Subjective Type

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

Solution

It is known that the shortest distance between the two lines,
$$\cfrac { x-{ x }_{ 1 } }{ a } =\cfrac { y-{ y }_{ 1 } }{ b } =\cfrac { z-{ z }_{ 1 } }{ c } $$ and $$\cfrac { x-{ x }_{ 2 } }{ a } =\cfrac { y-{ y }_{ 2 } }{ b } =\cfrac { z-{ z }_{ 2 } }{ c } $$ is given by,
$$d=\cfrac { \begin{vmatrix} { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \end{vmatrix} }{ \sqrt { { \left( { b }_{ 1 }{ c }_{ 2 }-{ b }_{ 2 }{ c }_{ 1 } \right) }^{ 2 }+{ \left( { c }_{ 1 }{ a }_{ 2 }-{ c }_{ 2 }{ a }_{ 1 } \right) }^{ 2 }+{ \left( { a }_{ 1 }{ b }_{ 2 }-{ a }_{ 2 }{ b }_{ 1 } \right) }^{ 2 } } } ....(1)$$
Comparing the given equations, we obtain
$${x}_{1}=-1, {y}_{1}=-1, {z}_{1}=-1$$
$${a}_{1}=7, {b}_{1}=-6, {c}_{1}=1$$
$${x}_{2}=3, {y}_{2}=5, {z}_{2}=7$$
$${a}_{2}=1, {b}_{2}=-2, {c}_{2}=1$$
Then, $$\begin{vmatrix} { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \end{vmatrix}=\begin{vmatrix} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix}$$
$$=4(-6+2)-6(7-1)+8(-14+6)$$
$$=-16-36-64=-116$$
$$\Rightarrow\sqrt { { \left( { b }_{ 1 }{ c }_{ 2 }-{ b }_{ 2 }{ c }_{ 1 } \right) }^{ 2 }+{ \left( { c }_{ 1 }{ a }_{ 2 }-{ c }_{ 2 }{ a }_{ 1 } \right) }^{ 2 }+{ \left( { a }_{ 1 }{ b }_{ 2 }-{ a }_{ 2 }{ b }_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( -6+2 \right) }^{ 2 }+{ \left( 1+7 \right) }^{ 2 }+{ \left( -14+6 \right) }^{ 2 } } =\sqrt { 16+36+64 } =\sqrt { 116 } =2\sqrt { 29 } $$
Substituting all the values in equation (1), we obtain
$$d=\cfrac{116}{2\sqrt 29}=\cfrac{-58}{\sqrt 29}=\cfrac{-2\times 29}{\sqrt 29}=-2\sqrt {29}$$
Since distance is always non-negative, the distance between the given lines is $$2\sqrt 29$$ units.

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