Question Number 103945 by bobhans last updated on 18/Jul/20

a cube ABCD.EFGH with length side  4 cm. Given point P is midpoint EF.  find the distance of line AP to line  HB.

Answered by bramlex last updated on 18/Jul/20

let : ax+by+cz = −1 be a  equation of plane HBP ′ where  BP ′ parallel to AP.   coordinates A(4,0,0) ; B(4,4,0)  ; P ′(4,6,4) and H(0,0,4)  (i)substitute B⇒4a+4b = −1  (ii) substitute H⇒4c = −1⇒c =−(1/4)  (iii) substitute P  ′⇒4a+6b+4c=−1⇒4a+6b = 0  substract (i) &(iii)   we get 2b = 1⇒b=(1/2) & a = −(3/4)  so eq of plane HBP ′ ⇒−(3/4)x+(1/2)y−(1/4)z+1=0  or −3x+2y−z+4 = 0  so the distance we desired is ((∣−12+4∣)/(√(9+4+1)))  = (8/(√(14))) = ((4(√(14)))/7) cm. ■

Commented bybramlex last updated on 18/Jul/20

Commented bybobhans last updated on 18/Jul/20

waw...nice and cooll

Answered by mr W last updated on 18/Jul/20

Commented bymr W last updated on 18/Jul/20

A(4,0,4)  P(4,2,0)  H(0,0,0)  B(4,4,4)  eqn. of AP=(4,0,4)+λ(0,1,−2)  eqn. of HB=(0,0,0)+μ(1,1,1)  a point on AP is S(4,λ,4−2λ)  a point on HB is T(μ,μ,μ)  let Φ=ST^2   Φ=(4−μ)^2 +(λ−μ)^2 +(4−2λ−μ)^2   (∂Φ/∂s)=2(λ−μ)+4(2λ+μ−4)=0  ⇒5λ+μ=8   ...(i)  (∂Φ/∂t)=−2(4−μ)−2(λ−μ)+2(2λ+μ−4)=0  ⇒λ+3μ=8   ...(ii)  ⇒λ=(8/7)  ⇒μ=((16)/7)  Φ_(min) =(4−((16)/7))^2 +((8/7)−((16)/7))^2 +(4−((16)/7)−((16)/7))^2 =((224)/(49))  min. distance from AP to HB is  ST_(min) =(√Φ_(min) )=(√((224)/(49)))=((4(√(14)))/7)

Commented bybobhans last updated on 18/Jul/20

what method it is sir?

Commented bybobhans last updated on 18/Jul/20

where become s(0,1,−2) sir

Commented bymr W last updated on 18/Jul/20

AP^(→) =(4−4,2−0,0−4)=(0,2,−4) or  (0,1,−2)

Commented bybramlex last updated on 18/Jul/20

waw...great your method sir.