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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry4 Marks
Question
Find the shortest distance between the following lines whose vector equations are:

$\overrightarrow{r}=\text{(1 - t)}\hat{\text{i}}+\text{(t - 2)}\hat{\text{j}}+\text{(3 - 2t)}\hat{\text{k}}$ and

$\overrightarrow{r}=\text{(s + 1)}\hat{\text{i}}+\text{(2s - 1)}\hat{\text{j}}-\text{(2s + 1)}\hat{\text{k}}$

 

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Answer

Equations of the lines are,
$\overrightarrow{r}=(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k})}+\text{t}(-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k})}\text{ and }$
$\overrightarrow{r}=(\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k})}+\text{s}(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k})}$
shortest distance =$\frac{\Big|(\overrightarrow{\text{a}_{2}}-\overrightarrow{\text{a}_{1}})\cdot(\overrightarrow{\text{b}_{1}}\times\overrightarrow{\text{b}_{2}})\Big|}{\Bigg|\overrightarrow{\text{b}_{1}}\times\overrightarrow{\text{b}_{2}}\Bigg|}$ where
$\overrightarrow{\text{a}_{1}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},\text{ }\overrightarrow{\text{a}_{2}}=\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}},\overrightarrow{\text{b}_{1}}=-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}},\text{ }\overrightarrow{\text{b}_{2}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}},$
$\overrightarrow{\text{a}_{2}}-\overrightarrow{\text{a}_{1}}=\hat{\text{j}}-4\hat{\text{k}},\text{ }\overrightarrow{\text{b}_{1}}\times\overrightarrow{\text{b}_{2}}=2\hat{\text{i}}-4\hat{\text{j}}-3\hat{\text{k}}$
$\therefore$ S.D. = $\frac{0-4+12}{\sqrt{29}}=\frac{8}{\sqrt{29}}$.

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