The solution vectors `vecr` of the equation `vecr xx hati=hatj+hatk and vecr xx hatj=hatk+hatj` represent two straight lines which are :

Options

A. Intersecting
B. Non coplanar
C. Coplanar
D. Non intersecting

Answer

To solve the given problem, we need to find the solution vectors \( \vec{r} \) for the equations \( \vec{r} \times \hat{i} = \hat{j} + \hat{k} \) and \( \vec{r} \times \hat{j} = \hat{k} + \hat{j} \). Let's break this down step by step.

### Step 1: Define the vector \( \vec{r} \)
Let \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \), where \( x, y, z \) are the components of the vector.

### Step 2: Solve the first equation
The first equation is:
\[ \vec{r} \times \hat{i} = \hat{j} + \hat{k} \]
Calculating the cross product:
\[ \vec{r} \times \hat{i} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{i} \]
Using the properties of the cross product:
\[ = y (\hat{j} \times \hat{i}) + z (\hat{k} \times \hat{i}) = y (-\hat{k}) + z \hat{j} = -y \hat{k} + z \hat{j} \]
Setting this equal to \( \hat{j} + \hat{k} \):
\[ -z \hat{k} + z \hat{j} = \hat{j} + \hat{k} \]
Now, comparing coefficients:
- For \( \hat{j} \): \( z = 1 \)
- For \( \hat{k} \): \( -y = 1 \) which gives \( y = -1 \)

Thus, from the first equation, we have:
\[ y = -1, \quad z = 1 \]

### Step 3: Write the equation of the line from the first equation
From the values we found, we can express \( \vec{r} \) as:
\[ \vec{r} = x \hat{i} - \hat{j} + \hat{k} \]
This represents a line parallel to the x-axis.

### Step 4: Solve the second equation
Now, we solve the second equation:
\[ \vec{r} \times \hat{j} = \hat{k} + \hat{j} \]
Calculating the cross product:
\[ \vec{r} \times \hat{j} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{j} \]
Using the properties of the cross product:
\[ = x (\hat{i} \times \hat{j}) + z (\hat{k} \times \hat{j}) = x \hat{k} - z \hat{i} \]
Setting this equal to \( \hat{k} + \hat{j} \):
\[ x \hat{k} - z \hat{i} = \hat{k} + \hat{j} \]
Now, comparing coefficients:
- For \( \hat{k} \): \( x = 1 \)
- For \( \hat{j} \): There is no \( \hat{j} \) term on the left side, so \( y \) must be 0.

Thus, from the second equation, we have:
\[ x = 1, \quad z = 1 \]

### Step 5: Write the equation of the line from the second equation
From the values we found, we can express \( \vec{r} \) as:
\[ \vec{r} = \hat{i} + 0 \hat{j} + \hat{k} \]
This represents a line parallel to the y-axis.

### Conclusion
We have two lines:
1. The first line is parallel to the x-axis: \( \vec{r} = x \hat{i} - \hat{j} + \hat{k} \)
2. The second line is parallel to the y-axis: \( \vec{r} = \hat{i} + 0 \hat{j} + \hat{k} \)

Since both lines intersect at the point \( (1, -1, 1) \), the solution is that the two lines are intersecting.