Notes On Shortest Distance Between Two Lines - CBSE Class 12 Maths
A pair of coplanar lines can either be parallel lines or intersecting lines. Consider two non-coplanar lines, like lines L1 and L2, lying on different faces of a cuboid. Two non-coplanar lines can be a pair of parallel lines. Two non-coplanar lines can be a pair of intersecting lines. Pair of non- coplanar lines may not be parallel and still do not intersect each other in space. Two lines that are neither parallel nor intersecting are called skew lines. Skew lines are always non-coplanar. The shortest distance between two intersecting lines is zero. Shortest Distance between a Pair of Skew Lines The shortest distance between two skew lines is the length of the shortest line segment that joins a point on one line to a point on the other line. The line segment is perpendicular to both the lines. Consider lines l1 and l2 with equations:   $\stackrel{\to }{\text{r}}$  =  $\stackrel{\to }{{\text{a}}_{\text{1}}}$  +  λ      and   $\stackrel{\to }{\text{r}}$  =   $\stackrel{\to }{{\text{a}}_{\text{2}}}$  +  λ         Let M and N be points on lines l1 and l2, having position vectors   $\stackrel{\to }{{\text{a}}_{\text{1}}}$  and  $\stackrel{\to }{{\text{a}}_{\text{2}}}$ , respectively. ⇒ $\stackrel{\to }{\text{MN}}$  =   $\stackrel{\to }{{\text{a}}_{\text{2}}}$    -    $\stackrel{\to }{{\text{a}}_{\text{1}}}$    ...(1) Let the shortest distance vector between l1 and l2 =  $\stackrel{\to }{\text{AB}}$       $\stackrel{\to }{\text{AB}}$   =  | $\stackrel{\to }{\text{AB}}$| $\stackrel{⏜}{\text{n}}$ Or  $\stackrel{\to }{\text{AB}}$  = d $\stackrel{⏜}{\text{n}}$ Where d = | $\stackrel{\to }{\text{AB}}$ |  $\stackrel{\to }{\text{AB}}$  ⊥ l1 and  $\stackrel{\to }{\text{AB}}$  ⊥ l2 ⇒   $\stackrel{\to }{\text{AB}}$  ⊥    and  $\stackrel{\to }{\text{AB}}$  ⊥      ⇒ $\stackrel{⏜}{\text{n}}$ = $\frac{\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}$ ⇒  $\stackrel{\to }{\text{AB}}$  = d  (  $\frac{\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}$ )    …(2) Let q be the angle between  $\stackrel{\to }{\text{MN}}$  and   $\stackrel{\to }{\text{AB}}$ . Thus, projection of  $\stackrel{\to }{\text{MN}}$  on  $\stackrel{\to }{\text{AB}}$  is:     $\stackrel{\to }{\text{AB}}$  = d = | $\stackrel{\to }{\text{MN}}$ | |cos θ|   …(3) cos q =  | |    …(4) From (1) and (2): cos θ =  $\frac{\text{d(}\frac{\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}\text{) (}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{)}}{\text{|d(}\frac{\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}\text{)| |}\stackrel{\to }{\text{MN}}\text{|}}$ ⇒ cos θ =  $\text{|}\frac{\text{(}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{).(}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{)}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}\text{|}$ ⇒  |$\stackrel{\to }{\text{MN}}$|  |Cos θ|  =   $\text{|}\frac{\text{(}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{).(}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{)}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}\text{|}$              …(5) From (3) and (5): Shortest distance between skew lines L1 and L2 (d) = $\text{|}\frac{\text{(}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{).(}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{)}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}\text{|}$ Cartesian form: The shortest distance between lines l1 and l2 given in Cartesian form as $\frac{\text{x -}{\text{x}}_{\text{1}}}{{\text{a}}_{\text{1}}}$  = $\frac{\text{y -}{\text{y}}_{\text{1}}}{{\text{b}}_{\text{1}}}$  = $\frac{\text{z -}{\text{z}}_{\text{1}}}{{\text{c}}_{\text{1}}}$  and $\frac{\text{x -}{\text{x}}_{\text{2}}}{{\text{a}}_{\text{2}}}$  = $\frac{\text{y -}{\text{y}}_{\text{2}}}{{\text{b}}_{\text{2}}}$  = $\frac{\text{z -}{\text{z}}_{\text{2}}}{{\text{c}}_{\text{2}}}$ , is given by: d =  $\left|\frac{\left|\begin{array}{ccc}{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}& {\text{y}}_{\text{2}}\text{-}{\text{y}}_{\text{1}}& {\text{z}}_{\text{2}}\text{-}{\text{z}}_{\text{1}}\\ {\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\end{array}\right|}{\sqrt{{\text{(}{{\text{b}}_{\text{1}}\text{c}}_{\text{2}}\text{-}{{\text{b}}_{\text{2}}\text{c}}_{\text{1}}\text{)}}^{\text{2}}\text{+}{\text{(}{{\text{c}}_{\text{1}}\text{a}}_{\text{2}}\text{-}{{\text{c}}_{\text{2}}\text{a}}_{\text{1}}\text{)}}^{\text{2}}\text{+}{\text{(}{{\text{a}}_{\text{1}}\text{b}}_{\text{2}}\text{-}{{\text{a}}_{\text{2}}\text{b}}_{\text{1}}\text{)}}^{\text{2}}}}\right|$                Shortest Distance between a Pair of Parallel Lines Consider lines l1 ∥ l2 with equations:   $\stackrel{\to }{\text{r}}$  =   $\stackrel{\to }{{\text{a}}_{\text{1}}}$  +  λ     and  $\stackrel{\to }{\text{r}}$  =   $\stackrel{\to }{{\text{a}}_{\text{2}}}$  +  μ    Let P and Q be points on lines l1 and l2, having position vectors  $\stackrel{\to }{{\text{a}}_{\text{1}}}$  and  $\stackrel{\to }{{\text{a}}_{\text{2}}}$ , respectively. ⇒ $\stackrel{\to }{\text{PQ}}$  =   $\stackrel{\to }{{\text{a}}_{\text{2}}}$   -  $\stackrel{\to }{{\text{a}}_{\text{1}}}$  ...(1) The shortest distance vector between l1 and l2 =  $\stackrel{\to }{\text{PR}}$   Where  $\stackrel{\to }{\text{PR}}$  ⊥ l1  and   $\stackrel{\to }{\text{PR}}$  ⊥ l2 Let  | $\stackrel{\to }{\text{PR}}$  | = d Let q be the angle between $\stackrel{\to }{\text{PQ}}$   and   $\stackrel{\to }{\text{b}}$  . ⇒   $\stackrel{\to }{\text{b}}$  ×   $\stackrel{\to }{\text{PQ}}$   =  |  $\stackrel{\to }{\text{b}}$  | |  $\stackrel{\to }{\text{PQ}}$  | sin θ $\stackrel{^}{\text{n}}$    …(2) In right-angled ∆ PQR: sin q =  $\frac{\left|\stackrel{\to }{\text{PR}}\right|}{\left|\stackrel{\to }{\text{PQ}}\right|}$   =   $\frac{\text{d}}{\left|\stackrel{\to }{\text{PQ}}\right|}$ ⇒ $\left|\stackrel{\to }{\text{PQ}}\right|$ sin q = d …(3) From (1), (2) and (3):   $\stackrel{\to }{\text{b}}$  x  ( $\stackrel{\to }{{\text{a}}_{\text{2}}}$ $\text{-}$ $\stackrel{\to }{{\text{a}}_{\text{1}}}$ )    =  |$\stackrel{\to }{\text{b}}$| d  $\stackrel{^}{\text{n}}$     …(4) Taking magnitude on both sides: | $\stackrel{\to }{\text{b}}$  x ( $\stackrel{\to }{{\text{a}}_{\text{2}}}$ $\text{-}$ $\stackrel{\to }{{\text{a}}_{\text{1}}}$ ) |   = | |$\stackrel{\to }{\text{b}}$| d  $\stackrel{^}{\text{n}}$   | ⇒ d =   $\frac{\left|\stackrel{\to }{\text{b}}\mathrm{\text{}}\text{}×\text{(}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{)}\right|}{\text{||}\stackrel{\to }{\text{b}}\text{|}\stackrel{^}{\text{n}}\text{|}}$ ∴ The euation for the shortest distance between parallel lines L1 and L2 (d)  = $\left|\frac{\stackrel{\to }{\text{b}}\text{}×\text{(}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{)}}{\left|\stackrel{\to }{\text{b}}\right|}\right|$  (Since $\left|\stackrel{⏜}{\text{n}}\right|$  =1 )

#### Summary

A pair of coplanar lines can either be parallel lines or intersecting lines. Consider two non-coplanar lines, like lines L1 and L2, lying on different faces of a cuboid. Two non-coplanar lines can be a pair of parallel lines. Two non-coplanar lines can be a pair of intersecting lines. Pair of non- coplanar lines may not be parallel and still do not intersect each other in space. Two lines that are neither parallel nor intersecting are called skew lines. Skew lines are always non-coplanar. The shortest distance between two intersecting lines is zero. Shortest Distance between a Pair of Skew Lines The shortest distance between two skew lines is the length of the shortest line segment that joins a point on one line to a point on the other line. The line segment is perpendicular to both the lines. Consider lines l1 and l2 with equations:   $\stackrel{\to }{\text{r}}$  =  $\stackrel{\to }{{\text{a}}_{\text{1}}}$  +  λ      and   $\stackrel{\to }{\text{r}}$  =   $\stackrel{\to }{{\text{a}}_{\text{2}}}$  +  λ         Let M and N be points on lines l1 and l2, having position vectors   $\stackrel{\to }{{\text{a}}_{\text{1}}}$  and  $\stackrel{\to }{{\text{a}}_{\text{2}}}$ , respectively. ⇒ $\stackrel{\to }{\text{MN}}$  =   $\stackrel{\to }{{\text{a}}_{\text{2}}}$    -    $\stackrel{\to }{{\text{a}}_{\text{1}}}$    ...(1) Let the shortest distance vector between l1 and l2 =  $\stackrel{\to }{\text{AB}}$       $\stackrel{\to }{\text{AB}}$   =  | $\stackrel{\to }{\text{AB}}$| $\stackrel{⏜}{\text{n}}$ Or  $\stackrel{\to }{\text{AB}}$  = d $\stackrel{⏜}{\text{n}}$ Where d = | $\stackrel{\to }{\text{AB}}$ |  $\stackrel{\to }{\text{AB}}$  ⊥ l1 and  $\stackrel{\to }{\text{AB}}$  ⊥ l2 ⇒   $\stackrel{\to }{\text{AB}}$  ⊥    and  $\stackrel{\to }{\text{AB}}$  ⊥      ⇒ $\stackrel{⏜}{\text{n}}$ = $\frac{\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}$ ⇒  $\stackrel{\to }{\text{AB}}$  = d  (  $\frac{\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}$ )    …(2) Let q be the angle between  $\stackrel{\to }{\text{MN}}$  and   $\stackrel{\to }{\text{AB}}$ . Thus, projection of  $\stackrel{\to }{\text{MN}}$  on  $\stackrel{\to }{\text{AB}}$  is:     $\stackrel{\to }{\text{AB}}$  = d = | $\stackrel{\to }{\text{MN}}$ | |cos θ|   …(3) cos q =  | |    …(4) From (1) and (2): cos θ =  $\frac{\text{d(}\frac{\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}\text{) (}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{)}}{\text{|d(}\frac{\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}\text{)| |}\stackrel{\to }{\text{MN}}\text{|}}$ ⇒ cos θ =  $\text{|}\frac{\text{(}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{).(}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{)}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}\text{|}$ ⇒  |$\stackrel{\to }{\text{MN}}$|  |Cos θ|  =   $\text{|}\frac{\text{(}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{).(}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{)}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}\text{|}$              …(5) From (3) and (5): Shortest distance between skew lines L1 and L2 (d) = $\text{|}\frac{\text{(}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{).(}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{)}}{\text{|}\stackrel{\to }{{\text{b}}_{\text{1}}}\text{}×\text{}\stackrel{\to }{{\text{b}}_{\text{2}}}\text{|}}\text{|}$ Cartesian form: The shortest distance between lines l1 and l2 given in Cartesian form as $\frac{\text{x -}{\text{x}}_{\text{1}}}{{\text{a}}_{\text{1}}}$  = $\frac{\text{y -}{\text{y}}_{\text{1}}}{{\text{b}}_{\text{1}}}$  = $\frac{\text{z -}{\text{z}}_{\text{1}}}{{\text{c}}_{\text{1}}}$  and $\frac{\text{x -}{\text{x}}_{\text{2}}}{{\text{a}}_{\text{2}}}$  = $\frac{\text{y -}{\text{y}}_{\text{2}}}{{\text{b}}_{\text{2}}}$  = $\frac{\text{z -}{\text{z}}_{\text{2}}}{{\text{c}}_{\text{2}}}$ , is given by: d =  $\left|\frac{\left|\begin{array}{ccc}{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}& {\text{y}}_{\text{2}}\text{-}{\text{y}}_{\text{1}}& {\text{z}}_{\text{2}}\text{-}{\text{z}}_{\text{1}}\\ {\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\end{array}\right|}{\sqrt{{\text{(}{{\text{b}}_{\text{1}}\text{c}}_{\text{2}}\text{-}{{\text{b}}_{\text{2}}\text{c}}_{\text{1}}\text{)}}^{\text{2}}\text{+}{\text{(}{{\text{c}}_{\text{1}}\text{a}}_{\text{2}}\text{-}{{\text{c}}_{\text{2}}\text{a}}_{\text{1}}\text{)}}^{\text{2}}\text{+}{\text{(}{{\text{a}}_{\text{1}}\text{b}}_{\text{2}}\text{-}{{\text{a}}_{\text{2}}\text{b}}_{\text{1}}\text{)}}^{\text{2}}}}\right|$                Shortest Distance between a Pair of Parallel Lines Consider lines l1 ∥ l2 with equations:   $\stackrel{\to }{\text{r}}$  =   $\stackrel{\to }{{\text{a}}_{\text{1}}}$  +  λ     and  $\stackrel{\to }{\text{r}}$  =   $\stackrel{\to }{{\text{a}}_{\text{2}}}$  +  μ    Let P and Q be points on lines l1 and l2, having position vectors  $\stackrel{\to }{{\text{a}}_{\text{1}}}$  and  $\stackrel{\to }{{\text{a}}_{\text{2}}}$ , respectively. ⇒ $\stackrel{\to }{\text{PQ}}$  =   $\stackrel{\to }{{\text{a}}_{\text{2}}}$   -  $\stackrel{\to }{{\text{a}}_{\text{1}}}$  ...(1) The shortest distance vector between l1 and l2 =  $\stackrel{\to }{\text{PR}}$   Where  $\stackrel{\to }{\text{PR}}$  ⊥ l1  and   $\stackrel{\to }{\text{PR}}$  ⊥ l2 Let  | $\stackrel{\to }{\text{PR}}$  | = d Let q be the angle between $\stackrel{\to }{\text{PQ}}$   and   $\stackrel{\to }{\text{b}}$  . ⇒   $\stackrel{\to }{\text{b}}$  ×   $\stackrel{\to }{\text{PQ}}$   =  |  $\stackrel{\to }{\text{b}}$  | |  $\stackrel{\to }{\text{PQ}}$  | sin θ $\stackrel{^}{\text{n}}$    …(2) In right-angled ∆ PQR: sin q =  $\frac{\left|\stackrel{\to }{\text{PR}}\right|}{\left|\stackrel{\to }{\text{PQ}}\right|}$   =   $\frac{\text{d}}{\left|\stackrel{\to }{\text{PQ}}\right|}$ ⇒ $\left|\stackrel{\to }{\text{PQ}}\right|$ sin q = d …(3) From (1), (2) and (3):   $\stackrel{\to }{\text{b}}$  x  ( $\stackrel{\to }{{\text{a}}_{\text{2}}}$ $\text{-}$ $\stackrel{\to }{{\text{a}}_{\text{1}}}$ )    =  |$\stackrel{\to }{\text{b}}$| d  $\stackrel{^}{\text{n}}$     …(4) Taking magnitude on both sides: | $\stackrel{\to }{\text{b}}$  x ( $\stackrel{\to }{{\text{a}}_{\text{2}}}$ $\text{-}$ $\stackrel{\to }{{\text{a}}_{\text{1}}}$ ) |   = | |$\stackrel{\to }{\text{b}}$| d  $\stackrel{^}{\text{n}}$   | ⇒ d =   $\frac{\left|\stackrel{\to }{\text{b}}\mathrm{\text{}}\text{}×\text{(}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{)}\right|}{\text{||}\stackrel{\to }{\text{b}}\text{|}\stackrel{^}{\text{n}}\text{|}}$ ∴ The euation for the shortest distance between parallel lines L1 and L2 (d)  = $\left|\frac{\stackrel{\to }{\text{b}}\text{}×\text{(}\stackrel{\to }{{\text{a}}_{\text{2}}}\text{-}\stackrel{\to }{{\text{a}}_{\text{1}}}\text{)}}{\left|\stackrel{\to }{\text{b}}\right|}\right|$  (Since $\left|\stackrel{⏜}{\text{n}}\right|$  =1 )

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