QJEEM24Ma57da: JEE Main 2024 Mathematics - Class 12 JEEM24Ma57da Solution: Let the line of the shortest distance between the lines…

Question

Verified Solution • 2025–26 Session moderate

Let the line of the shortest distance between the lines

\begin{aligned}&amp; \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\\\&amp; \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 ____________.

Accepted Answer

\begin{array}{ll}L_1 \equiv \vec{r}=(1,2,3)+\lambda(1,-1,1) &amp; \left(\vec{r}=\vec{a}_1+\lambda \vec{b}_1\right) \\\\ L_2 \equiv \vec{r}=(4,5,6)+\mu(1,1,-1) &amp; \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}&amp; P \equiv\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right) \&amp; Q \equiv\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right) \\\\&amp; \alpha=\frac{5}{2}, \beta=\frac{4}{2}, \gamma=\frac{12}{2} \\\\&amp; 2(\alpha+\beta+\gamma)=21\end{aligned}

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