Find the vector equation of the line perpendicular to the lines ` (x-1)/( 2) = ( y-1)/(2) = (z-3)/( 4) and ( x+1)/( 4)= ( y)/(3) = z-1 ` and which passes through the point (2,1,2)

Answer

To find the vector equation of the line perpendicular to the given lines and passing through the point (2, 1, 2), we will follow these steps:

### Step 1: Identify the direction vectors of the given lines

The first line is given by:
\[ \frac{x-1}{2} = \frac{y-1}{2} = \frac{z-3}{4} \]
From this, we can identify the direction vector \( \mathbf{A} \) as:
\[ \mathbf{A} = 2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k} \]

The second line is given by:
\[ \frac{x+1}{4} = \frac{y}{3} = z-1 \]
From this, we can identify the direction vector \( \mathbf{B} \) as:
\[ \mathbf{B} = 4\mathbf{i} + 3\mathbf{j} + 1\mathbf{k} \]

### Step 2: Find the cross product of the direction vectors

To find a vector that is perpendicular to both lines, we calculate the cross product \( \mathbf{A} \times \mathbf{B} \).

\[ \mathbf{A} = \begin{pmatrix} 2 \\ 2 \\ 4 \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} \]

The cross product \( \mathbf{A} \times \mathbf{B} \) is calculated using the determinant:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 4 \\ 4 & 3 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 2 & 4 \\ 3 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 4 \\ 4 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 2 \\ 4 & 3 \end{vmatrix} \]
\[ = \mathbf{i} (2 \cdot 1 - 4 \cdot 3) - \mathbf{j} (2 \cdot 1 - 4 \cdot 4) + \mathbf{k} (2 \cdot 3 - 2 \cdot 4) \]
\[ = \mathbf{i} (2 - 12) - \mathbf{j} (2 - 16) + \mathbf{k} (6 - 8) \]
\[ = -10\mathbf{i} + 14\mathbf{j} - 2\mathbf{k} \]

Thus, the vector perpendicular to both lines is:
\[ \mathbf{C} = -10\mathbf{i} + 14\mathbf{j} - 2\mathbf{k} \]

### Step 3: Write the vector equation of the line

The vector equation of a line passing through a point \( \mathbf{A} \) and parallel to a vector \( \mathbf{C} \) is given by:
\[ \mathbf{r} = \mathbf{A} + \lambda \mathbf{C} \]

The point through which the line passes is \( (2, 1, 2) \), which can be expressed as:
\[ \mathbf{A} = 2\mathbf{i} + 1\mathbf{j} + 2\mathbf{k} \]

Substituting \( \mathbf{A} \) and \( \mathbf{C} \):
\[ \mathbf{r} = (2\mathbf{i} + 1\mathbf{j} + 2\mathbf{k}) + \lambda (-10\mathbf{i} + 14\mathbf{j} - 2\mathbf{k}) \]

### Final Expression

Thus, the vector equation of the required line is:
\[ \mathbf{r} = (2 - 10\lambda)\mathbf{i} + (1 + 14\lambda)\mathbf{j} + (2 - 2\lambda)\mathbf{k} \]