Consider three vectors `p=i+j+k, q=2i+4j-k and r=i+j+3k`. If p, q and r denotes the position vector of three non-collinear points, then the equation of the plane containing these points is

Options

A. `2x-3y+1=0`
B. `x-3y+2z=0`
C. `3x-y+z-3=0`
D. `3x-y-2=0`

Answer

To find the equation of the plane containing the points represented by the position vectors \( \mathbf{p} \), \( \mathbf{q} \), and \( \mathbf{r} \), we can follow these steps:

### Step 1: Identify the position vectors
The position vectors are given as:
- \( \mathbf{p} = \mathbf{i} + \mathbf{j} + \mathbf{k} \)
- \( \mathbf{q} = 2\mathbf{i} + 4\mathbf{j} - \mathbf{k} \)
- \( \mathbf{r} = \mathbf{i} + \mathbf{j} + 3\mathbf{k} \)

### Step 2: Calculate the vectors \( \mathbf{PQ} \) and \( \mathbf{PR} \)
To find the vectors \( \mathbf{PQ} \) and \( \mathbf{PR} \):
- \( \mathbf{PQ} = \mathbf{q} - \mathbf{p} \)
\[ \mathbf{PQ} = (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) - (\mathbf{i} + \mathbf{j} + \mathbf{k}) = (2-1)\mathbf{i} + (4-1)\mathbf{j} + (-1-1)\mathbf{k} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} \]

- \( \mathbf{PR} = \mathbf{r} - \mathbf{p} \)
\[ \mathbf{PR} = (\mathbf{i} + \mathbf{j} + 3\mathbf{k}) - (\mathbf{i} + \mathbf{j} + \mathbf{k}) = (1-1)\mathbf{i} + (1-1)\mathbf{j} + (3-1)\mathbf{k} = 0\mathbf{i} + 0\mathbf{j} + 2\mathbf{k} = 2\mathbf{k} \]

### Step 3: Find the normal vector to the plane
The normal vector \( \mathbf{n} \) to the plane can be found using the cross product of \( \mathbf{PQ} \) and \( \mathbf{PR} \):
\[ \mathbf{n} = \mathbf{PQ} \times \mathbf{PR} \]
Calculating the cross product:
\[ \mathbf{PQ} = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}, \quad \mathbf{PR} = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} \]
Using the determinant to calculate the cross product:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & -2 \\ 0 & 0 & 2 \end{vmatrix} = \mathbf{i}(3 \cdot 2 - 0 \cdot -2) - \mathbf{j}(1 \cdot 2 - 0 \cdot -2) + \mathbf{k}(1 \cdot 0 - 3 \cdot 0) \]
\[ = 6\mathbf{i} - 2\mathbf{j} + 0\mathbf{k} = 6\mathbf{i} - 2\mathbf{j} \]

### Step 4: Write the equation of the plane
The equation of the plane can be expressed as:
\[ 6(x - x_0) - 2(y - y_0) + 0(z - z_0) = 0 \]
Using point \( \mathbf{p}(1, 1, 1) \):
\[ 6(x - 1) - 2(y - 1) = 0 \]
Expanding this gives:
\[ 6x - 6 - 2y + 2 = 0 \implies 6x - 2y - 4 = 0 \implies 3x - y - 2 = 0 \]

### Final Answer
The equation of the plane containing the points represented by the vectors \( \mathbf{p} \), \( \mathbf{q} \), and \( \mathbf{r} \) is:
\[ 3x - y - 2 = 0 \]