Find the length and equations of the line of the shortest distance between the lines given by
$\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ and $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$

Question

Find the length and equations of the line of the shortest distance between the lines given by2x+1​=1y−1​=−3z−9​ and 2x−3​=−7y+15​=5z−9​

Filo · Accepted Answer

Given lines are 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 thatThe 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​=bSo 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 thatThe direction ratio's of line joining (x1​,y1​,z1​) and (x2​,y2​,z2​) is The direction ratio's of shortest distance line is As we know thatIf two lines having direction ratio's as  and  are perpendicular then a1​a2​+b1​b2​+c1​c2​=0As 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=−2Substituting 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 distanceSo 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

Question Text	Find the length and equations of the line of the shortest distance between the lines given by $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ and $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$
Answer Type	Text solution:1
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Find the length and equations of the line of the shortest distance between the lines given by
$\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ and $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$

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Find the length and equations of the line of the shortest distance between the lines given by2x+1​=1y−1​=−3z−9​ and 2x−3​=−7y+15​=5z−9​

Text solutionVerified

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Find the length and equations of the line of the shortest distance between the lines given by
$\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ and $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$