Find the shortest distance between lines r =6 i +2 j +2 k + ( i -2 j +2 k ) and r =-4 i - k + (3 i -2 j -2 k ).

The equations of two lines are r =6 i +2 j +2 k + ( i -2 j +2 k ) ...(1) r =-4 i - k + (3 i -2 j -2 k ) .....(2) Comparing these equations with r = a_1 + b_1 and r = a_2 + b_2 , we get, a_1 =6 i +2 j +2 k , b_1 = i -2 j +2 k a_2 =-4 i - k , b_2 =3 i -2 j -2 k Let 'S' be the point on line (1) with position vector a_1 and T be the point on line (2) with position vector a_2 so that ST = a_2 - a_1 =-10 i -2 j -3 k Now, b_1 b_2 = i & j & k 1&-2&2 3&-2&-2 = i -2&2 -2&-2 - j 1&2 3&-2 + k 1&-2 3&-2 =(4+4) i -(-2-6) j +(-2+6) k =8 i +8 j +4 k | b_1 b_2 |=64+64+16=144=12 Let PQ be the S.D. vector between given lines. Therefore, it is parallel to b_1 b_2 . If n is a unit vector along PQ , then n = b_1 b_2 | b_1 b_2 | =1 12 (8 i +8 j +4 k ) =1 3 (2 i +2 j + k ) Now S.D. = Projection of ST on PQ = Projection of ST on n = ST . n = (-10 i -2 j -3 k ).1 3 (2 i +2 j + k ) =1 3 [(-10)(2)+(-2)(2)+(-3)(1)] =1 3 (-20-4-3)=-27 3 =-9=9 units.(in magnitude)

Find the shortest distance between lines r =6 i +2 j +2 k + ( i -…

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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 $\overrightarrow{\text{ST}}\ \text{on}\ \vec{\text{n}}=\overrightarrow{\text{ST}}.\vec{\text{n}}$ $=\Big(-10\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\Big).\frac{1}{3}\Big(2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\Big)$ $=\frac{1}{3}[(-10)(2)+(-2)(2)+(-3)(1)]$ $=\frac{1}{3}(-20-4-3)=-\frac{27}{3}=-9=9\ \text{units}.(\text{in magnitude})$