Navjyot Kalra
Last Activity: 10 Years ago
Sol. In ∆ABC , tan θ = x/2 and in ∆DCE, tanθ = (2-x) / 4 tan θ = (x/2) = (2 - x)/4 = 4x
⇒ 4 – 2x = 4x
⇒ 6 x = 4 ⇒ x = 2/3 ft
(a) In ∆ABC, AC = √(AB^2+BC^2 ) = 2/3 √10 ft
(b) In ∆CDE, DE = 1 – (2/3) = 4/3 ft
CD = 4 ft. So, CE = √(CD^2+DE^2 ) = 4/3 √10 ft
(c) Here the displacement vector r ⃗ = 7 i ̂ + 4 j ̂ + 3k ̂
(b) the components of the displacement vector are 7 ft, 4 ft, and 3 ft.