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
$${ L }_{ 1 }:\cfrac { x+1 }{ 3 } =\cfrac { y+2 }{ 1 } =\cfrac { z+1 }{ 2 } $$
$$ { L }_{ 2 }:\cfrac { x-2 }{ 1 } =\cfrac { y+2 }{ 2 } =\cfrac { z-3 }{ 3 } $$
$$ (x_{ 1 },y_{ 1 },z_{ 1 })\equiv (-1,-2,-1)$$
$$ (x_{ 2 },y_{ 2 },z_{ 2 })\equiv (2,-2,3)$$
$$ (l_{ 1 },m_{ 1 },n_{ 1 })\equiv (3,1,2)$$
$$ (l_{ 2 },m_{ 2 },n_{ 2 })\equiv (1,2,3)$$
$$\vec { { b }_{ 1 } } ={ l }_{ 1 }\hat { i } +{ m }_{ 1 }\hat { j } +{ n }_{ 1 }\hat { k } $$
$$ \vec { { b }_{ 2 } } ={ l }_{ 2 }\hat { i } +{ m }_{ 2 }\hat { j } +{ n }_{ 2 }\hat { k } $$
$$\therefore $$ Shortest distance$$=(\vec { a_{ 2 } } -\vec { { a }_{ 1 } } )\cdot \cfrac { \vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } } }{ |\vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } } | } $$
$$\cfrac { \begin{bmatrix} x_{ 2 }-x_{ 1 } & y_{ 2 }-y_{ 1 } & z_{ 2 }-z_{ 1 } \\ l_{ 1 } & m_{ 1 } & n_{ 1 } \\ l_{ 2 } & m_{ 2 } & n_{ 2 } \end{bmatrix} }{ \sqrt { (m_{ 1 }n_{ 2 }-n_{ 1 }m_{ 2 })^{ 2 }+(m_{ 2 }l_{ 1 }-l_{ 2 }m_{ 1 })^{ 2 }+(n_{ 1 }l_{ 2 }-l_{ 1 }n_{ 2 })^{ 2 } } } $$
$$=\cfrac { 17 }{ 5\sqrt { 3 } } units$$
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 :
Find the shortest distance between the following pair of lines. $$\bar{r}=(\bar{i}+2\bar{j}+\bar{k})+\lambda(2\bar{i}-\bar{j}+3\bar{k})$$ & $$\bar{r}=(\bar{i}-3\bar{j}-\bar{k})+\mu (3\bar{i}+2\hat{j}-5\bar{k})$$.
The shortest distance between the $$z$$-axis and the line $$x+ y + 2z - 3 = 0 , 2x + 3y + 4z - 4 $$, is
The shortest distance between the lines $$\dfrac {x}{2} = \dfrac {y}{2} = \dfrac {z}{1}$$ and $$\dfrac {x + 2}{-1} = \dfrac {y - 4}{8} = \dfrac {z - 5}{4}$$ lies in the interval:
The shortest distance between the lines $$x = y + 2 = 6z - 6$$ and $$x + 1 = 2y = -12z$$ is
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 $$\vec { r } =\hat { 2i } -\hat { j } -\hat { k } +\mu \left( \hat { 2i } +\hat { j } +2\hat { k } \right) $$
Find the shortest distance between lines $$\cfrac{x+1}{7}=\cfrac{y+1}{-6}=\cfrac{z+1}{1}$$ and $$\cfrac{x-3}{1}=\cfrac{y-5}{-2}=\cfrac{z-7}{1}$$
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 $$\vec {r} = (4\hat {i} + 5\hat {j} + 6\hat {k}) + \mu (2\hat {i} + 3\hat {j} + \hat {k})$$.
Find the shortest distance between the lines whose vector equations are : $$\vec { r } =(1-p)\hat { i } +(p-2)\hat { j } +(3-2p)\hat { k } $$ $$\vec { r } =(q+1)\hat { i } +(2q-1)\hat { j } -(2q-1)\hat { k } $$
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 $$\vec { r } =-4\hat { i } -\hat { k } +\mu \left(3 \hat { i } -2\hat { j } -2\hat { k } \right) $$