Let A and B be two distinct points on the line L: x-63=y-72=z-7-2.

Question

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

Accepted Answer

$$L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$$

Point $A(3 \lambda+6,2 \lambda+7,7-2 \lambda)$

$$B(3 \mu+6,2 \mu+7,7-2 \mu)$$

Let $P(3 k+6,2 k+7,7-2 k)$ be foot of perpendicular from $P^{\prime}(1,2,3)$

$$\begin{aligned}
& \therefore \quad P P^{\prime}\langle 3,2,-2\rangle=0 \
& 3(3 k+5)+(2 k+5) 2+(4-2 k)(-2)=0 \
& 9 k+15+4 k+10-8+4 k=0 \
& 17 k+17=0 \
& \Rightarrow k=-1 \qquad \therefore \quad P(3,5,9)\
& |\overrightarrow{A P}|=2 \sqrt{17}
\end{aligned}$$

$$\begin{aligned}
& \Rightarrow(3 \lambda+3)^2+(2 \lambda+2)^2+(-2-2 \lambda)^2=17 \times 4 \
& =17(\lambda+1)^2=17 \times 4 \
& \Rightarrow \lambda+1= \pm 2 \Rightarrow \lambda=1 \text { or } \lambda=-3 \
& \therefore A(9,9,5), B(-3,1,13) \
& \overrightarrow{O A} \cdot \overrightarrow{O B}=(9 \hat{i}+9 \hat{j}+5 \hat{k}) \cdot(-3 \hat{i}+\hat{j}+13 \hat{k}) \
& =-27+9+65=47
\end{aligned}$$

Solutions · 1