Shortest Distance between two non Intersecting Line:

Two lines are called as non intersecting lines if they do not lie in same plane. The straight line which is perpendicular to each of non-intersecting lines is called as line of shortest distance. And the length of shortest distance line intercepted between 2 lines is called as length of shortest distance.

Method: Let the equation of 2 non intersecting lines be

Any point on line (1) is P (x₁ + l₁r₁, y₁ + m₁r₁, z₁ + n₁r₁) and on line (2) is Q (x₂ + l₂r₂, y₂ + m₂r₂, z₂ + n₂r₂).

Let PQ be the line of shortest distance. The direction ratios of will be [(l₁r₁ + x₁- x₂- l₂r₂), (m₁r₁ + y₁- y₂- m₂r₂), (n₁r₁ + z₁- z₂- n₂r₂)] This line is perpendicular to both the given line. By using condition of perpendicularity we obtain 2 equations in r₁ and r₂.

So by solving these, values of r₁ and r₂ can be found. And subsequently the points P and Q can be found. The distance PQ is the shortest distance.

The shortest distance can be found by .

Note:

• If any straight line can be given in general form then it can be transformed into symmetrical form and we can proceed further.

Example: Find shortest distance between the lines      . Also find the equation of line of shortest distance.

Solution :     Given lines are                 ......(1)

                                                                 ......(2)

                        Any point on line (1) is P (3r₁ + 3, 8 - r₁, r₁ + 3) and on line (2) is

                        Q (-3 - 3r₂, 2r₂ - 7, 4r₂ + 6).

                        If PQ is line of shortest distance, then the direction ratios of PQ is

                        = (3r₁ + 3) - (-3 - 3r₂), (8 - r₁) - (2r₂ - 7), (r₁+ 3) - (4r₂ + 6)

                        That is 3r₁ + 3r₂ + 6, -r₁ - 3r₂ + 15, r₁ - 4r₂ - 3

                        As PQ is perpendicular to liens (1) and (2)

                        ∴3(3r₁ + 3r₂ + 6) - 1(-r₁ - 2r₂ + 15) + 1(r₁ - 4r₂ + 3) = 0

                        => 11r₁ + 7r₂ = 0                                                                    ......(3)

                        and -3(3r₁ + 3r₂ + 6) + 2(-r₁ - 2r₂ + 15) + 4(r₁ - 4r₂ + 3) = 0

                        that is 7r₁ + 11r₂ = 0                                                             ......(4)

                        On solving equations (3) and (4), we get r₁ = r₂= 0.

                        So, point P (3, 8,3) and Q (-3, -7, 6)

                        ∴ Length of shortest distance PQ = 

                        Direction ratios of shortest distance line is 2, 5, -1

                  ∴ Equation of shortest distance line x-3/2 = y-8/5 = z-3/-1.

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