Find the shortest distance between the lines⃗ = 3i+ 2j− 4k + (i+ 2

1.	Find the shortest distance between the lines⃗ = 3i+ 2j− 4k + (i+ 2j+ 2k) ⃗ = 5i− 2j+ (3i+ 2j+ 6k)If the lines intersect find their point of intersection
Answer» We have a1 = 3i + 2j − 4k b1 = i + 2j+ 2k a2 = 5i − 2j b2 = 3i + 2j + 6k a2 − a1= 2I − 4j + 4k b1 × b2 = (12 − 4)i − (6 − 6)j + (2 − 6)k b1 × b2 = 8i+ 0j − 4k = 8i − 4k ∵ (b1× b2). (a2 − a1 ) = 16 − 16 = 0 ∴ The lines are intersecting and the shortest distance between the lines is 0. Now for point of intersection 3i + 2j − 4k + (i + 2j + 2k) = 5i− 2j+ (3i + 2j + 6k) ⟹ 3 + λ = 5 + 3μ_____(1) 2 + 2λ = −2 + 2μ______(2) −4 + 2λ = 6μ_________(3) Solving (1) AD (2) we get, μ = −2 , λ = −4 Substituting in equation of line we get R = 5i − 2j + (−2)(3i + 2j − 6k) = − i − 6j − 12k Point of intersection is (−1, −6, −12).

Find the shortest distance between the lines⃗ = 3i+ 2j− 4k + (i+ 2j+ 2k) ⃗ = 5i− 2j+ (3i+ 2j+ 6k)If the lines intersect find their point of intersection

Answer»

We have

a1 = 3i + 2j − 4k

b1 = i + 2j+ 2k

a2 = 5i − 2j

b2 = 3i + 2j + 6k

a2 − a1= 2I − 4j + 4k

b1 × b2 = (12 − 4)i − (6 − 6)j + (2 − 6)k

b1 × b2 = 8i+ 0j − 4k = 8i − 4k

∵ (b1× b2). (a2 − a1 ) = 16 − 16 = 0

∴ The lines are intersecting and the shortest distance between the lines is 0.

Now for point of intersection

3i + 2j − 4k + (i + 2j + 2k) = 5i− 2j+ (3i + 2j + 6k)

⟹ 3 + λ = 5 + 3μ_____(1)

2 + 2λ = −2 + 2μ______(2)

−4 + 2λ = 6μ_________(3)

Solving (1) AD (2) we get, μ = −2 , λ = −4

Substituting in equation of line we get

R = 5i − 2j + (−2)(3i + 2j − 6k) = − i − 6j − 12k

Point of intersection is (−1, −6, −12).