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MODEL PAPER 2025 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMODEL PAPER 20255 Marks
Question
Find the shortest distance between the lines $I_1$ and $l_2$ $\vec{r}=(-\hat{\imath}-\hat{\jmath}-\hat{k})+\lambda(7 \hat{\imath}-6 \hat{\jmath}+\hat{k})$ and $\vec{r}=(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})+\mu(\hat{\imath}-2 \hat{\jmath}+\hat{k})$where $\lambda$ and $\mu$ are parameters.
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Answer

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Given that equation of lines are
$\vec{r} =(-\hat{\imath}-\hat{\jmath}-\hat{k})+\lambda(7 \hat{\imath}-6 \hat{\jmath}+\hat{k})\ldots(i)$
$\vec{r} =(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})+\mu(\hat{\imath}-2 \hat{\jmath}+\hat{k})\ldots(ii)$
The given lines are non $-$  parallel lines as vectors $7 \hat{\imath}-6 \hat{\jmath}+\hat{k}$ and $\hat{\imath}-2 \hat{\jmath}+\hat{k}$ are not parallel.
There is a unique line segment $P Q$ ( $P$ lying on line $(i)$ and $Q$ on the other line $(i i) ),$ which is at right angles to both the lines $P Q$ is the shortest distance between the lines.
Hence, the shortest possible distance between the lines $=P Q$.
Let the position vector of the point $P$ lying on the line $r=(-\hat{\imath}-\hat{\jmath}-\hat{k})+\lambda(7 \hat{\imath}-6 \hat{\jmath}+\hat{k})$ where  $\lambda$  is a scalar, is $(7 \lambda-1) \hat{\imath}-(6 \lambda+1) \hat{\jmath}+(\lambda-1) \hat{k}, $ for some $\lambda$ and the position vector of the point $Q$ lying on the line $r=(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})+\mu(\hat{\imath}-2 \hat{\jmath}+\hat{k})$ where  $\mu^{\prime}$  is a scalar, is $(\mu+3) \hat{\imath}+(-2 \mu+5) \hat{\jmath}+(\mu+7) \hat{k},$ for some $\mu$. Now, the vector
$\overline{P Q}=\overline{O Q}-\overline{O P}=(\mu+3-7 \lambda+1) \hat{\imath}+(-2 \mu+5+6 \lambda+1) \hat{\jmath}+(\mu+7-\lambda+1) \hat{k}$
i.e., $\overline{P Q}=(\mu-7 \lambda+4) \hat{\imath}+(-2 \mu+6 \lambda+6) \hat{\jmath}+(\mu-\lambda+8) \hat{k} ; ($where  $'O\ '$  is the origin$),$ is perpendicular to both the lines,
so the vector $\overline{P Q}$ is perpendicular to both the vectors $7 \hat{\imath}-6 \hat{\jmath}+\hat{k}$ and $\hat{\imath}-2 \hat{\jmath}+\hat{k}$.
$\Rightarrow(\mu-7 \lambda+4) \cdot 7+(-2 \mu+6 \lambda+6) \cdot(-6)+(\mu-\lambda+8) \cdot 1=0$
$\ (\mu-7 \lambda+4) \cdot 1+(-2 \mu+6 \lambda+6) \cdot(-2)+(\mu-\lambda+8) \cdot 1=0$
$\Rightarrow 2 0 \mu- 8 6 \lambda= 0 $
$ \Rightarrow1 0 \mu- 4 3 \lambda= 0 \ 6 \mu-20 \lambda=0 $
$\Rightarrow 3 \mu-10 \lambda=0$
On solving the above equations, we get $\mu=\lambda=0$
So, the position vector of the points $P$ and $Q$ are $-\hat{\imath}-\hat{\jmath}-\hat{k}$ and $3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k}$ respectively.
$\overline{P Q}=4 \hat{\imath}+6 \hat{\jmath}+8 \hat{k}$ and 
$|\overline{P Q}|=\sqrt{4^2+6^2+8^2}$
$=\sqrt{116}=2 \sqrt{29} \text { units. }$

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