Find the image of the point `(3hati-hatj+11hatk)` in the line `vecr = 2 hatj + 3hatk+lambda(2hati+3hatj+4hatk)`.

Answer

To find the image of the point \( \vec{P} = 3\hat{i} - \hat{j} + 11\hat{k} \) in the line given by \( \vec{r} = 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k}) \), we can follow these steps:

### Step 1: Identify the Point and the Line
The point \( \vec{P} \) can be expressed in coordinate form as \( (3, -1, 11) \). The line can be expressed in parametric form:
- \( x = 2\lambda \)
- \( y = 2 + 3\lambda \)
- \( z = 3 + 4\lambda \)

### Step 2: Set Up the Midpoint Condition
Let the image of the point \( \vec{P} \) be \( \vec{A} = (a, b, c) \). According to the midpoint theorem, the midpoint \( M \) between points \( \vec{P} \) and \( \vec{A} \) lies on the line. Therefore, we have:
\[ M = \left( \frac{3 + a}{2}, \frac{-1 + b}{2}, \frac{11 + c}{2} \right) \]
This midpoint \( M \) must satisfy the parametric equations of the line.

### Step 3: Equate the Midpoint to the Line
From the line equations, we have:
1. \( \frac{3 + a}{2} = 2\lambda \)
2. \( \frac{-1 + b}{2} = 2 + 3\lambda \)
3. \( \frac{11 + c}{2} = 3 + 4\lambda \)

### Step 4: Solve for \( a, b, c \) in terms of \( \lambda \)
From the first equation:
\[ 3 + a = 4\lambda \implies a = 4\lambda - 3 \]
From the second equation:
\[ -1 + b = 4 + 6\lambda \implies b = 6\lambda + 5 \]
From the third equation:
\[ 11 + c = 6 + 8\lambda \implies c = 8\lambda - 5 \]

### Step 5: Find Direction Ratios of the Line Segment
The direction ratios of the line segment \( \vec{PA} \) from \( \vec{P} \) to \( \vec{A} \) are:
- \( a - 3 = 4\lambda - 6 \)
- \( b + 1 = 6\lambda + 6 \)
- \( c - 11 = 8\lambda - 16 \)

### Step 6: Set Up the Perpendicular Condition
Since the line segment \( \vec{PA} \) is perpendicular to the line, their direction ratios must satisfy the dot product condition:
\[ (2)(4\lambda - 6) + (3)(6\lambda + 6) + (4)(8\lambda - 16) = 0 \]

### Step 7: Simplify and Solve for \( \lambda \)
Expanding the equation:
\[ 8\lambda - 12 + 18\lambda + 18 + 32\lambda - 64 = 0 \]
Combining like terms:
\[ 58\lambda - 58 = 0 \implies \lambda = 1 \]

### Step 8: Substitute \( \lambda \) Back to Find \( a, b, c \)
Substituting \( \lambda = 1 \):
\[ a = 4(1) - 3 = 1 \]
\[ b = 6(1) + 5 = 11 \]
\[ c = 8(1) - 5 = 3 \]

### Step 9: Write the Image Point
Thus, the image of the point \( \vec{P} \) is:
\[ \vec{A} = 1\hat{i} + 11\hat{j} + 3\hat{k} \]

### Final Answer
The image of the point \( (3\hat{i} - \hat{j} + 11\hat{k}) \) in the line is \( \vec{A} = \hat{i} + 11\hat{j} + 3\hat{k} \).

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