ISC (Commerce) Class 12CISCE
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# Shortest Distance Between Two Lines

#### description

• Distance between two skew lines
• Distance between parallel lines

#### notes

If two lines in space intersect at a point, then the shortest distance between them is zero. Also, if two lines in space are parallel, then the shortest distance between them will be the perpendicular distance, i.e. the length of the perpendicular drawn from a point on one line onto the other line. Further, in a space, there are lines which are neither intersecting nor parallel. In fact, such pair of lines are non coplanar and are called skew lines.Fig.

The line GE that goes diagonally across the ceiling and the line DB passes through one corner of the ceiling directly above A and goes diagonally down the wall. These lines are skew because they are not parallel and also never meet.
By the shortest distance between two lines we mean the join of a point in one line with one point on the other line so that the length of the segment so obtained is the smallest.
For skew lines, the line of the shortest distance will be perpendicular to both the lines.

Distance between two skew lines:
We now determine the shortest distance between two skew lines in the following way: Let l_1 and l_2  be two skew lines with equations in fig.

vec r = vec a _1 + lambda vec b_1     ...(1)
and vec r = vec a _2 + mu vec b _2    ...(2)
Take any point S on l_1with position vector vec a_1  and T on l_2, with position vector vec a_2.  Then the magnitude of the shortest distance vector will be equal to that of the projection of ST along the direction of the line of shortest distance.
If vec (PQ) is the shortest distance vector between l_1 and l_2 , then it being perpendicular to both vec b_1 and vec b_2 , the unit vector hat n along vec (PQ) would  therefore be
hat n = (vec b_1 xx vec b_2)/ |vec b_1 xx vec b_2|   ...(3)
Then vec (PQ) = d . hat n
where, d is the magnitude of the shortest distance vector. Let θ be the angle between vec (ST)  and vec (PQ). Then
PQ = ST  |cos θ|
But cos θ = |(vec (PQ) . vec (ST))/(|vec (PQ)| |vec (ST)|)|

= |(d . hat n (vec a_2 - vec a_1)) / (d  ST)|    (since vec (ST) = vec a_2 - vec a_1)

= |((vec b _1 xx vec b_2) . (vec a_2 - vec a_1)) /( ST    |vec b_1 xx vec b_2|)| [From (3)]
Hence, the required shortest distance is
d = PQ =ST |cos θ|
or d= |((vec b_1 xx vec b_2) . (vec a_2 xx vec a_1))/(|vec b_1 xx vec b_2|)|

The shortest distance between the lines
l_1 : (x - x_1)/a_1 = (y - y_1)/b_1 = (z - z_1)/c_1

and l_2 : (x - x_2)/a_2 = (y - y_2)/b_2 = (z - z_2)/c_2

is ||(x_2-x_1 , y_2 - y_1 , z_2 - z_1),(a_1 , b_1 , c_1), (a_2 , b_2 , c_2)|/ sqrt ((b_1c_2 - b_2c_1)^2 + (c_1a_2 - c_2a_1)^2 + (a_1b_2 - a_2b_1)^2) |

Distance between parallel lines:
If two lines l_1 and  l_2 are parallel, then they are coplanar.  Let the lines be given by
vec r = vec a_1 + lambda vec b   ...(1)
and vec r = vec a_2 + mu vec b  ...(2)
where , vec a_1 is the position vector of a point S on l_1 and vec a_2 is the position vector of a point T on l_2 in following fig.

As l_1, l_2 are coplanar, if the foot of the perpendicular from T on the line l_1 is P, then the distance between the lines l_1 and l_2 =  |TP|.
Let θ be the angle between the vectors vec (ST) and vec b. Then

vec b xx vec (ST) = ( |vec b|  |vec (ST)|  sin θ)  hat n     ...(3)

where hat n is the unit vector perpendicular to the plane of the lines l_1 and l_2 But vec (ST) = vec a_2 -vec a_1

Therefore, from (3), we get
vec b xx (vec a_2 - vec a_1) = |vec b|  PT  hat n  (since PT =ST sin θ )
i.e. |vec b xx (vec a_2 - vec a_1)| = |vec b| PT . 1  (as |hat n| = 1)
Hence, the distance between the given parallel lines is
d = |vec (PT)| = |(vec b xx (vec a_2 - vec a _1))/|vec b||

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Three Dimensional Geometry Part 4 - Straight Line [00:32:13]
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