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Definition, Theorem, Proof, Solved Example Problems, Solution - Shortest distance between two straight lines | 12th Mathematics : UNIT 6 : Applications of Vector Algebra

Chapter: 12th Mathematics : UNIT 6 : Applications of Vector Algebra

Shortest distance between two straight lines

We have just explained how the point of intersection of two lines are found and we have also studied how to determine whether the given two lines are parallel or not.

Shortest distance between two straight lines

We have just explained how the point of intersection of two lines are found and we have also studied how to determine whether the given two lines are parallel or not.

 

Definition 6.6

Two lines are said to be coplanar if they lie in the same plane.

Note

If two lines are either parallel or intersecting, then they are coplanar.

Definition 6.7

Two lines in space are called skew lines if they are not parallel and do not intersect

Note

If two lines are skew lines, then they are non coplanar. If the lines are not parallel and intersect, the distance between them is zero. If they are parallel and non-intersecting, the distance is determined by the length of the line segment perpendicular to both the parallel lines. In the same way, the shortest distance between two skew lines is defined as the length of the line segment perpendicular to both the skew lines. Two lines will either be parallel or skew.


 



Proof

The given two parallel lines    s and   t are denoted by L1 and L2 respectively. Let A and B be the points on L1 and L2 whose position vectors are  and  respectively. The two given lines are parallel to .

Let AD be a perpendicular to the two given lines. If θ is the acute angle between  and , then


But, from the right angle triangle ABD ,


 


Proof

The two skew lines  =  + s and  =  + t are denoted by L1 and L2 respectively. 

Let A and C be the points on L1 and L2 with position vectors  and  respectively.

From the given equations of skew lines, we observe that L1 is parallel to the vector  and L2 is parallel to the vector . So,  ×  is perpendicular to the lines L1 and L2.


Let SD  be the line segment perpendicular to both the lines L1 and L2. Then the vector  is perpendicular to the vectors  and  and therefore it is parallel to the vector  ×  .

So,  is a unit vector in the direction of . Then, the shortest distance |  | is the absolute value of the projection of  on . That is,

δ = |  | =| . (Unit vector in the direction of )| 


Remark

(i) It follows from theorem (6.14) that two straight lines  =  + s and  =  + t  intersect each other (that is, coplanar) if ( - ) . ( × ) = 0 

(2) If two lines  intersect each other  (that is, coplanar), then we have


 

Example 6.34

Find the parametric form of vector equation of a straight line passing through the point of intersection of the straight lines  and perpendicular to both straight lines.

Solution

The Cartesian equations of the straight line  = (iˆ + 3 ˆj k ) + t(2iˆ + 3 ˆj + 2k ) is


Then any point on this line is of the form (2s +1, 3s + 3, 2s -1)          ... (1)

The Cartesian equation of the second line is (x – 2)/1 = (y – 4)/2 = (z + 3)/4 = t  (say)

Then any point on this line is of the form (t + 2, 2t + 4, 4t - 3)

If the given lines intersect, then there must be a common point. Therefore, for some s, t R, we have (2s +1, 3s + 3, 2s −1) = (t + 2, 2t + 4, 4t − 3) .

Equating the coordinates of x, y and z we get

2s t = 1, 3s − 2t = 1 and s − 2t = −1.

Solving the first two of the above three equations, we get s = 1 and t = 1. These values of s and t satisfy the third equation. So, the lines are intersecting.

Now, using the value of s in (1) or the value of t in (2), the point of intersection (3, 6,1) of these two straight lines is obtained.

If we take = 2iˆ + 3ˆj + 2kˆ  and  = iˆ + 2ˆj + 4kˆ  then  is a vector perpendicular to both the given straight lines. Therefore, the required straight line passing through (3, 6,1) and perpendicular to both the given straight lines is the same as the straight line passing through (3, 6,1) and parallel to 8iˆ − 6 ˆj + kˆ . Thus, the equation of the required straight line is


 

Example 6.35

Determine whether the pair of straight lines  = (2ˆi + 6ˆj + 3ˆk ) + t(2ˆi + 3ˆj + 4ˆk ) ,  = (2ˆj − 3ˆk ) + s(ˆi + 2ˆj + 3ˆk ) are parallel. Find the shortest distance between them.

Solution

Comparing the given two equations with

 =  + s and  =  + t 

we have  = 2ˆi + 6ˆj + 3ˆk,  = 2ˆi + 3ˆj + 4ˆk,  = 2ˆj − 3ˆk,  = ˆi + 2ˆj + 3ˆk

Clearly,  is not a scalar multiple of  . So, the two vectors are not parallel and hence the two lines are not parallel.

The shortest distance between the two straight lines is given by


Therefore, the distance between the two given straight lines is zero.Thus, the given lines intersect each other.

 

Example 6.36

Find the shortest distance between the two given straight lines  = (2ˆi + 3ˆj + 4ˆk ) + t(−2ˆi + ˆj − 2ˆk ) and 

Solution

The parametric form of vector equations of the given straight lines are


Clearly,  is a scalar multiple of , and hence the two straight lines are parallel. We know that the shortest distance between two parallel straight lines is given by d =


 

Example 6.37

Find the coordinates of the foot of the perpendicular drawn from the point (−1, 2, 3) to the straight line  = (ˆi − 4ˆj + 3ˆk ) + t(2ˆi + 3ˆj + ˆk ) . Also, find the shortest distance from the given point to the straight line.

Solution

Comparing the given equation  = (ˆi - 4ˆj + 3ˆk ) + t(2ˆi + 3ˆj + ˆk ) with  =  + t , we get a = ˆi - 4ˆj + 3ˆk , and  = 2ˆi + 3ˆj + ˆk . We denote the given point (-1, 2, 3) by D , and the point (1, -4, 3) on the straight line by A . If F is the foot of the perpendicular from D to the straight line, then F is of the form (2t +1, 3t - 4, t + 3) and  = (2t + 2)iˆ + (3t - 6) ˆj + tkˆ.


Since  is perpendicular to  , we have

 . = 0 2(2t + 2) + 3(3t - 6) +1(t) = 0

t = 1

Therefore, the coordinate of F is  (3,-1, 4)

Now, the perpendicular distance from the given point to the given line is

DF = |  |= √[42+(-3)2+12] =  √26 units.

 

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