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Model Paper 6 — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 65 Marks
Question
Find the shortest distance between the given lines. $\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$, $\vec{r}=(3 \hat{i}+3 \hat{j}-5 \hat{k})+\mu(-2 \hat{i}+3 \hat{j}+8 \hat{k})$
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Answer

Given
$\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$
$\vec{r}=(3 \hat{i}+3 \hat{j}-5 \hat{k})+\mu(-2 \hat{i}+3 \hat{j}+8 \hat{k})$
Here, we have
$\vec{a}_1=\hat{\imath}+2 \hat{j}-4 \hat{k}$
$\overrightarrow{b_1}=2 \hat{\imath}+3 \hat{j}+6 \hat{k}$
$\overrightarrow{a_2}=3 \hat{i}+3 \hat{j}-5 \hat{k}$
$\overrightarrow{b_2}=-2 \hat{\imath}+3 \hat{j}+8 \hat{k}$
Thus,
$\overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k}\\2 & 3 & 6\\-2 & 3 & 8\end{array}\right|$
$=\hat{i}(24-18)-\hat{j}(16+12)+\hat{k}(6-6)$
$\overrightarrow{b_1} \times \overrightarrow{b_2}=6 \hat{i}-28 \hat{j}+0 \hat{k}$
$\therefore\left|b_1 \times \overrightarrow{b_2}\right|=\sqrt{6^2+(-28)^2+0^2}$
$=\sqrt{36+784+9}$
$=\sqrt{820}$
$\overrightarrow{a_2}-\overrightarrow{a_1}=(3-1) \hat{i}+(3-2) \hat{j}+(-5+4) \hat{k}$
$\therefore \overrightarrow{a_2}-\overrightarrow{a_1}=2 \hat{\imath}+\hat{j}-\hat{k}$
Now, we have
$\begin{array}{l}\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=(6 \hat{\imath}-28 \hat{\jmath}+0 \hat{k}) \cdot(2 \hat{\imath}+\hat{\jmath}-\hat{k})\\=(6 \times 2)+((-28) \times 1)+(0 \times(-1))\\=12-28+0 \\=-16\end{array}$
Thus, the shortest distance between the given lines is
$ d =\left|\frac{\left(\overrightarrow{ b _1} \times \overrightarrow{ b _2}\right)\cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right| \\ \Rightarrow d =\left|\frac{-16}{\sqrt{820}}\right|$
$\therefore d=\frac{16}{\sqrt{820}} \text { units }$

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