Class 12

Math

3D Geometry

Three Dimensional Geometry

Find the length and equations of the line of the shortest distance between the lines given by

$2x+1 =1y−1 =−3z−9 $ and $2x−3 =−7y+15 =5z−9 $

$1x+1 =1y−1 =−3z−9 ⋯(1)$

$2x−3 =−7y+15 =5z−9 ⋯(2)$

The vector form of direction ratio's of line $(1)$ is $2i^+j^ −3k^$

The vector form of direction ratio's of line $(2)$ is $2i^−7j^ +5k^$

As we know that

The line of shortest distance will be perpendicular to both lines so the direction ratio's of shortest distance line will be cross product of direction ratio's of both the lines

$(2i^+j^ −3k^)×(2i^−7j^ +5k^)=∣∣ i^22 j^ 1−7 k^−35 ∣∣ =i^(5−21)−j^ (10+6)+k^(−14−2)=−16i^−16j^ −16k^$

So direction of shortest distance is $−16i^−16j^ −16k^$

Let $1x+1 =1y−1 =−3z−9 =a$

$2x−3 =−7y+15 =5z−9 =b$

So the point of contact of first line and shortest distance line is $(2a−1,a+1,−3a+9)$

The point of contact of second line and shortest distance line is $(2b+3,−7b−15,5b+9)$

As we know that

The direction ratio's of line joining $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ is $<x_{1}−x_{2},y_{1}−y_{2},z_{1}−z_{2}>$

The direction ratio's of shortest distance line is $<2a−2b−4,a+7b+16,−3a−5b>$

As we know that

If two lines having direction ratio's as $<a_{1},b_{1},c_{1}>$ and $<a_{2},b_{2},c_{2}>$ are perpendicular then $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$

As line of shortest distance is perpendicular to both lines

$2(2a−2b−4)+(a+7b+16)−3(−3a−5b)=0⟹7a+9b=−4⋯(3)$

$2(2a−2b−4)−7(a+7b+16)+5(−3a−5b)=0⟹18a+78b=−120⋯(4)$

Solving $(3)$ and $(4)$ gives $a=2,b=−2$

Substituting $a=2,b=−2$ we get the common points as $(3,3,3)$ and $(−1,−1,−1)$

Distance between these two points will be the shortest distance

So shortest distance is $(3+1)_{2}+(3+1)_{2}+(3+1)_{2} =16+16+16 =43 $

The equation of shortest distance line is $−16x−3 =−16y−3 =−16z−4 ⟹x−3=y−3=z−3⟹x=y=z$