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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
Find the shortest distance between the lines
$\vec{\text{r}}=6\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}+\lambda\big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=-4\hat{\text{i}}-\hat{\text{k}}+\mu\big(3\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$
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Answer

$\vec{\text{r}}=6\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}+\lambda\big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=-4\hat{\text{i}}-\hat{\text{k}}+\mu\big(3\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}\big)$
Comparing the given equations with the equations $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2,$ we get
$\vec{\text{a}}_1=6\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{a}}_2=-4\hat{\text{i}}-\hat{\text{k}}$
$\vec{\text{b}}_1=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}_2=3\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
$\therefore\vec{\text{a}}_2-\vec{\text{a}}_1=-10\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
and $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-2&2\\3&-2&-2\end{vmatrix}$
$=8\hat{\text{i}}+8\hat{\text{j}}+4\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{8^2+8^2+4^2}$
$=\sqrt{64+64+16}$
$=\sqrt{144}$
$=12$
and $\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)$
$=\big(-10\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big).\big(8\hat{\text{i}}+8\hat{\text{j}}+4\hat{\text{k}}\big)$
$=-80-16-12$
$=-108$
The shortest distence between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2.\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$\Rightarrow\text{d}=\Big|\frac{-108}{12}\Big|$
$=9$

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