Three Dimensional Geometry Question 15
Question 15 - 2024 (29 Jan Shift 2)
Let $O$ be the origin, and $M$ and $N$ be the points on the lines $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ and $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ respectively such that $MN$ is the shortest distance between the given lines. Then $\overrightarrow{OM} \cdot \overrightarrow{ON}$ is equal to
Show Answer
Answer (9)
Solution
$L _1: \frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}=\lambda \quad \operatorname{drs}(4,1,3)=b _1$
$M(4 \lambda+5, \lambda+4,3 \lambda+5)$
$L _2: \frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}=\mu$
$N(12 \mu-8,5 \mu-2,9 \mu-11)$
$\overrightarrow{MN}=(4 \lambda-12 \mu+13, \lambda-5 \mu+6,3 \lambda-9 \mu+16)$.
Now
$\overrightarrow{b} _1 \times \overrightarrow{b} _2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 3 \\ 12 & 5 & 9\end{array}\right|=-6 \hat{i}+8 \hat{k}$.
Equation (1) and (2)
$\therefore \frac{4 \lambda-12 \mu+13}{-6}=\frac{\lambda-5 \mu+6}{0}=\frac{3 \lambda-9 \mu+16}{8}$
I and II
$\lambda-5 \mu+6=0$…
I and III
$\lambda-3 \mu+4=0$.
Solve (3) and (4) we get
$\lambda=-1, \mu=1$
$\therefore M(1,3,2)$
$N(4,3,-2)$
$\therefore \overrightarrow{OM} \cdot \overrightarrow{ON}=4+9-4=9$
