Let A be the point (1,2,3) and B be a point on the line `(x-1)/-2=(y+1)/3=(z-5)/4=k` Then value of k such that line AB is perpendicular to the plane `4x+9y-18z=6` is

Options

A. `-2/5`
B. `1/5`
C. `2/5`
D. no such value of k is possible

Answer

To solve the problem, we need to find the value of \( k \) such that the line segment \( AB \) is perpendicular to the plane defined by the equation \( 4x + 9y - 18z = 6 \).

### Step 1: Identify Points A and B
Let \( A = (1, 2, 3) \) be the given point. The point \( B \) lies on the line defined by the equation:
\[ \frac{x-1}{-2} = \frac{y+1}{3} = \frac{z-5}{4} = k \]
From this, we can express the coordinates of point \( B \) in terms of \( k \):
- For \( x \):
\[ x = -2k + 1 \]
- For \( y \):
\[ y = 3k - 1 \]
- For \( z \):
\[ z = 4k + 5 \]
Thus, the coordinates of point \( B \) can be expressed as:
\[ B = (-2k + 1, 3k - 1, 4k + 5) \]

### Step 2: Find Direction Ratios of Line AB
The direction ratios of the line segment \( AB \) can be calculated as follows:
\[ AB = B - A = (-2k + 1 - 1, 3k - 1 - 2, 4k + 5 - 3) \]
This simplifies to:
\[ AB = (-2k, 3k - 3, 4k + 2) \]

### Step 3: Identify Normal Vector of the Plane
The normal vector \( \vec{n} \) of the plane \( 4x + 9y - 18z = 6 \) can be identified from the coefficients of \( x, y, z \):
\[ \vec{n} = (4, 9, -18) \]

### Step 4: Condition for Perpendicularity
For the line \( AB \) to be perpendicular to the plane, the direction ratios of \( AB \) must be parallel to the normal vector of the plane. This gives us the condition:
\[ \frac{-2k}{4} = \frac{3k - 3}{9} = \frac{4k + 2}{-18} \]

### Step 5: Solve the Ratios
1. From the first ratio:
\[ \frac{-2k}{4} = \frac{3k - 3}{9} \]
Cross-multiplying gives:
\[ -2k \cdot 9 = 4(3k - 3) \]
This simplifies to:
\[ -18k = 12k - 12 \]
Rearranging gives:
\[ -18k - 12k = -12 \implies -30k = -12 \implies k = \frac{12}{30} = \frac{2}{5} \]

2. To verify, we can check the second ratio:
\[ \frac{-2k}{4} = \frac{4k + 2}{-18} \]
Cross-multiplying gives:
\[ -2k \cdot (-18) = 4(4k + 2) \]
This simplifies to:
\[ 36k = 16k + 8 \]
Rearranging gives:
\[ 36k - 16k = 8 \implies 20k = 8 \implies k = \frac{8}{20} = \frac{2}{5} \]

### Final Answer
Thus, the value of \( k \) such that line \( AB \) is perpendicular to the plane is:
\[ \boxed{\frac{2}{5}} \]