Question

Let \(Q(a,b,c)\) be the image of the point \(P(3,2,1)\) in the line \[ \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{1}. \] The distance of \(Q\) from the line \[ \frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2} \] is:

• A. \(8\)
• B. \(7\)
• C. \(6\)
• D. \(5\)

Hint:

Reflection of a point in a line is easily found using vector projection and symmetry about the foot of the perpendicular.

Answer 1

The Correct Option is B

Concept:
The image of a point in a line is obtained by reflecting the point about the line.
If \( H \) is the foot of the perpendicular from \( P \) onto the line, then the image point is
\[ Q = 2H - P. \]
The distance of a point from a line is given by:
\[ \text{Distance}=\frac{|\vec{BP}\times\vec{d}|}{|\vec{d}|}, \] where \( \vec{d} \) is the direction vector of the line and \( B \) is any point on the line.

Step 1: Line of reflection

Given line:
\[ \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{1} \]
A point on the line:
\[ A(1,2,1) \]
Direction vector:
\[ \vec{d}_1=(1,2,1) \]

Step 2: Foot of perpendicular from \( P \) onto the line

\[ \vec{AP}=(3-1,\,2-2,\,1-1)=(2,0,0) \]
Parameter of projection:
\[ t=\frac{\vec{AP}\cdot\vec{d}_1}{|\vec{d}_1|^2} =\frac{2}{1^2+2^2+1^2} =\frac{2}{6}=\frac{1}{3} \]
Thus,
\[ H=A+t\vec{d}_1 =\left(1+\frac{1}{3},\,2+\frac{2}{3},\,1+\frac{1}{3}\right) =\left(\frac{4}{3},\frac{8}{3},\frac{4}{3}\right) \]

Step 3: Image point \( Q \)

\[ Q=2H-P \]
\[ Q=\left(\frac{8}{3}-3,\,\frac{16}{3}-2,\,\frac{8}{3}-1\right) =\left(-\frac{1}{3},\,\frac{10}{3},\,\frac{5}{3}\right) \]

Step 4: Distance of \( Q \) from the second line

Second line:
\[ \frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2} \]
Point on line:
\[ B(9,9,5) \]
Direction vector:
\[ \vec{d}_2=(3,2,-2) \]
Vector:
\[ \vec{BQ}=Q-B=\left(-\frac{28}{3},-\frac{17}{3},-\frac{10}{3}\right) \]

Step 5: Compute distance

\[ \vec{BQ}\times\vec{d}_2= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k} \\ -28 & -17 & -10 \\ 3 & 2 & -2 \end{vmatrix} =14\hat{i}-86\hat{j}-5\hat{k} \]
\[ |\vec{BQ}\times\vec{d}_2| =\sqrt{14^2+86^2+5^2} =\sqrt{7617} \]
Since \( \vec{BQ} \) was scaled by \( 3 \), actual magnitude:
\[ |\vec{BQ}\times\vec{d}_2|=\frac{\sqrt{7617}}{3} \]
\[ |\vec{d}_2|=\sqrt{3^2+2^2+(-2)^2}=\sqrt{17} \]
\[ \text{Distance}=\frac{\sqrt{7617}}{3\sqrt{17}}\approx 7 \]

Final Answer:
\[ \boxed{7} \]