Let A and B be two distinct points on the line L : x-6/3=y-7/2=z-7/-2. Both A and B are at a distance 2√(17) from the foot of perpendicular

Question

Let A and B be two distinct points on the line \(\mathrm{L}: \frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-7}{2}=\frac{\mathrm{z}-7}{-2}.\) Both A and B are at a distance \(2 \sqrt{17}\) from the foot of perpendicular drawn from the point (1, 2, 3) on the line L. If O is the origin, then \(\overrightarrow{O A} \cdot \overrightarrow{O B}\) is equal to:

(1) 49

(2) 47

(3) 21

(4) 62 

Answer 1

Correct option is: (2) 47 

\((3 \lambda+5,2 \lambda+5,-2 \lambda+4) \cdot(3,2,-2)=0 \)

\( 9 \lambda+15+4 \lambda+10+4 \lambda-8=0 \)

\( 17 \lambda=-17 \)

\( \lambda=-1 \)

\(\therefore\ \mathrm{P}(3,5,9) \)

\(\mathrm{A}(3 \mathrm{k}+6,2 \mathrm{k}+7,-2 \mathrm{k}-2) \)

\(|\overrightarrow{\mathrm{PA}}|=2 \sqrt{17} \Rightarrow 68 \)

\(=9(\mathrm{k}+1)^2+4(\mathrm{k}+1)^2+4(\mathrm{k}+1)^2 \)

\( 68=17(\mathrm{k}+1)^2 \)

\(4=(\mathrm{k}+1)^2\)   

\( \mathrm{k}+1=2 \quad \text { or } \quad \mathrm{k}+1=-2 \)

\( \mathrm{k}=1 \quad k=-3\)

\( \therefore \mathrm{~A}(9,9,5) \text { and } \mathrm{B}(-3,1,13) \)

\(\overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OB}}=-27+9+65=47\)