Question

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

• A. \(3\)
• B. \(5\)
• C. \(6\)
• D. \(7\)

Hint:

For image of a point in a line: first find the foot of the perpendicular, then reflect using \(Q=2M-P\).

Answer 1

The Correct Option is D

Step 1: Image of a point in a line
The given line \(L_1\) passes through: \[ A(1,0,1) \] with direction vector: \[ \vec{d}_1=\langle 1,2,1\rangle \] Let \(M\) be the foot of the perpendicular from \(P(3,2,1)\) onto line \(L_1\). Then the image point \(Q\) is given by: \[ Q=2M-P \]
Step 2: Find foot of perpendicular \(M\)
Vector form: \[ \vec{AM}=\lambda \vec{d}_1 \Rightarrow M(1+\lambda,\,2\lambda,\,1+\lambda) \] Condition for perpendicularity: \[ (\vec{PM})\cdot \vec{d}_1=0 \] \[ (1+\lambda-3,\;2\lambda-2,\;1+\lambda-1)\cdot(1,2,1)=0 \] \[ (\lambda-2)+2(2\lambda-2)+(\lambda)=0 \] \[ 6\lambda-6=0 \Rightarrow \lambda=1 \] \[ \Rightarrow M(2,2,2) \]
Step 3: Find image point \(Q\)
\[ Q=2M-P=2(2,2,2)-(3,2,1)=(1,2,3) \]
Step 4: Distance of point \(Q\) from second line
Second line \(L_2\) passes through: \[ B(9,9,5) \] with direction vector: \[ \vec{d}_2=\langle 3,2,-2\rangle \] Vector: \[ \vec{BQ}=(1-9,\;2-9,\;3-5)=(-8,-7,-2) \] Distance of a point from a line: \[ \text{Distance}=\frac{|\vec{BQ}\times\vec{d}_2|}{|\vec{d}_2|} \] \[ \vec{BQ}\times\vec{d}_2= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ -8 & -7 & -2\\ 3 & 2 & -2 \end{vmatrix} =18\hat{i}-22\hat{j}+5\hat{k} \] Magnitude: \[ |\vec{BQ}\times\vec{d}_2|=\sqrt{18^2+22^2+5^2}=\sqrt{833} \] \[ |\vec{d}_2|=\sqrt{3^2+2^2+(-2)^2}=\sqrt{17} \] \[ \text{Distance}=\frac{\sqrt{833}}{\sqrt{17}}=\sqrt{49}=7 \] \[ \boxed{7} \]