Let `vec a=hati+2hatj-hatk , vec b=hati-hatj , vec c=hati-hatj-hatk` ,and `vecrxxveca = vecc xx veca , vecr*vecb=0` . Find `vecr*veca`

Answer

To solve the problem, we need to find \( \vec{r} \cdot \vec{a} \) given the vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) and the conditions \( \vec{r} \times \vec{a} = \vec{c} \times \vec{a} \) and \( \vec{r} \cdot \vec{b} = 0 \).

### Step 1: Define the vectors
We have:
\[ \vec{a} = \hat{i} + 2\hat{j} - \hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} \]
\[ \vec{c} = \hat{i} - \hat{j} - \hat{k} \]

### Step 2: Use the cross product condition
From the condition \( \vec{r} \times \vec{a} = \vec{c} \times \vec{a} \), we can rearrange it to:
\[ \vec{r} \times \vec{a} - \vec{c} \times \vec{a} = \vec{0} \]
This implies:
\[ \vec{r} \times \vec{a} = \vec{c} \times \vec{a} \]

### Step 3: Express \( \vec{r} \)
Since the cross product is zero, we can express \( \vec{r} \) in terms of \( \vec{a} \) and \( \vec{c} \):
\[ \vec{r} = \lambda \vec{a} + \vec{c} \]
for some scalar \( \lambda \).

### Step 4: Use the dot product condition
Now, we use the second condition \( \vec{r} \cdot \vec{b} = 0 \):
\[ (\lambda \vec{a} + \vec{c}) \cdot \vec{b} = 0 \]
Expanding this gives:
\[ \lambda (\vec{a} \cdot \vec{b}) + \vec{c} \cdot \vec{b} = 0 \]

### Step 5: Calculate \( \vec{a} \cdot \vec{b} \) and \( \vec{c} \cdot \vec{b} \)
Calculating \( \vec{a} \cdot \vec{b} \):
\[ \vec{a} \cdot \vec{b} = (1)(1) + (2)(-1) + (-1)(0) = 1 - 2 + 0 = -1 \]

Calculating \( \vec{c} \cdot \vec{b} \):
\[ \vec{c} \cdot \vec{b} = (1)(1) + (-1)(-1) + (-1)(0) = 1 + 1 + 0 = 2 \]

### Step 6: Substitute back into the equation
Substituting these values back into the equation:
\[ \lambda (-1) + 2 = 0 \]
This simplifies to:
\[ -\lambda + 2 = 0 \implies \lambda = 2 \]

### Step 7: Substitute \( \lambda \) back to find \( \vec{r} \)
Now substituting \( \lambda \) back:
\[ \vec{r} = 2\vec{a} + \vec{c} \]
Calculating \( \vec{r} \):
\[ \vec{r} = 2(\hat{i} + 2\hat{j} - \hat{k}) + (\hat{i} - \hat{j} - \hat{k}) \]
\[ = (2\hat{i} + 4\hat{j} - 2\hat{k}) + (\hat{i} - \hat{j} - \hat{k}) \]
\[ = (2 + 1)\hat{i} + (4 - 1)\hat{j} + (-2 - 1)\hat{k} \]
\[ = 3\hat{i} + 3\hat{j} - 3\hat{k} \]

### Step 8: Calculate \( \vec{r} \cdot \vec{a} \)
Now we find \( \vec{r} \cdot \vec{a} \):
\[ \vec{r} \cdot \vec{a} = (3\hat{i} + 3\hat{j} - 3\hat{k}) \cdot (\hat{i} + 2\hat{j} - \hat{k}) \]
Calculating this gives:
\[ = 3(1) + 3(2) + (-3)(-1) = 3 + 6 + 3 = 12 \]

### Final Answer
Thus, the value of \( \vec{r} \cdot \vec{a} \) is:
\[ \boxed{12} \]