfrac{1}{2}\,\hat{k}
= -\tfrac{1}{2}\bigl(\hat{i} + \hat{j} + \hat{k}\bigr).$
Step 6: Find $\overrightarrow{c} \cdot \overrightarrow{b}$
Compute the dot product:
$\overrightarrow{c} \cdot \overrightarrow{b} = \Bigl(-\tfrac{1}{2}\bigl(\hat{i} + \hat{j} + \hat{k}\bigr)\Bigr)
\;\cdot\; \bigl(\hat{i} - \hat{j} + \hat{k}\bigr).$
Distribute inside the dot product:
$= -\tfrac{1}{2} \Bigl[\,
(\hat{i}\cdot\hat{i}) \;+\; (\hat{j}\cdot(-\hat{j})) \;+\; (\hat{k}\cdot\hat{k})\\
\quad+\; \text{(cross terms that vanish if the directions differ)}\, \Bigr].$
So we get:
$= -\tfrac{1}{2} \bigl[\,1 - 1 + 1\bigr] = -\tfrac{1}{2} \times 1 = -\tfrac{1}{2}.$
Therefore,
$\boxed{-\tfrac{1}{2}}$
is the value of $\overrightarrow{c} \cdot \overrightarrow{b}.$
