Equations of the line which passe through the point with position vector `(2, 1, 0)` and perpendicular to the plane containing the vectors `i+j and j+k ` is

Options

A. `r=(2, 1, 0)+t(1, -1, 1)`
B. `r=(2, 1, 0)+t(-1, 1, 1)`
C. `r=(2, 1, 0)+t(1, 1, -1)`
D. `r=(2, 1, 0)+t(1, 1, 1)`

Answer

To find the equation of the line that passes through the point with position vector \( \mathbf{r_0} = (2, 1, 0) \) and is perpendicular to the plane containing the vectors \( \mathbf{u} = \mathbf{i} + \mathbf{j} \) and \( \mathbf{v} = \mathbf{j} + \mathbf{k} \), we can follow these steps:

### Step 1: Find the normal vector to the plane
The normal vector \( \mathbf{n} \) to the plane can be found using the cross product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \).

\[ \mathbf{u} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \quad \mathbf{v} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \]

The cross product \( \mathbf{n} = \mathbf{u} \times \mathbf{v} \) is calculated using the determinant:

\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ \mathbf{n} = \mathbf{i} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} \]

Calculating the minors:

\[ \mathbf{n} = \mathbf{i} (1 \cdot 1 - 0 \cdot 1) - \mathbf{j} (1 \cdot 1 - 0 \cdot 0) + \mathbf{k} (1 \cdot 1 - 1 \cdot 0) \]

\[ \mathbf{n} = \mathbf{i} (1) - \mathbf{j} (1) + \mathbf{k} (1) = \mathbf{i} - \mathbf{j} + \mathbf{k} \]

Thus, the normal vector is:

\[ \mathbf{n} = (1, -1, 1) \]

### Step 2: Write the equation of the line
The equation of the line can be expressed in vector form as:

\[ \mathbf{r} = \mathbf{r_0} + t \mathbf{b} \]

where \( \mathbf{r_0} = (2, 1, 0) \) is the position vector of the point through which the line passes, \( t \) is a parameter, and \( \mathbf{b} \) is the direction vector of the line, which is parallel to the normal vector \( \mathbf{n} \).

Thus, we have:

\[ \mathbf{b} = (1, -1, 1) \]

Substituting into the equation of the line:

\[ \mathbf{r} = (2, 1, 0) + t(1, -1, 1) \]

### Step 3: Write the final equation
The final equation of the line in vector form is:

\[ \mathbf{r} = (2 + t, 1 - t, t) \]

### Summary
The equation of the line that passes through the point \( (2, 1, 0) \) and is perpendicular to the plane containing the vectors \( \mathbf{i} + \mathbf{j} \) and \( \mathbf{j} + \mathbf{k} \) is given by:

\[ \mathbf{r} = (2, 1, 0) + t(1, -1, 1) \]