Let `P(1, 7, sqrt2)` be a point and the equation of the line L is `(x-1)/(sqrt2)=(y-7)/(1)=(z-sqrt2)/(-1)`. If PQ is the distnace of the plane `sqrt2x+y-z=1` from the point P measured along a line inclined at an angle `60^(@)` with L, then the length of PQ is

Options

A. 3 units
B. `3sqrt2` units
C. 6 units
D. 8 units

Answer

To solve the problem, we need to find the distance \( PQ \) from the point \( P(1, 7, \sqrt{2}) \) to the plane \( \sqrt{2}x + y - z = 1 \) along a line that is inclined at an angle of \( 60^\circ \) with the line \( L \).

### Step-by-Step Solution:

1. **Identify the Direction Ratios of Line L**:
The equation of the line \( L \) is given as:
\[ \frac{x-1}{\sqrt{2}} = \frac{y-7}{1} = \frac{z-\sqrt{2}}{-1} \]
From this, we can extract the direction ratios of line \( L \):
\[ \text{Direction Ratios} = (\sqrt{2}, 1, -1) \]

2. **Equation of the Line through P Inclined at 60° to L**:
The line through point \( P(1, 7, \sqrt{2}) \) that is inclined at \( 60^\circ \) with line \( L \) will have a direction vector that can be derived from the direction ratios of \( L \). If \( \mathbf{d} = (\sqrt{2}, 1, -1) \), we can find the new direction vector \( \mathbf{d}' \) using the angle \( 60^\circ \).

The cosine of the angle between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by:
\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \]
Here, \( \theta = 60^\circ \), so \( \cos 60^\circ = \frac{1}{2} \).

3. **Finding the Perpendicular Distance PR**:
The perpendicular distance \( PR \) from point \( P(1, 7, \sqrt{2}) \) to the plane \( \sqrt{2}x + y - z = 1 \) can be calculated using the formula for the distance from a point to a plane:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}} \]
Here, \( A = \sqrt{2}, B = 1, C = -1, D = 1 \), and \( (x_1, y_1, z_1) = (1, 7, \sqrt{2}) \).

Plugging in the values:
\[ PR = \frac{|\sqrt{2}(1) + 1(7) - (-1)(\sqrt{2}) - 1|}{\sqrt{(\sqrt{2})^2 + 1^2 + (-1)^2}} \]
Simplifying the numerator:
\[ = \frac{|\sqrt{2} + 7 + \sqrt{2} - 1|}{\sqrt{2 + 1 + 1}} = \frac{|2\sqrt{2} + 6|}{\sqrt{4}} = \frac{2\sqrt{2} + 6}{2} = \sqrt{2} + 3 \]

4. **Finding PQ**:
From the relationship established earlier, \( PQ = 2 \times PR \):
\[ PQ = 2(\sqrt{2} + 3) = 2\sqrt{2} + 6 \]

### Final Answer:
The length of \( PQ \) is \( 2\sqrt{2} + 6 \).