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 *L*_{1} and *L*_{2} respectively. Let *A *and
*B *be the points on *L*_{1} and *L*_{2} whose
position vectors are * *and * *respectively. The two
given lines are parallel to .

Let *AD *be a perpendicular to the two given lines. If *Î¸ *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 *L*_{1}
and *L*_{2} respectively.

Let *A *and *C *be the points on *L*_{1}
and *L*_{2} with position vectors * *and * *respectively.

From the given equations of skew lines, we observe that *L*_{1}
is parallel to the vector * *and *L*_{2} is parallel
to the vector . So, * *Ã— * *is perpendicular
to the lines L_{1 }and L_{2}.

Let *SD *be the line segment perpendicular to both the
lines *L*_{1} and *L*_{2}. Then the vector * *is perpendicular to the vectors * *and and therefore it is
parallel to the vector * *Ã— * *.

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

Î´ = | | =| . (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 ( - ) . ( Ã— ) = 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 * *=
(*i*Ë† + 3 Ë†*j *âˆ’ *k *) + *t*(2*i*Ë† + 3 Ë†*j *+ 2*k
*) is

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

The Cartesian equation of the second line is (*x* â€“ 2)/1 = (*y* â€“ 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 *âˆˆ **R**, we have (2*s *+1, 3*s *+ 3, 2*s *âˆ’1)
= (*t *+ 2, 2*t *+ 4, 4*t *âˆ’ 3) .

Equating the coordinates of *x*, *y *and *z *we
get

2*s *âˆ’ *t *= 1, 3*s *âˆ’ 2*t *= 1 and *s *âˆ’
2*t *= âˆ’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 = 2*i*Ë†
+ 3Ë†*j* + 2*k*Ë† and = *i*Ë† + 2Ë†*j* + 4*k*Ë† 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 8*i*Ë† âˆ’ 6 Ë†*j *+ *k*Ë† .
Thus, the equation of the required straight line is

**Example 6.35**

Determine whether the pair of straight lines * *=
(2Ë†*i *+ 6Ë†*j *+ 3Ë†*k *) + *t*(2Ë†*i *+ 3Ë†*j *+ 4Ë†*k *) , * *= (2Ë†*j *âˆ’ 3Ë†*k *) + *s*(Ë†*i *+ 2Ë†*j *+ 3Ë†*k *) are parallel. Find
the shortest distance between them.

**Solution**

Comparing the given two equations with

= + s and = + t

we have * *= 2Ë†*i *+ 6Ë†*j *+ 3Ë†*k*, * *= 2Ë†*i *+ 3Ë†*j *+ 4Ë†*k*, * *= 2Ë†*j *âˆ’ 3Ë†*k*, * *= Ë†*i *+ 2Ë†*j *+ 3Ë†*k*

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 * *= (2Ë†*i *+ 3Ë†*j *+ 4Ë†*k *) + *t*(âˆ’2Ë†*i *+ Ë†*j *âˆ’ 2Ë†*k *) 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 (âˆ’1, 2, 3) to the straight line * *= (Ë†*i *âˆ’ 4Ë†*j *+ 3Ë†*k *) + *t*(2Ë†*i *+ 3Ë†*j *+ Ë†*k *) . Also, find the
shortest distance from the given point to the straight line.

**Solution**

Comparing the given equation = (Ë†*i* - 4Ë†*j* + 3Ë†*k* ) + t(2Ë†*i* + 3Ë†*j* + Ë†*k* ) with = + t , we get
a = Ë†*i* - 4Ë†*j* + 3Ë†*k* , and = 2Ë†*i* + 3Ë†*j* + Ë†*k* . 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)*i*Ë† + (3t - 6) Ë†*j* + t*k*Ë†.

Since is perpendicular to , we have

. = 0 â‡’ 2(2t + 2) + 3(3t - 6)
+1(t) = 0

â‡’ 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 = | |= âˆš[4^{2}+(-3)^{2}+1^{2}]
= âˆš26 units.

Tags : Definition, Theorem, Proof, Solved Example Problems, Solution , 12th Mathematics : UNIT 6 : Applications of Vector Algebra

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

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