Let the line of the shortest distance between the lines
$$
\begin{aligned}
& \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\\\
& \mathrm{L}_2: \overrightarrow{\mathrm{r}}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})
\end{aligned}
$$
intersect $\mathrm{L}_1$ and $\mathrm{L}_2$ at $\mathrm{P}$ and $\mathrm{Q}$ respectively. If $(\alpha, \beta, \gamma)$ is the mid point of the line segment $\mathrm{PQ}$, then $2(\alpha+\beta+\gamma)$ is equal to ____________.
Answer (integer)
21
Solution
$\begin{array}{ll}L_1 \equiv \vec{r}=(1,2,3)+\lambda(1,-1,1) & \left(\vec{r}=\vec{a}_1+\lambda \vec{b}_1\right) \\\\ L_2 \equiv \vec{r}=(4,5,6)+\mu(1,1,-1) & \left(\vec{r}=\vec{a}_2+\lambda \vec{b}_2\right)\end{array}$
<br/><br/>$P \equiv(\lambda+1,-\lambda+2, \lambda+3)$
<br/><br/>$Q \equiv(\mu+4, \mu+5,-\mu+6)$
<br/><br/>$\overrightarrow{P Q}=(\mu-\lambda+3, \mu+\lambda+3,-\mu-\lambda+3)$
<br/><br/>$\overrightarrow{P Q} \cdot \vec{b}_1=0 \Rightarrow 3 \lambda+\mu=3$ ..........(i)
<br/><br/>$\overrightarrow{P Q} \cdot \vec{b}_2=0 \Rightarrow 3 \mu+\lambda=3$ ..........(ii)
<br/><br/>From (i) and (ii),
<br/><br/>$$
\begin{aligned}
& P \equiv\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right) \& Q \equiv\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right) \\\\
& \alpha=\frac{5}{2}, \beta=\frac{4}{2}, \gamma=\frac{12}{2} \\\\
& 2(\alpha+\beta+\gamma)=21
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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