1
Question
Find the shortest distance between lines x+17=y+1−6=z+11 and x−31=y−5−2=z−71
Open in App
Solution
It is known that the shortest distance between the two lines,
x−x1a=y−y1b=z−z1c and x−x2a=y−y2b=z−z2c is given by,
d=∣∣
∣∣x2−x1y2−y1z2−z1a1b1c1a2b2c2∣∣
∣∣√(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2....(1)
Comparing the given equations, we obtain
x1=−1,y1=−1,z1=−1
a1=7,b1=−6,c1=1
x2=3,y2=5,z2=7
a2=1,b2=−2,c2=1
Then,
∣∣
∣∣x2−x1y2−y1z2−z1a1b1c1a2b2c2∣∣
∣∣=∣∣
∣∣4687−611−21∣∣
∣∣
=4(−6+2)−6(7−1)+8(−14+6)
=−16−36−64=−116
⇒√(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2=√(−6+2)2+(1+7)2+(−14+6)2=√16+36+64=√116=2√29
Substituting all the values in equation (1), we obtain
d=1162√29=−58√29=−2×29√29=−2√29
Since distance is always non-negative, the distance between the given lines is 2√29 units.
x−x1a=y−y1b=z−z1c and x−x2a=y−y2b=z−z2c is given by,
d=∣∣ ∣∣x2−x1y2−y1z2−z1a1b1c1a2b2c2∣∣ ∣∣√(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2....(1)
Comparing the given equations, we obtain
x1=−1,y1=−1,z1=−1
a1=7,b1=−6,c1=1
x2=3,y2=5,z2=7
a2=1,b2=−2,c2=1
Then, ∣∣ ∣∣x2−x1y2−y1z2−z1a1b1c1a2b2c2∣∣ ∣∣=∣∣ ∣∣4687−611−21∣∣ ∣∣
=4(−6+2)−6(7−1)+8(−14+6)
=−16−36−64=−116
⇒√(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2=√(−6+2)2+(1+7)2+(−14+6)2=√16+36+64=√116=2√29
Substituting all the values in equation (1), we obtain
d=1162√29=−58√29=−2×29√29=−2√29
Since distance is always non-negative, the distance between the given lines is 2√29 units.
Suggest Corrections
0
Similar questions
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
