Three Dimensional Geometry Question 2
Question 2 - 2024 (01 Feb Shift 1)
Let the line of the shortest distance between the lines
$L_{1}: \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and
$\mathrm{L}_{2}: \overrightarrow{\mathrm{r}}=(4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$
intersect $\mathrm{L}{1}$ and $\mathrm{L}{2}$ at $\mathrm{P}$ and $\mathrm{Q}$ respectively. If $(\alpha, \beta, \gamma)$ is the midpoint of the line segment $\mathrm{PQ}$, then
$2(\alpha+\beta+\gamma)$ is equal to
Show Answer
Answer (21)
Solution
$$ \mathrm{A}(1+\lambda, 2-\lambda, 3+\lambda) $$
$$ \mathrm{B}(4+\mu, 5+\mu, 6-\mu) $$
$\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\left(\mathrm{DR}\right.$ ’s of $\left.L_{1}\right)$
$\overrightarrow{\mathrm{d}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\left(\mathrm{DR}\right.$ ’s of $\left.L_{2}\right)$
$\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 1 & -1 & 1 \ 1 & 1 & -1\end{array}\right|$
$=0 \hat{i}+2 \hat{j}+2 \hat{k}$ (DR’s of Line perpendicular to
$\mathrm{L}{1}$ and $\mathrm{L}{2}$ )
DR of $A B$ line
$$ \begin{aligned} & =(0,2,2)=(3+\mu-\lambda, 3+\mu+\lambda, 3-\mu-\lambda) \ & \frac{3+\mu-\lambda}{0}=\frac{3+\mu+\lambda}{2}=\frac{3-\mu-\lambda}{2} \end{aligned} $$
Solving above equation we get $\mu=-\frac{3}{2}$ and $\lambda=\frac{3}{2}$
point $\mathrm{A}=\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right)$
$\mathrm{B}=\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right)$
Point of $\mathrm{AB}=\left(\frac{5}{2}, 2,6\right)=(\alpha, \beta, \gamma)$
$2(\alpha+\beta+\gamma)=5+4+12=21$
