The direction ratios of the line `I_1` passing through `P(1, 3, 4)` and perpendicular to line `I_2 (x-1)/(2)=(y-2)/(3)=(z-3)/(4)` (where `I_1 and I_2` are coplanar) is

Options

A. `14, 8, 1`
B. `-14, 8, -1`
C. `14, -8, -1`
D. `-14, -8, 1`

Answer

To find the direction ratios of the line \( I_1 \) that passes through the point \( P(1, 3, 4) \) and is perpendicular to the line \( I_2 \) given by the equations \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \), we can follow these steps:

### Step 1: Identify the direction ratios of line \( I_2 \)
The line \( I_2 \) is given in symmetric form. The direction ratios of \( I_2 \) can be extracted directly from the coefficients in the equation:
\[ \text{Direction ratios of } I_2 = (2, 3, 4) \]

### Step 2: Set up the condition for perpendicularity
For two lines to be perpendicular, the dot product of their direction ratios must equal zero. Let the direction ratios of line \( I_1 \) be \( (a, b, c) \). The condition for perpendicularity is:
\[ 2a + 3b + 4c = 0 \quad \text{(Equation 1)} \]

### Step 3: Set up the condition for coplanarity
The lines \( I_1 \) and \( I_2 \) are coplanar. For two lines to be coplanar, the determinant formed by their direction ratios and the vector connecting points on the two lines must equal zero.

The position vector of point \( P(1, 3, 4) \) is \( (1, 3, 4) \) and a point on line \( I_2 \) can be taken as \( (1, 2, 3) \). The vector connecting these points is:
\[ (1 - 1, 3 - 2, 4 - 3) = (0, 1, 1) \]

Now, we can form the determinant:
\[ \begin{vmatrix} 0 & 1 & 1 \\ 2 & 3 & 4 \\ a & b & c \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Calculating the determinant:
\[ 0 \cdot (3c - 4b) - 1 \cdot (2c - 4a) + 1 \cdot (2b - 3a) = 0 \]
This simplifies to:
\[ -2c + 4a + 2b - 3a = 0 \]
Rearranging gives:
\[ (4a - 3a) + 2b - 2c = 0 \quad \Rightarrow \quad a + 2b - 2c = 0 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \( 2a + 3b + 4c = 0 \) (Equation 1)
2. \( a + 2b - 2c = 0 \) (Equation 2)

From Equation 2, we can express \( a \) in terms of \( b \) and \( c \):
\[ a = 2c - 2b \]

Substituting this into Equation 1:
\[ 2(2c - 2b) + 3b + 4c = 0 \]
This simplifies to:
\[ 4c - 4b + 3b + 4c = 0 \quad \Rightarrow \quad 8c - b = 0 \quad \Rightarrow \quad b = 8c \]

### Step 6: Substitute back to find \( a, b, c \)
Substituting \( b = 8c \) back into the expression for \( a \):
\[ a = 2c - 2(8c) = 2c - 16c = -14c \]

### Step 7: Choose a value for \( c \)
Let’s choose \( c = -1 \) (for simplicity):
\[ b = 8(-1) = -8, \quad a = -14(-1) = 14 \]

### Conclusion
Thus, the direction ratios of line \( I_1 \) are:
\[ \text{Direction ratios of } I_1 = (14, -8, -1) \]