JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 29)
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 ____________.
$$ \begin{aligned} & \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\\\ & \mathrm{L}_2: \overrightarrow{\mathrm{r}}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k}) \end{aligned} $$
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
21
Explanation
$\begin{array}{ll}L_1 \equiv \vec{r}=(1,2,3)+\lambda(1,-1,1) & \left(\vec{r}=\vec{a}_1+\lambda \vec{b}_1\right) \\\\ L_2 \equiv \vec{r}=(4,5,6)+\mu(1,1,-1) & \left(\vec{r}=\vec{a}_2+\lambda \vec{b}_2\right)\end{array}$
$P \equiv(\lambda+1,-\lambda+2, \lambda+3)$
$Q \equiv(\mu+4, \mu+5,-\mu+6)$
$\overrightarrow{P Q}=(\mu-\lambda+3, \mu+\lambda+3,-\mu-\lambda+3)$
$\overrightarrow{P Q} \cdot \vec{b}_1=0 \Rightarrow 3 \lambda+\mu=3$ ..........(i)
$\overrightarrow{P Q} \cdot \vec{b}_2=0 \Rightarrow 3 \mu+\lambda=3$ ..........(ii)
From (i) and (ii),
$$ \begin{aligned} & P \equiv\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right) \& Q \equiv\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right) \\\\ & \alpha=\frac{5}{2}, \beta=\frac{4}{2}, \gamma=\frac{12}{2} \\\\ & 2(\alpha+\beta+\gamma)=21 \end{aligned} $$
$P \equiv(\lambda+1,-\lambda+2, \lambda+3)$
$Q \equiv(\mu+4, \mu+5,-\mu+6)$
$\overrightarrow{P Q}=(\mu-\lambda+3, \mu+\lambda+3,-\mu-\lambda+3)$
$\overrightarrow{P Q} \cdot \vec{b}_1=0 \Rightarrow 3 \lambda+\mu=3$ ..........(i)
$\overrightarrow{P Q} \cdot \vec{b}_2=0 \Rightarrow 3 \mu+\lambda=3$ ..........(ii)
From (i) and (ii),
$$ \begin{aligned} & P \equiv\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right) \& Q \equiv\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right) \\\\ & \alpha=\frac{5}{2}, \beta=\frac{4}{2}, \gamma=\frac{12}{2} \\\\ & 2(\alpha+\beta+\gamma)=21 \end{aligned} $$
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