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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 lines $\vec{\text{r}}=6\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}+\lambda\Big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\Big)\ \text{and}\ \vec{\text{r}}=-4\hat{\text{i}}-\hat{\text{k}}+\mu\Big(3\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}\Big).$
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Answer

The equations of two lines are $\vec{\text{r}}=6\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}+\lambda\Big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\Big)\ \ ...(1)$ $\vec{\text{r}}=-4\hat{\text{i}}-\hat{\text{k}}+\mu\Big(3\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}\Big)\ \ \ \ .....(2)$ Comparing these equations with $\vec{\text{r}}=\vec{\text{a}_1}+\lambda\vec{\text{b}_1}\ \text{and}\ \vec{\text{r}}=\vec{\text{a}_2}+\mu\vec{\text{b}_2},$ we get, $\vec{\text{a}_1}=6\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}_1}=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$ $\vec{\text{a}_2}=-4\hat{\text{i}}-\hat{\text{k}},\ \vec{\text{b}_2}=3\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$ Let 'S' be the point on line (1) with position vector $\vec{\text{a}_1}$ and T be the point on line (2) with position vector $\vec{\text{a}_2}$ so that
$\overrightarrow{\text{ST}}=\vec{\text{a}_2}-\vec{\text{a}_1}=-10\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$ Now, $\vec{\text{b}_1}\times\vec{\text{b}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-2&2\\3&-2&-2\end{vmatrix}$ $=\hat{\text{i}}\begin{vmatrix}-2&2\\-2&-2\end{vmatrix}-\hat{\text{j}}\begin{vmatrix}1&2\\3&-2\end{vmatrix}+\hat{\text{k}}\begin{vmatrix}1&-2\\3&-2\end{vmatrix}$ $=(4+4)\hat{\text{i}}-(-2-6)\hat{\text{j}}+(-2+6)\hat{\text{k}}=8\hat{\text{i}}+8\hat{\text{j}}+4\hat{\text{k}}$ $\therefore\ \ \Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|=\sqrt{64+64+16}=\sqrt{144}=12$ Let $\overrightarrow{\text{PQ}}$ be the S.D. vector between given lines. Therefore, it is parallel to $\vec{\text{b}_1}\times\vec{\text{b}_2}.$ If $\vec{\text{n}}$ is a unit vector along $\overrightarrow{\text{PQ}},$ then $\vec{\text{n}}=\frac{\vec{\text{b}_1}\times\vec{\text{b}_2}}{\big|\vec{\text{b}_1}\times\vec{\text{b}_2}\big|}=\frac{1}{12}(8\hat{\text{i}}+8\hat{\text{j}}+4\hat{\text{k}})$ $=\frac{1}{3}(2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})$ Now S.D. = Projection of $\overrightarrow{\text{ST}}\ \text{on}\ \overrightarrow{\text{PQ}}$ = Projection of $\overrightarrow{\text{ST}}\ \text{on}\ \vec{\text{n}}=\overrightarrow{\text{ST}}.\vec{\text{n}}$ $=\Big(-10\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\Big).\frac{1}{3}\Big(2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\Big)$ $=\frac{1}{3}[(-10)(2)+(-2)(2)+(-3)(1)]$ $=\frac{1}{3}(-20-4-3)=-\frac{27}{3}=-9=9\ \text{units}.(\text{in magnitude})$

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