Image of the point P with position vector `7hat(i) - hat(j) + 2hat(k)` in the line whose vector equation is `vec(r) = 9hat(i) + 5hat(j) + 5hat(k) + lambda(hat(i) + 3hat(j) + 5hat(k))` has the position .

Options

A. (–9, 5, 2)
B. (9, 5, –2)
C. (9, –5, –2)
D. none of these

Answer

To find the image of the point \( P \) with position vector \( \vec{p} = 7\hat{i} - \hat{j} + 2\hat{k} \) in the line given by the vector equation

\[ \vec{r} = 9\hat{i} + 5\hat{j} + 5\hat{k} + \lambda(\hat{i} + 3\hat{j} + 5\hat{k}), \]

we can follow these steps:

### Step 1: Identify the direction ratios of the line
The direction ratios of the line can be extracted from the coefficients of \( \lambda \) in the vector equation. The direction ratios are \( \hat{i} + 3\hat{j} + 5\hat{k} \), which gives us:

\[ \text{Direction ratios} = (1, 3, 5). \]

### Step 2: Determine the coordinates of point \( P \)
The coordinates of point \( P \) from its position vector \( \vec{p} = 7\hat{i} - \hat{j} + 2\hat{k} \) are:

\[ P(7, -1, 2). \]

### Step 3: Set up the image point \( Q(a, b, c) \)
Let the image of point \( P \) be point \( Q \) with coordinates \( (a, b, c) \). The direction ratios of the line segment \( PQ \) will be:

\[ (a - 7, b + 1, c - 2). \]

### Step 4: Use the perpendicularity condition
Since \( PQ \) is perpendicular to the line, we can use the dot product condition:

\[ (a - 7) \cdot 1 + (b + 1) \cdot 3 + (c - 2) \cdot 5 = 0. \]

Expanding this gives:

\[ a - 7 + 3b + 3 + 5c - 10 = 0, \]

which simplifies to:

\[ a + 3b + 5c = 14. \quad \text{(Equation 1)} \]

### Step 5: Find the midpoint of segment \( PQ \)
The midpoint \( M \) of segment \( PQ \) can be expressed as:

\[ M\left(\frac{7 + a}{2}, \frac{-1 + b}{2}, \frac{2 + c}{2}\right). \]

### Step 6: Parametrize the line
The line can be parametrized as follows:

\[ x = 9 + \lambda, \quad y = 5 + 3\lambda, \quad z = 5 + 5\lambda. \]

### Step 7: Equate the midpoint to the line
Since the midpoint \( M \) lies on the line, we can set up the following equations:

1. \(\frac{7 + a}{2} = 9 + \lambda\)
2. \(\frac{-1 + b}{2} = 5 + 3\lambda\)
3. \(\frac{2 + c}{2} = 5 + 5\lambda\)

### Step 8: Solve for \( a, b, c \) in terms of \( \lambda \)
From the first equation:

\[ 7 + a = 18 + 2\lambda \implies a = 2\lambda + 11. \]

From the second equation:

\[ -1 + b = 10 + 6\lambda \implies b = 6\lambda + 11. \]

From the third equation:

\[ 2 + c = 10 + 10\lambda \implies c = 10\lambda + 8. \]

### Step 9: Substitute into Equation 1
Substituting \( a, b, c \) into Equation 1:

\[ (2\lambda + 11) + 3(6\lambda + 11) + 5(10\lambda + 8) = 14. \]

Expanding this gives:

\[ 2\lambda + 11 + 18\lambda + 33 + 50\lambda + 40 = 14. \]

Combining like terms:

\[ 70\lambda + 84 = 14 \implies 70\lambda = -70 \implies \lambda = -1. \]

### Step 10: Calculate \( a, b, c \)
Now substituting \( \lambda = -1 \):

\[ a = 2(-1) + 11 = 9, \]
\[ b = 6(-1) + 11 = 5, \]
\[ c = 10(-1) + 8 = -2. \]

### Conclusion
Thus, the coordinates of the image point \( Q \) are:

\[ Q(9, 5, -2). \]