Subjective Type

Find the shortest distance between the lines whose vector equations are $$\vec {r} = (\hat {i} + 2\hat {j} + 3\hat {k}) + \lambda (\hat {i} - 3\hat {j} + 2\hat {k})$$ and $$\vec {r} = (4\hat {i} + 5\hat {j} + 6\hat {k}) + \mu (2\hat {i} + 3\hat {j} + \hat {k})$$.

Solution

The given lines are $$\vec { r } =\left( \hat { i } +2\hat { j } +3\hat {

k } \right) +\lambda \left( \hat { i } -3\hat { j } +2\hat { k }

\right) $$ and
$$\vec { r } =4\hat { i } +5\hat { j } +6\hat { k } +\mu \left( \hat { 2i } +3\hat { j } +\hat { k } \right) $$
It is known that the shortest distance between the lines, $$\vec { r }

=\vec { { a }_{ 1 } } +\lambda \vec { { b }_{ 1 } } $$ and $$\vec { r }

=\vec { { a }_{ 2 } } +\mu \vec { { b }_{ 2 } } $$ is given by,
$$d=\left|

\cfrac { \left( \vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } }

\right) \left( \vec { { a }_{ 1 } } -\vec { { a }_{ 2 } } \right) }{

\left| \vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } } \right| }

\right| .....(1)$$
Comparing the given equations with $$\vec {

r } =\vec { { a }_{ 1 } } +\lambda \vec { { b }_{ 1 } } $$ and $$\vec {

r } =\vec { { a }_{ 1 } } +\mu \vec { { b }_{ 1 } } $$, we obtain
$$\vec { { a }_{ 1 } } =\hat { i } +2\hat { j } +3\hat { k } $$
$$\vec { { b }_{ 1 } } =\hat { i } -3\hat { j } +2\hat { k } $$
$$\vec { { a }_{ 2 } } =4\hat { i } +5\hat { j } +6\hat { k } $$
$$\vec { { b }_{ 2 } } =\hat { 2i } +3\hat { j } +\hat { k } $$
$$\vec

{ { a }_{ 2 } } -\vec { { a }_{ 1 } } =(4\hat { i } +5\hat { j } +6\hat

{ k } )-(\left( \hat { i } +2\hat { j } +3\hat { k } \right) =3\hat { i

} +3\hat { j } +3\hat { k } $$
$$\vec { { b }_{ 1 } } \times \vec { {

b }_{ 2 } } =\sqrt { { \left( -9 \right) }^{ 2 }+{ \left( 3 \right)

}^{ 2 }+{ \left( 9 \right) }^{ 2 } } =\sqrt { 18+9+81 } =\sqrt { 171 }

=3\sqrt { 19 } $$
$$\left( \vec { { b }_{ 1 } } \times \vec { { b

}_{ 2 } } \right) .\left( \vec { { a }_{ 1 } } -\vec { { a }_{ 2 } }

\right) =\left( -9\hat { i } +3\hat { j } +9\hat { k } \right) .\left(

3\hat { i } +3\hat { j } +3\hat { k } \right) $$
$$=-9\times 3+3\times 3+9\times 3=9$$
Substituting all the values in equation (1), we obtain
$$d=\left| \cfrac { 9 }{ 3\sqrt { 19 } } \right| =\cfrac { 3 }{ \sqrt { 19 } } $$
Therefore, the shortest distance between the two given lines is $$\cfrac{3}{\sqrt {19}}$$ units.

SIMILAR QUESTIONS

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Consider the line $$\displaystyle {L}_{1}:\dfrac{{x}+1}{3}=\dfrac{{y}+2}{1}=\dfrac{{z}+1}{2},\ \displaystyle {L}_{2}:\dfrac{{x}-2}{1}=\dfrac{{y}+2}{2}=\dfrac{{z}-3}{3}$$ The shortest distance between $$L_1$$ and $$L_2$$ is

3D Geometry

If the shortest distance between the lines $$\displaystyle \frac{x-1}{\alpha} = \frac{y+1}{-1} = \frac{z}{1}, \,\,\, (\alpha \neq -1)$$ and $$x + y + z + 1 = 0 = 2x - y + z + 3$$ is $$\displaystyle \frac{1}{\sqrt{3}}$$, then a value of $$\alpha$$ is :

3D Geometry

Find the shortest distance between the following pair of lines. $$\bar{r}=(\bar{i}+2\bar{j}+\bar{k})+\lambda(2\bar{i}-\bar{j}+3\bar{k})$$ & $$\bar{r}=(\bar{i}-3\bar{j}-\bar{k})+\mu (3\bar{i}+2\hat{j}-5\bar{k})$$.

3D Geometry

The shortest distance between the $$z$$-axis and the line $$x+ y + 2z - 3 = 0 , 2x + 3y + 4z - 4 $$, is

3D Geometry

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3D Geometry

The shortest distance between the lines $$x = y + 2 = 6z - 6$$ and $$x + 1 = 2y = -12z$$ is

3D Geometry

Find the shortest distance between the lines $$\vec { r } =\left( \hat { i } +2\hat { j } +\hat { k } \right) +\lambda \left( \hat { i } -\hat { j } +\hat { k } \right) $$ and $$\vec { r } =\hat { 2i } -\hat { j } -\hat { k } +\mu \left( \hat { 2i } +\hat { j } +2\hat { k } \right) $$

3D Geometry

Find the shortest distance between lines $$\cfrac{x+1}{7}=\cfrac{y+1}{-6}=\cfrac{z+1}{1}$$ and $$\cfrac{x-3}{1}=\cfrac{y-5}{-2}=\cfrac{z-7}{1}$$

3D Geometry

Find the shortest distance between the lines whose vector equations are : $$\vec { r } =(1-p)\hat { i } +(p-2)\hat { j } +(3-2p)\hat { k } $$ $$\vec { r } =(q+1)\hat { i } +(2q-1)\hat { j } -(2q-1)\hat { k } $$

3D Geometry

Find the shortest distance between lines $$\vec { r } =6\hat { i } +2\hat { j } +2\hat { k } +\lambda \left( \hat { i } -2\hat { j } +2\hat { k } \right) $$ and $$\vec { r } =-4\hat { i } -\hat { k } +\mu \left(3 \hat { i } -2\hat { j } -2\hat { k } \right) $$

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