Q.

Let P be the point of intersection of the lines x–21=y–45=z–21 and x–32=y–23=z–32. Then, the shortest distance of P from the line 4x = 2y = z is          [2024]

1 147  
2 5147  
3 6147  
4 3147  

Ans.

(4)

L1 : x–21=y–45=z–21=λ (say)

L2 : x–32=y–23=z–32=μ (say)

For point of intersection P,

(λ+2,5λ+4,λ+2)=(2μ+3,3μ+2,2μ+3)

⇒ λ+2=2μ+3,5λ+4=3μ+2,λ+2=2μ+3

⇒ λ=–1,μ=–1

∴  Point P is (1, –1, 1)

Distance of point P from L3 : 4x = 2y = z

L3 : x14=y12=z1

Any point on L3 be R(α4,α2,α)

D.r.'s of PR : (α4–1,α2+1,α–1)       ∵  PR⊥(14,12,1)

⇒ (α4–1)14+12(α2+1)+(α–1)=0

⇒ α16–14+α4+12+α–1=0 ⇒ 2116α=34 ⇒ α=47

∴  R(17,27,47)

Now, RP=(17–1)2+(27+1)2+(47–1)2

=3649+8149+949=1267=3147.