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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}}=(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})+\lambda(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\ \text{and}\ $
$\vec{\text{r}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}+\mu(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})$
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Answer

Comparing the given equations with
$\vec{\text{r}}=\vec{\text{a}_1}+\lambda\vec{\text{b}_1}\ \text{and}\ \vec{\text{r}}=\vec{\text{a}_2}+\lambda\vec{\text{b}_2},$ we get
$\vec{\text{a}_1}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}},\ \ \vec{\text{b}_1}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}} \ \ \text{and}$
$\vec{\text{a}_2}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}},\ \ \vec{\text{b}_2}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
Since, the shortest distance between the two skew lines is given by
$\text{d}=\frac{\Big|\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)\Big|}{\Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|}\ \ \ .....(\text{i})$
Here, $\vec{\text{a}_2}-\vec{\text{a}_1}=\Big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\Big)-\Big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\Big)=\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{b}_1}\times\vec{\text{b}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-1&1\\2&1&2\end{vmatrix}$
$=(-2-1)\hat{\text{i}}-(2-2)\hat{\text{j}}+(1+2)\hat{\text{k}}=-3\hat{\text{i}}+3\hat{\text{k}}$
$\Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|=\sqrt{(-3)^2+(0)^2+(3)^2}=\sqrt{18}=3\sqrt{2}$
$\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)=\Big(\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\Big).\Big(-3\hat{\text{i}}+3\hat{\text{k}}\Big)$
=1 × (-3) + (-3 × 0) + (-2 × 3) = -9
Putting these values in eq. (i)
Shortest distance $(\text{d})=\frac{|-9|}{3\sqrt{2}}=\frac{9}{3\sqrt{2}}=\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}.$

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