If the line `(x-1)/(2)=(y-3)/(a)=(z+1)/(3)` lies in the plane `bx+2y+3z-4=0`, then find a and b.

Options

A. `a=11//2,b=1`
B. `a=-5//2,b=-7`
C. `a=-11//2,b=1`
D. `a=1,b=-11//2`

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) given that the line

\[ \frac{x-1}{2} = \frac{y-3}{a} = \frac{z+1}{3} \]

lies in the plane

\[ bx + 2y + 3z - 4 = 0. \]

### Step 1: Identify a point on the line

From the equation of the line, we can identify a point on the line. When \( t = 0 \):

\[ x = 1, \quad y = 3, \quad z = -1. \]

Thus, the point \( P(1, 3, -1) \) lies on the line.

### Step 2: Substitute the point into the plane equation

Since the line lies in the plane, the point \( P(1, 3, -1) \) must satisfy the plane equation. Substitute \( x = 1 \), \( y = 3 \), and \( z = -1 \) into the plane equation:

\[ b(1) + 2(3) + 3(-1) - 4 = 0. \]

### Step 3: Simplify the equation

Now, simplify the equation:

\[ b + 6 - 3 - 4 = 0, \]

which simplifies to:

\[ b + 6 - 7 = 0, \]

or

\[ b - 1 = 0. \]

### Step 4: Solve for \( b \)

From the equation above, we find:

\[ b = 1. \]

### Step 5: Find the direction ratios of the line

The direction ratios of the line are \( (2, a, 3) \).

### Step 6: Use the normal vector of the plane

The normal vector of the plane \( bx + 2y + 3z - 4 = 0 \) is given by \( (b, 2, 3) = (1, 2, 3) \).

### Step 7: Set up the dot product condition

Since the line lies in the plane, the direction ratios of the line must be perpendicular to the normal vector of the plane. Therefore, we can set up the dot product:

\[ (2, a, 3) \cdot (1, 2, 3) = 0. \]

### Step 8: Calculate the dot product

Calculating the dot product gives:

\[ 2 \cdot 1 + a \cdot 2 + 3 \cdot 3 = 0. \]

This simplifies to:

\[ 2 + 2a + 9 = 0. \]

### Step 9: Solve for \( a \)

Now, simplifying gives:

\[ 2a + 11 = 0, \]

which leads to:

\[ 2a = -11, \]

thus:

\[ a = -\frac{11}{2}. \]

### Final Answer

The values of \( a \) and \( b \) are:

\[ a = -\frac{11}{2}, \quad b = 1. \]