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. (a)`2x-3y+1=0`
B. (b)`x-3y+2z=0`
C. (c)`3x-y+z-3=0`
D. (d)`3x-y-2=0`

Answer

To find the equation of the plane containing the points represented by the position vectors \( \mathbf{p} = \mathbf{i} + \mathbf{j} + \mathbf{k} \), \( \mathbf{q} = 2\mathbf{i} + 4\mathbf{j} - \mathbf{k} \), and \( \mathbf{r} = \mathbf{i} + \mathbf{j} + 3\mathbf{k} \), we can follow these steps:

### Step 1: Identify the position vectors
We have the following position vectors:
- \( \mathbf{p} = (1, 1, 1) \)
- \( \mathbf{q} = (2, 4, -1) \)
- \( \mathbf{r} = (1, 1, 3) \)

### Step 2: Find two vectors in the plane
To find the equation of the plane, we need two vectors that lie in the plane. We can find these vectors by subtracting the position vectors:
- \( \mathbf{pq} = \mathbf{q} - \mathbf{p} = (2 - 1, 4 - 1, -1 - 1) = (1, 3, -2) \)
- \( \mathbf{pr} = \mathbf{r} - \mathbf{p} = (1 - 1, 1 - 1, 3 - 1) = (0, 0, 2) \)

### Step 3: Find the normal vector to the plane
The normal vector \( \mathbf{n} \) to the plane can be found by taking the cross product of the vectors \( \mathbf{pq} \) and \( \mathbf{pr} \):
\[ \mathbf{n} = \mathbf{pq} \times \mathbf{pr} \]

Calculating the cross product:
\[ \mathbf{pq} = (1, 3, -2), \quad \mathbf{pr} = (0, 0, 2) \]
Using the determinant formula for the cross product:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & -2 \\ 0 & 0 & 2 \end{vmatrix} = \mathbf{i}(3 \cdot 2 - 0 \cdot -2) - \mathbf{j}(1 \cdot 2 - 0 \cdot -2) + \mathbf{k}(1 \cdot 0 - 3 \cdot 0) \]
\[ = 6\mathbf{i} - 2\mathbf{j} + 0\mathbf{k} = (6, -2, 0) \]

### Step 4: Write the equation of the plane
The equation of a plane can be expressed as:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane (we can use point \( \mathbf{p} \)) and \( (a, b, c) \) are the components of the normal vector \( \mathbf{n} \).

Using \( \mathbf{p} = (1, 1, 1) \) and \( \mathbf{n} = (6, -2, 0) \):
\[ 6(x - 1) - 2(y - 1) + 0(z - 1) = 0 \]
Expanding this gives:
\[ 6x - 6 - 2y + 2 = 0 \implies 6x - 2y - 4 = 0 \]
Dividing through by 2:
\[ 3x - y - 2 = 0 \]

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