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