Direction ratios of the line perpendicular to the lines `(x-3)/(2) = (y+7)/( -3) = (z-2)/(1) and (x+2)/(1) = (y+3)/( 2) = ( z-5)/(-2)` are

Options

A. `lt 4, -5 , 7 gt`
B. ` lt -4, 5, 7 gt`
C. `lt 4, 5, -7 gt`
D. `lt 4, 5, 7 gt`

Answer

To find the direction ratios of the line that is perpendicular to the given lines, we will follow these steps:

### Step 1: Identify the direction ratios of the given lines
The equations of the lines are given in symmetric form:
1. For the first line: \((x-3)/(2) = (y+7)/(-3) = (z-2)/(1)\)
- The direction ratios (DR) of the first line are \(2, -3, 1\).

2. For the second line: \((x+2)/(1) = (y+3)/(2) = (z-5)/(-2)\)
- The direction ratios (DR) of the second line are \(1, 2, -2\).

### Step 2: Represent the direction ratios as vectors
- Let \( \mathbf{B_1} = 2\mathbf{i} - 3\mathbf{j} + 1\mathbf{k} \) for the first line.
- Let \( \mathbf{B_2} = 1\mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \) for the second line.

### Step 3: Calculate the cross product of the two direction vectors
To find a vector that is perpendicular to both lines, we will compute the cross product \( \mathbf{B_1} \times \mathbf{B_2} \).

\[ \mathbf{B_1} \times \mathbf{B_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 1 \\ 1 & 2 & -2 \end{vmatrix} \]

### Step 4: Compute the determinant
Calculating the determinant, we have:

\[ \mathbf{B_1} \times \mathbf{B_2} = \mathbf{i} \left((-3)(-2) - (1)(2)\right) - \mathbf{j} \left((2)(-2) - (1)(1)\right) + \mathbf{k} \left((2)(2) - (-3)(1)\right) \]

Calculating each component:
- For \( \mathbf{i} \): \( 6 - 2 = 4 \)
- For \( \mathbf{j} \): \( -(-4 - 1) = 5 \)
- For \( \mathbf{k} \): \( 4 + 3 = 7 \)

Thus, we have:
\[ \mathbf{B_1} \times \mathbf{B_2} = 4\mathbf{i} + 5\mathbf{j} + 7\mathbf{k} \]

### Step 5: Write the direction ratios
The direction ratios of the line that is perpendicular to both given lines are \(4, 5, 7\).

### Final Answer
The direction ratios of the line perpendicular to the given lines are \(4, 5, 7\).

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