If `vec(r )=x hat(i)+y hat(j)+x hat(k)`, find : `(vec(r )xx hat(i)).(vec(r )xx hat(j))+xy`.

Answer

To solve the problem, we need to find the expression \((\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + xy\), where \(\vec{r} = x \hat{i} + y \hat{j} + x \hat{k}\).

### Step-by-Step Solution:

1. **Define the Vector**:
\[ \vec{r} = x \hat{i} + y \hat{j} + x \hat{k} \]

2. **Calculate \(\vec{r} \times \hat{i}\)**:
\[ \vec{r} \times \hat{i} = (x \hat{i} + y \hat{j} + x \hat{k}) \times \hat{i} \]
Using the properties of the cross product:
- \(\hat{i} \times \hat{i} = 0\)
- \(\hat{j} \times \hat{i} = -\hat{k}\)
- \(\hat{k} \times \hat{i} = \hat{j}\)

Therefore,
\[ \vec{r} \times \hat{i} = 0 + y(-\hat{k}) + x\hat{j} = x\hat{j} - y\hat{k} \]

3. **Calculate \(\vec{r} \times \hat{j}\)**:
\[ \vec{r} \times \hat{j} = (x \hat{i} + y \hat{j} + x \hat{k}) \times \hat{j} \]
Using the properties of the cross product:
- \(\hat{i} \times \hat{j} = \hat{k}\)
- \(\hat{j} \times \hat{j} = 0\)
- \(\hat{k} \times \hat{j} = -\hat{i}\)

Therefore,
\[ \vec{r} \times \hat{j} = x\hat{k} + 0 - x\hat{i} = x\hat{k} - x\hat{i} \]

4. **Calculate the Dot Product**:
Now we need to compute \((\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j})\):
\[ (x\hat{j} - y\hat{k}) \cdot (x\hat{k} - x\hat{i}) \]
Expanding this using the distributive property of the dot product:
\[ = x\hat{j} \cdot x\hat{k} - x\hat{j} \cdot x\hat{i} - y\hat{k} \cdot x\hat{k} + y\hat{k} \cdot x\hat{i} \]
- \(\hat{j} \cdot \hat{k} = 0\)
- \(\hat{j} \cdot \hat{i} = 0\)
- \(\hat{k} \cdot \hat{k} = 1\)
- \(\hat{k} \cdot \hat{i} = 0\)

Therefore, we have:
\[ = 0 - 0 - yx + 0 = -yx \]

5. **Combine with \(xy\)**:
Now we add \(xy\) to the result:
\[ -yx + xy = 0 \]

### Final Result:
Thus, the final answer is:
\[ (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + xy = 0 \]