Let the equation of lines be as L1:r1=a1+λb1 and L2:r2=a2+λb2 such that a1−a2=2i^+4j^+4k^ and b1×b2=8i^−4k^. Then the shortest distance between L1 and L2 is
Solution
✅ Correct Option: 3
To find the shortest distance (d) between two skew lines, we use the standard formula:
d=∣b1×b2∣(a2−a1)⋅(b1×b2)
Identify the given values
From the problem, we have:
a1−a2=2i^+4j^+4k^
Note: For the formula, a2−a1=−(a1−a2)=−2i^−4j^−4k^
b1×b2=8i^−4k^ (or 8i^+0j^−4k^)
Calculate the Dot Product
(a2−a1)⋅(b1×b2)=(−2)(8)+(−4)(0)+(−4)(−4)
=−16+0+16=0
Calculate the Magnitude of the Cross Product
∣b1×b2∣=82+02+(−4)2=64+16=80=45
Calculate Shortest Distance
d=450=0
The shortest distance between the lines L1 and L2 is 0. This implies that the two lines are intersecting.