which of the following point lies on plane contaning lines `vecr=hati+lambda(hati+hatj+hatk)` and `vecr = -hatj+mu(-hati-2hatj+hatk)`

Options

A. `(1,3,6)`
B. `(1,-3,6)`
C. `(-2,1,2)`
D. `(1,3,1)`

Answer

To determine which point lies on the plane containing the given lines, we will follow these steps:

### Step 1: Identify the Direction Vectors of the Lines
The lines are given in vector form:
1. \( \vec{r} = \hat{i} + \lambda(\hat{i} + \hat{j} + \hat{k}) \)
2. \( \vec{r} = -\hat{j} + \mu(-\hat{i} - 2\hat{j} + \hat{k}) \)

From the first line, we can extract the direction vector:
- For Line 1, the direction vector \( \vec{b_1} = \hat{i} + \hat{j} + \hat{k} \).

From the second line, we can extract the direction vector:
- For Line 2, the direction vector \( \vec{b_2} = -\hat{i} - 2\hat{j} + \hat{k} \).

### Step 2: Calculate the Normal Vector of the Plane
The normal vector \( \vec{n} \) of the plane can be found using the cross product of the two direction vectors:
\[ \vec{n} = \vec{b_1} \times \vec{b_2} \]

Calculating the cross product:
\[ \vec{b_1} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \vec{b_2} = \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix} \]
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ -1 & -2 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i} \begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ -1 & -2 \end{vmatrix} \]
\[ = \hat{i}(1 \cdot 1 - 1 \cdot (-2)) - \hat{j}(1 \cdot 1 - 1 \cdot (-1)) + \hat{k}(1 \cdot (-2) - 1 \cdot (-1)) \]
\[ = \hat{i}(1 + 2) - \hat{j}(1 + 1) + \hat{k}(-2 + 1) \]
\[ = 3\hat{i} - 2\hat{j} - \hat{k} \]

### Step 3: Write the Equation of the Plane
Using the point-normal form of the plane equation:
\[ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \]
where \( \vec{a} \) is a point on the plane. We can take \( \vec{a} = \hat{i} \) (from Line 1 when \( \lambda = 0 \)).

The equation becomes:
\[ \vec{r} \cdot (3\hat{i} - 2\hat{j} - \hat{k}) = \hat{i} \cdot (3\hat{i} - 2\hat{j} - \hat{k}) \]
Calculating the right-hand side:
\[ \hat{i} \cdot (3\hat{i} - 2\hat{j} - \hat{k}) = 3 \]
Thus, the equation of the plane is:
\[ 3x - 2y - z = 3 \]

### Step 4: Check Which Points Lie on the Plane
Now we will check each given point to see if it satisfies the plane equation \( 3x - 2y - z - 3 = 0 \).

1. **Point (1, 3, 6)**:
\[ 3(1) - 2(3) - 6 - 3 = 3 - 6 - 6 - 3 = -12 \quad (\text{not on the plane}) \]

2. **Point (1, -3, 6)**:
\[ 3(1) - 2(-3) - 6 - 3 = 3 + 6 - 6 - 3 = 0 \quad (\text{on the plane}) \]

3. **Point (-2, 1, 2)**:
\[ 3(-2) - 2(1) - 2 - 3 = -6 - 2 - 2 - 3 = -13 \quad (\text{not on the plane}) \]

4. **Point (1, 3, 1)**:
\[ 3(1) - 2(3) - 1 - 3 = 3 - 6 - 1 - 3 = -7 \quad (\text{not on the plane}) \]

### Conclusion
The point that lies on the plane is **(1, -3, 6)**.