find the foot of the perpendicular from P(1,2,3) on the line x-6/3 = y-7/2=z-7/-2 - Maths - Three Dimensional Geometry

Question

find the foot of the perpendicular from P(1,2,3) on the
line x-6/3 = y-7/2=z-7/-2

Manish Bharti · Accepted Answer

$L e t \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2} = λ$
Coordinates of any point on the given line will be $(3 λ + 6, 2 λ + 7, - 2 λ + 7)$
Let A be the foot of perpendicular on given line.
$d . r' s o f A P = 3 λ + 6 - 1, 2 λ + 7 - 2, - 2 λ + 7 - 3 = 3 λ + 5, 2 λ + 5, - 2 λ + 4 d . r' s o f g i v e n l i n e = 3, 2, - 2 S i n c e A P ⊥ g i v e n l i n e, 3 (3 λ + 5) + 2 (2 λ + 5) - 2 (- 2 λ + 4) = 0 9 λ + 15 + 4 λ + 10 + 4 λ - 8 = 0 17 λ = - 17 \Rightarrow λ = - 1$
Therefore, coordinate of A are ( -3+6 , -2+7 , +2+7) or ( 3, 5, 9 )