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Step-by-Step Solution
Step 1: Express the given vectors
Let
\overrightarrow{a} = \hat{i} + 2\hat{j} - \hat{k}, \quad
\overrightarrow{b} = \hat{i} - \hat{j}, \quad
\overrightarrow{c} = \hat{i} - \hat{j} - \hat{k}.
Step 2: Use the cross product condition
The problem states:
\overrightarrow{r} \times \overrightarrow{a}
= \overrightarrow{c} \times \overrightarrow{a}.
Rewrite as:
\overrightarrow{r} \times \overrightarrow{a}
- \overrightarrow{c} \times \overrightarrow{a} = \mathbf{0}
\quad \Longrightarrow \quad
(\overrightarrow{r} - \overrightarrow{c}) \times \overrightarrow{a} = \mathbf{0}.
Since the cross product with \overrightarrow{a} is zero, the vector \overrightarrow{r} - \overrightarrow{c} must be parallel to \overrightarrow{a} . Thus,
\overrightarrow{r} - \overrightarrow{c} = \lambda\,\overrightarrow{a}
\quad \Longrightarrow \quad
\overrightarrow{r} = \lambda\,\overrightarrow{a} + \overrightarrow{c}.
Step 3: Use the dot product condition with \overrightarrow{b}
We are given:
\overrightarrow{r} \cdot \overrightarrow{b} = 0.
Substitute \overrightarrow{r} = \lambda\,\overrightarrow{a} + \overrightarrow{c} :
(\lambda\,\overrightarrow{a} + \overrightarrow{c}) \cdot \overrightarrow{b} = 0
\quad \Longrightarrow \quad
\lambda\,(\overrightarrow{a}\cdot \overrightarrow{b}) + (\overrightarrow{c}\cdot \overrightarrow{b})=0.
Next, calculate each dot product explicitly:
\overrightarrow{a}\cdot \overrightarrow{b}
= (\hat{i} + 2\hat{j} - \hat{k})
\cdot (\hat{i} - \hat{j})
= 1 \cdot 1 + 2 \cdot (-1) + (-1) \cdot 0
= 1 - 2 = -1.
\overrightarrow{c}\cdot \overrightarrow{b}
= (\hat{i} - \hat{j} - \hat{k})
\cdot (\hat{i} - \hat{j})
= 1\cdot 1 + (-1)\cdot(-1) + (-1)\cdot 0
= 1 + 1
= 2.
Putting these into the equation:
\lambda\,(-1) + 2 = 0
\quad \Longrightarrow \quad
-\lambda + 2 = 0
\quad \Longrightarrow \quad
\lambda = 2.
Step 4: Write the vector \overrightarrow{r}
With \lambda = 2 ,
\overrightarrow{r} = 2\,\overrightarrow{a} + \overrightarrow{c}.
Step 5: Find the desired dot product \overrightarrow{r}\cdot \overrightarrow{a}
\overrightarrow{r} \cdot \overrightarrow{a}
= (2\,\overrightarrow{a} + \overrightarrow{c}) \cdot \overrightarrow{a}
= 2(\overrightarrow{a}\cdot \overrightarrow{a}) + \overrightarrow{c}\cdot \overrightarrow{a}.
Step 5a: Compute \overrightarrow{a}\cdot \overrightarrow{a}
\overrightarrow{a}\cdot \overrightarrow{a}
= (\hat{i} + 2\hat{j} - \hat{k})
\cdot (\hat{i} + 2\hat{j} - \hat{k})
= 1^2 + 2^2 + (-1)^2
= 1 + 4 + 1
= 6.
Therefore,
2(\overrightarrow{a}\cdot \overrightarrow{a}) = 2 \times 6 = 12.
Step 5b: Compute \overrightarrow{c}\cdot \overrightarrow{a}
\overrightarrow{c}\cdot \overrightarrow{a}
= (\hat{i} - \hat{j} - \hat{k})
\cdot (\hat{i} + 2\hat{j} - \hat{k})
= 1 \cdot 1 + (-1)\cdot 2 + (-1)\cdot(-1)
= 1 - 2 + 1
= 0.
Step 6: Final calculation
\overrightarrow{r} \cdot \overrightarrow{a}
= 12 + 0
= 12.
Thus, the required value of \overrightarrow{r}\cdot \overrightarrow{a} is 12.
Answer: 12
