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
$L _2: \overrightarrow{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})$
intersect $L _1$ and $L _2$ at $P$ and $Q$ respectively. If $(\alpha, \beta, \gamma)$ is the midpoint of the line segment $PQ$, then
$2(\alpha+\beta+\gamma)$ is equal to
Show Answer
Answer (21)
Solution
$$ A(1+\lambda, 2-\lambda, 3+\lambda) $$
$$ B(4+\mu, 5+\mu, 6-\mu) $$
$\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}\left(DR\right.$ ’s of $\left.L _1\right)$
$\overrightarrow{d}=\hat{i}+\hat{j}-\hat{k}\left(DR\right.$ ’s of $\left.L _2\right)$
$\overrightarrow{b} \times \overrightarrow{d}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right|$
$=0 \hat{i}+2 \hat{j}+2 \hat{k}$ (DR’s of Line perpendicular to
$L _1$ and $L _2$ )
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 $A=\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right)$
$B=\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right)$
Point of $AB=\left(\frac{5}{2}, 2,6\right)=(\alpha, \beta, \gamma)$
$2(\alpha+\beta+\gamma)=5+4+12=21$
