Question

Perpendicular distance between line vector r = ( 2, – 2, 3) + k(1, – 1, 4), k ∈ R and x + 5y + z = 5 is ____ 

(A) (10/3)

(B) [10/(3√3)]

(C) (10/√3)

(D) 10   

Answer 1

The correct option (B) [10/(3√3)]  

Explanation:

line vector r = ( 2, – 2, 3) + k(1, – 1, 4)

plane vector r ∙ (1, 5, 1) = 5

vector r = (1, – 1, 4), n = (1, 5, 1)

ℓ ∙ n = (1, – 1, 4) ∙ (1, 5, 1) = 1 – 5 + 4 = 0

∴ line is parallel to plane.

perpendicular distance from (2, – 2, 3) to plane x + 5y + z – 5 = 0.

∴ p   = [{|2 + (– 2)(5) + 3 – 5|}/√(1 + 25 + 1)]

= [10/√(27)]

= [10/(3√3)] unit.