Find the equations of the straight line passing through point (2,3,-1) and is perpendicular t `(x-1)/(2) = (y+1)/1 = (z-3)/(-3) and (x-3)/(1) = (y+2)/(1) = (z-1)/(1) `

Answer

To find the equations of the straight line passing through the point (2, 3, -1) and perpendicular to the two given lines, we can follow these steps:

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

The first line is given by the equation:
\[ \frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-3} \]
From this, we can extract the direction vector \( \mathbf{b_1} \) as:
\[ \mathbf{b_1} = (2, 1, -3) \]

The second line is given by the equation:
\[ \frac{x-3}{1} = \frac{y+2}{1} = \frac{z-1}{1} \]
From this, we can extract the direction vector \( \mathbf{b_2} \) as:
\[ \mathbf{b_2} = (1, 1, 1) \]

### Step 2: Set up the equations for the direction vector of the required line.

Let the direction vector of the required line be \( \mathbf{b} = (b_1, b_2, b_3) \). Since this line is perpendicular to both given lines, we can use the dot product to set up the following equations:

1. \( \mathbf{b} \cdot \mathbf{b_1} = 0 \):
\[ 2b_1 + 1b_2 - 3b_3 = 0 \quad \text{(Equation 1)} \]

2. \( \mathbf{b} \cdot \mathbf{b_2} = 0 \):
\[ 1b_1 + 1b_2 + 1b_3 = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations.

From Equation 1:
\[ 2b_1 + b_2 - 3b_3 = 0 \quad \text{(1)} \]

From Equation 2:
\[ b_1 + b_2 + b_3 = 0 \quad \text{(2)} \]

We can express \( b_2 \) from Equation (2):
\[ b_2 = -b_1 - b_3 \]
Substituting this into Equation (1):
\[ 2b_1 + (-b_1 - b_3) - 3b_3 = 0 \]
This simplifies to:
\[ 2b_1 - b_1 - b_3 - 3b_3 = 0 \implies b_1 - 4b_3 = 0 \implies b_1 = 4b_3 \]

Let \( b_3 = k \), then:
\[ b_1 = 4k \]
Substituting \( b_1 \) back into Equation (2):
\[ 4k + b_2 + k = 0 \implies b_2 = -5k \]

Thus, the direction vector \( \mathbf{b} \) can be expressed as:
\[ \mathbf{b} = (4k, -5k, k) \]

### Step 4: Write the equation of the line in vector form.

Using the point \( (2, 3, -1) \) and the direction vector \( \mathbf{b} \):
\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]
Where \( \mathbf{a} = (2, 3, -1) \) and \( \mathbf{b} = (4k, -5k, k) \):
\[ \mathbf{r} = (2, 3, -1) + \lambda (4k, -5k, k) \]

### Step 5: Convert to Cartesian form.

To express the line in Cartesian form, we can write:
\[ \frac{x - 2}{4k} = \frac{y - 3}{-5k} = \frac{z + 1}{k} \]

### Final Result:

The equations of the line in Cartesian form are:
\[ \frac{x - 2}{4} = \frac{y - 3}{-5} = \frac{z + 1}{1} \]